Some new integrals involving

Some trigonometric integrals involving log Γ( x) and the digamma function Donal F. Connon [email protected] 19 May 2010 Abstract This paper considers various integrals where the integrand includes the log gamma function (or its derivative, the digamma function ψ ( x)) multiplied by a trigonometric or hyperbolic function. Some apparently new integrals and series are evaluated; these include 1

∫ log Γ( x) cos pπ x dx = 0

+

sin pπ [γ + log(2π )] sin pπ + 2 pπ 4 pπ

⎡ ⎛ p⎞ ⎛ p ⎞⎤ ⎢ψ ⎜ 2 ⎟ + ψ ⎜ − 2 ⎟ ⎥ ⎝ ⎠⎦ ⎣ ⎝ ⎠

2(1 − cos pπ )[γ + log(2π )] ⎡ 1 π ⎛ pπ ⎞ ⎤ 2(1 − cos pπ ) ∞ log n − cot ∑ ⎜ ⎟⎥ + ⎢ 2 p2 4 p 2 2 π2 π2 ⎝ 2 ⎠⎦ n =1 4n − p ⎣

1

∫ log Γ( x) sin pπ x dx = 0

−

(1 − cos pπ )[γ + log(2π )] (1 − cos pπ ) ⎡ ⎛ p ⎞ ⎛ p ⎞⎤ + ψ ⎜ ⎟ +ψ ⎜ − ⎟ ⎥ ⎢ 2 pπ 4 pπ ⎝ 2 ⎠⎦ ⎣ ⎝2⎠

2sin pπ [γ + log(2π )] ⎡ 1 π ⎛ pπ ⎞ ⎤ 2sin pπ − cot ⎜ ⎟⎥ − ⎢ 2 2 π 4p π2 ⎝ 2 ⎠⎦ ⎣2p

∞

log n 2 − p2

∑ 4n n =1

⎡1 ⎤ sin pπ Si( pπ ) 2(1 − cos pπ ) ∞ Ci (2nπ ) log Γ ( x ) cos p π x dx = log(2 π ) − 1 ∑ 2 2 ∫0 ⎢⎣ 2 ⎥⎦ pπ + pπ + π2 n =1 4n − p 1

+ 1

p sin pπ

π

2

1 si (2nπ ) 2 − p2 n =1 ∞

∑ n 4n

Ci (2nπ ) nπ n =1 ∞

∫ log Γ( x + 1) cot π x dx = ∑ 0

where si ( x) and Ci ( x) are the sine and cosine integrals.

CONTENTS

Page

1. A connection with the sine and cosine integrals Representations of log Γ( x) in terms of the sine and cosine integrals Applications of Nielsen’s representation of log Γ( x)

2 27 43

2. Another approach to the log Γ( x) integrals An observation by Glasser

66 91

3. Some applications of the Fourier series for the Hurwitz zeta function Another proof of Hurwitz’s formula for the Fourier series expansion of the Hurwitz zeta function

95 98

1

4. The ∫ log Γ( x) cos pπ x dx integral

111

0

1

5. The ∫ log Γ( x) sin pπ x dx integral

122

6. Some integrals involving log sin π x

128

0

Appendices A. Some series for log Γ( x) B. Fourier coefficients for log x C. Some aspects of the multiple gamma functions D. Table of relevant integrals and series

136 139 143 145

1. A connection with the sine and cosine integrals

The frequently thumbed table of integrals compiled by Gradshteyn and Ryzhik [33] contains only a handful of definite integrals involving log Γ( x) . Of these, one of the apparently more complex examples is the following formula recorded in [33, p.650, 6.443.5] which is stated to be valid for a > 0 1

1

∫ log Γ( x + a) sin 2kπ x dx = − 2kπ [log a + cos(2kπ a) Ci (2kπ a) − sin(2kπ a) si (2kπ a)] 0

but, as shown below, the corrected integral which is in fact valid for a ≥ 0 is Proposition 1.1 1

(1.1) ∫ log Γ( x + a ) sin 2kπ x dx = − 0

1 [log a − cos(2kπ a) Ci (2kπ a) − sin(2kπ a) si (2kπ a)] 2kπ

2

where si ( x) and Ci ( x) are the sine and cosine integrals defined [33, p.878] by ∞

sin t dt t x

si( x) = − ∫

(1.2)

and for x > 0 ∞

cos t cos t − 1 Ci ( x) = − ∫ dt = γ + log x + ∫ dt t t x 0

(1.3)

x

where γ is Euler’s constant. Note that Si ( x) is a slightly different sine integral which is defined in [33, p.878] and also in [1, p.231] by x

sin t dt t 0

Si ( x) = ∫

(1.4) We have

∞

x

∞

sin t sin t sin t si ( x) = − ∫ dt = ∫ dt − ∫ dt t t t x 0 0

and using the well-known integral from Fourier series analysis (1.5)

π 2

∞

sin t dt t 0

=∫

we therefore see that the two sine integrals are intimately related by (1.6)

si ( x) = Si ( x) −

π 2

Proof:

We start with equation (A.8) which is derived in Appendix A to this paper (1.7)

∞ x+a⎤ ⎡ log Γ( x + a ) = − log( x + a) − γ ( x + a) + ∑ ⎢log n − log ( n + a + x ) + n ⎥⎦ n =1 ⎣

and multiply this by sin pπ x and integrate to obtain

3

1

1

1

0

0

0

∫ log Γ( x + a) sin pπ x dx = − ∫ log( x + a) sin pπ x dx − γ ∫ ( x + a) sin pπ x dx 1

∞ x+a⎤ ⎡ sin pπ x dx + ∑ ∫ ⎢log n − log ( n + a + x ) + n ⎥⎦ n =1 0 ⎣

where we have assumed that the interchange of the order of integration and summation is valid. The three component integrals are dealt with in turn below. We have using integration by parts

∫ log(a +x) sin pπ x dx = −

log(a + x) cos pπ x 1 + pπ pπ

∫

cos pπ x dx a+x

and with the substitution t = a + x we get 1 pπ

∫

cos pπ x cos pπ a cos pπ t + sin pπ a sin pπ t dx = ∫ dt a+x pπ t = cos pπ a ∫

=

cos pπ t sin pπ t dt + sin pπ a ∫ dt pπ t pπ t

cos pπ a cos u sin pπ a sin u du + du ∫ pπ u pπ ∫ u

and reference to the definitions of the sine and cosine integrals shows that this is equivalent to =

cos pπ a Ci (u ) + sin pπ a Si (u ) pπ

Hence we obtain

∫ log(a +x) sin pπ x dx =

cos pπ a Ci[ pπ (a + x )] + sin pπ a Si[ pπ (a + x )] − log(a + x) cos pπ x pπ

which may be easily verified by differentiation since

d sin x Si ( x) = dx x

and

d cos x Ci ( x) = dx x 4

The definite integral becomes 1

∫ log(a +x) sin pπ x dx

(1.8)

0

=

−

cos pπ a {Ci [ pπ (a + 1)] − Ci [ pπ a]} + sin pπ a {Si [ pπ ( a + 1)] − Si [ pπ a]} pπ log(a + 1) cos pπ − log a pπ

We also have the definite integral (where k is an integer) 1

∫ log(a +x) sin 2kπ x dx

(1.9)

0

=

−

cos 2kπ a {Ci [2kπ (a + 1)] − Ci [2kπ a ]} + sin 2kπ a {si [2kπ (a + 1)] − si [2kπ a]} 2 kπ log(a + 1) − log a 2 kπ

where we note from (1.6) that Si (u ) − Si (v) = si (u ) − si (v) . We also have the elementary integral

∫ ( x + a) sin pπ x dx =

sin pπ x ( x + a ) cos pπ x − p 2π 2 pπ

and therefore 1

∫ ( x + a) sin pπ x dx = 0

sin pπ (1 + a) cos pπ − a − p 2π 2 pπ

With p = 2k we get 1

1

∫ ( x + a) sin 2kπ x dx = − 2kπ 0

We see by letting a → n + a in (1.8) that

5

1

∫ log(n + a +x) sin 2kπ x dx =

cos 2kπ (a + n ) {Ci [2kπ (a + n + 1)] − Ci [2kπ (a + n )]} 2 kπ

0

+

−

=

−

−

sin 2kπ (a + n ) {si [2kπ (a + n + 1)] − si [2kπ (a + n )]} 2 kπ log(a + n + 1) − log(a + n) 2 kπ

cos 2kπ a {Ci [2kπ (a + n + 1)] − Ci [2kπ (a + n )]} 2kπ

sin 2kπ a {si [2kπ (a + n + 1)] − si [2kπ (a + n )]} 2 kπ log(a + n + 1) − log(a + n) 2 kπ

Due to telescoping we see that cos 2kπ a 2kπ

∑ {Ci [2kπ (a + n + 1)] − Ci [2kπ (a + n )]} = −

sin 2kπ a 2 kπ

∑ {si [2kπ (a + n + 1)] − si [2kπ (a + n )]} = −

∞

n =1

∞

n =1

cos 2kπ a Ci [2kπ (a + 1)] 2kπ

sin 2kπ a si [2kπ ( a + 1)] 2 kπ

We also see that 1 2 kπ

N

⎡

1⎤

1

∑ ⎢⎣log(a + n + 1) − log(a + n) − n ⎥⎦ = 2kπ [log( N + 1 + a) − H n =1

N

− log(1 + a ) ]

and as N → ∞ we obtain 1 2 kπ

∞

⎡

1⎤

∑ ⎢⎣log(a + n + 1) − log(a + n) − n ⎥⎦ = − n =1

1

γ + log(1 + a ) 2 kπ

Finally, we note that the integral involving ∫ log n sin 2kπ x dx vanishes and we have 0

thereby obtained

6

1

∫ log Γ( x + a) sin 2kπ x dx 0

=−

+

+

cos 2kπ a {Ci [2kπ (a + 1)] − Ci [2kπ a]} + sin 2kπ a {si [2kπ (a + 1)] − si [2kπ a ]} 2 kπ log(a + 1) − log a cos 2kπ a Ci [2kπ (a + 1)] sin 2kπ a si [2kπ (a + 1)] + − 2 kπ 2 kπ 2 kπ

γ 2 kπ

−

γ + log(a + 1) 2 kπ

which simplifies to 1

1

∫ log Γ( x + a) sin 2kπ x dx = − 2kπ [log a − cos(2kπ a) Ci (2kπ a) − sin(2kπ a) si (2kπ a)] 0

This therefore shows that one of the signs in (1.1) is recorded incorrectly in Gradshteyn and Ryzhik [33, p.650, 6.443.5] (and also in “Integrals and Series”, Volume 2, p.60 by Prudnikov et al. [48]) and, in this regard, I note that both Havil [37, p.126] and Elizalde [29] define Ci ( x) as the negative of (1.3); it seems that this lack of consistency in the definition of Ci ( x) is likely to be the source of the error. The definitions used in this paper correspond with those employed by Nielsen [44] and Nörlund [46] (except that those authors use the notation ci ( x) for Ci ( x)) . The similarity between the integral definitions in (1.2) and (1.3) does indeed support the notation ci ( x) instead of Ci ( x) which, unfortunately, has been used in this paper and elsewhere. Equation (1.1) also applies in the limit as a → 0 because y ⎡ cos t − 1 ⎤ dt ⎥ lim[cos y Ci ( y ) − log y ] = lim ⎢γ cos y + log y[cos y − 1] + cos y ∫ y →0 y →0 t 0 ⎣⎢ ⎦⎥

so that (1.9.1)

lim[cos y Ci ( y ) − log y ] = γ y →0

since, applying L’Hôpital’s rule, we see that ⎡ cos y − 1 ⎤ lim[log y (cos y − 1)] = lim ⎢ y log y y →0 y →0 y ⎥⎦ ⎣

7

⎡ cos y − 1 ⎤ = lim [ y log y ] lim ⎢ y →0 y →0 y ⎥⎦ ⎣ = − lim [ y log y ] lim [sin y ] = 0 y →0

y →0

and we therefore obtain the well known Fourier coefficients [33, p.650, 6.443.3] 1

(1.9.2)

∫ log Γ( x) sin 2kπ x dx = 0

γ + log 2kπ 2 kπ

With a = 1/ 2 we have 1

(1.10)

⎛

1⎞

1

∫ log Γ ⎝⎜ x + 2 ⎠⎟ sin 2kπ x dx = 2kπ ⎡⎣log 2 + (−1)

k

0

Ci (kπ ) ⎤⎦

□ Using the definition of Ci ( x) in (1.3) we see that ax

Ci (ax) − cos(ax) log x = γ + log ax − cos(ax) log x + ∫ 0

cos t − 1 dt t ax

= γ + log a − log x [ cos(ax) − 1] + ∫ 0

We consider the limit lim[cos(ax) − 1]log x = lim x log x x →0

x →0

cos(ax) − 1 x cos(ax) − 1 x →0 x

= lim x log x lim x →0

and using L’Hôpital’s rule we obtain lim x →0

cos(ax) − 1 a sin(ax) = − lim x → 0 x 1

which shows that (1.10.1)

lim[cos(ax) − 1]log x = 0 x →0

8

cos t − 1 dt t

and then taking the limit as x → 0 we obtain

lim [Ci(ax) − cos(ax) log x ] = γ + log a

(1.10.2)

x →0

Hence by letting a → 0 in (1.8) we obtain 1

∫ log x sin pπ x dx =

(1.10.3)

0

Ci ( pπ ) − γ − log pπ pπ

1

∫ log x sin 2kπ x dx =

(1.10.4)

0

Ci (2kπ ) − γ − log 2kπ 2 kπ

With a = 1 in (1.1) we have 1

∫ log Γ( x + 1) sin 2kπ x dx =

(1.11)

0

Ci (2kπ ) 2 kπ

which is (indirectly) reported with the correct sign in [33, p.650]. An alternative proof of (1.11) is shown below. We have 1

1

1

0

0

0

∫ log Γ( x + 1) sin 2kπ x dx = ∫ log x sin 2kπ x dx + ∫ log Γ( x) sin 2kπ x dx and equation (1.11) results by combining (1.10) and (1.10.4). □ 1

We now consider the integral ∫ log Γ( x + a ) sin(2k + 1)π x dx . 0

Proposition 1.2 1

∫ log Γ( x) sin(2k + 1)π x dx = 0

1 (2k + 1)π

k −1 ⎡ ⎛π ⎞ 1 1 ⎤ log 2 + + ⎢ ⎜ ⎟ ⎥ ∑ j =1 2 j + 1 ⎦ ⎣ ⎝ 2 ⎠ 2k + 1

Proof

We multiply (1.7) by sin(2k + 1)π x and integrate to obtain

9

1

1

1

0

0

0

∫ log Γ( x + a) sin(2k + 1)π x dx = − ∫ log( x + a) sin(2k + 1)π x dx − γ ∫ ( x + a) sin(2k + 1)π x dx 1

∞ x+a⎤ ⎡ + ∑ ∫ ⎢log n − log ( n + a + x ) + sin(2k + 1)π x dx n ⎥⎦ n =1 0 ⎣

These three integrals are dealt with in turn below. Using (1.8) we get 1

∫ log(a +x) sin(2k + 1)π x dx

(1.12)

0

=

+

+

cos(2k + 1)π a {Ci [(2k + 1)π ( a + 1)] − Ci [(2k + 1)π a]} (2k + 1)π sin(2k + 1)π a {si [(2k + 1)π (a + 1)] − si [(2k + 1)π a ]} (2k + 1)π log(a + 1) + log a (2k + 1)π

and 1

∫ log(n + a +x) sin(2k + 1)π x dx 0

=

+

+

cos[(2k + 1)(n + a)π ]{Ci [(2k + 1)π (n + a + 1)] − Ci [(2k + 1)π (n + a )]} (2k + 1)π sin[(2k + 1)(n + a )π ]{si [(2k + 1)π (n + a + 1)] − si [(2k + 1)(n + a )π ]} (2k + 1)π log(n + a + 1) + log(n + a ) (2k + 1)π

Since cos[(2k + 1)(n + a )π ] = (−1)n cos(2k + 1)aπ sin[(2k + 1)(n + a )π ] = (−1) n sin(2k + 1)aπ

10

we obtain 1

∫ log(n + a +x) sin(2k + 1)π x dx 0

=

cos[(2k + 1)aπ ](−1) n {Ci [(2k + 1)π (n + a + 1)] − Ci [(2k + 1)π (n + a)]}

+

+

(2k + 1)π sin[(2k + 1)aπ ](−1) n {si [(2k + 1)π (n + a + 1)] − si [(2k + 1)( n + a )π ]} (2k + 1)π log(n + a + 1) + log(n + a) (2k + 1)π

We saw above that 1

∫ ( x + a) sin pπ x dx = 0

sin pπ (1 + a ) cos pπ − a − p 2π 2 pπ

and hence we have 1

2a + 1

∫ ( x + a) sin(2k + 1)π x dx = (2k + 1)π 0

Since

∞

∞

n =1

n =1

∑ (−1)n [ xn+1 − xn ] = − x1 − 2∑ (−1)n xn we have

(1.13.1)

∞

∑ (−1) {Ci [(2k + 1)π (a + n + 1)] − Ci [(2k + 1)π (a + n )]} n

n =1

∞

= − Ci [(2k + 1)π (a + 1)] − 2∑ (−1) n Ci [(2k + 1)π (a + n )] n =1

and (1.13.2)

∞

∑ (−1) {si [(2k + 1)π (a + n + 1)] − si [(2k + 1)π (a + n )]} n

n =1

∞

= − si [(2k + 1)π (a + 1)] − 2∑ (−1) n si [(2k + 1)π (a + n )] n =1

11

We will see later in (1.70) that

ψ (a) = log a −

(1.13.3)

∞ 1 + 2∑ [cos(2nπ a)Ci (2nπ a) + sin(2nπ a) si (2nπ a)] 2a n =1

and we have [52, p.14] (by differentiating the corresponding expression for the log gamma function) 1 j =1 a + j − 1 k

ψ (a + k ) = ψ (a) + ∑ k −1

1 j =0 a + j

= ψ (a) + ∑

= ψ (a ) + H k(1) (a) where H n( m ) (a) is the generalised harmonic number function defined by k −1

1 m j =0 (a + j )

H k( m ) (a ) = ∑

With a = 1/ 2 we obtain [52, p.20] k −1 1 ⎛ 2k + 1 ⎞ (1.14) ψ ⎜ ⎟ = −γ − 2 log 2 + 2∑ ⎝ 2 ⎠ j =0 2 j + 1

and we then see from (1.13.3) that ∞ 1 ⎛ 2k + 1 ⎞ ⎛ 2k + 1 ⎞ = log − + 2 cos[(2k + 1)nπ ]Ci[(2k + 1)nπ ] ∑ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ 2k + 1 n =1

ψ⎜

or equivalently ∞

(1.15) 2∑ (−1) n Ci[(2k + 1)nπ ] = −γ − log 2 − log ( 2k + 1) + n =1

We next need to consider the series 1 ∞ ⎡ 2a + 1 ⎤ 2 log n − log(n + a + 1) − log(n + a ) + ∑ ⎢ 2k + 1 n =1 ⎣ n ⎥⎦ Using (1.7)

12

k −1 1 1 + 2∑ 2k + 1 j =0 2 j + 1

x+a⎤ ⎡ log Γ( x + a ) + log( x + a) + γ ( x + a ) = ∑ ⎢log n − log ( n + a + x ) + n ⎥⎦ n =1 ⎣ ∞

which, with x → 1 + x , becomes ∞ 1+ x + a ⎤ ⎡ log Γ(1 + x + a) + log(1 + x + a) + γ (1 + x + a) = ∑ ⎢log n − log ( n + a + 1 + x ) + n ⎥⎦ n =1 ⎣

Adding these two equations together and noting the functional equation log Γ(1 + x + a) = log( x + a ) + log Γ( x + a) results in

2 log Γ( x + a) + 2 log( x + a) + log(1 + x + a) + γ ( x + a) + γ = ∞ 2a + 1 + 2 x ⎤ ⎡ = ∑ ⎢ 2 log n − log ( n + a + 1 + x ) − log ( n + a + x ) + ⎥⎦ n n =1 ⎣

and with x = 0 we see that 2 log Γ(1 + a) + log(1 + a) + γ (1 + a)

(1.16)

2a + 1 ⎤ ⎡ = ∑ ⎢ 2 log n − log(n + a + 1) − log(n + a ) + n ⎥⎦ n =1 ⎣ ∞

With a = 0 we obtain the familiar formula ∞

⎡

1⎤

∑ ⎢⎣log n − log(n + 1) + n ⎥⎦ = γ n =1

We therefore obtain by combining the various integrals 1

∫ log Γ( x + a) sin(2k + 1)π x dx 0

=−

−

cos(2k + 1)π a {Ci [(2k + 1)π (a + 1)] − Ci [(2k + 1)π a ]} (2k + 1)π

sin(2k + 1)π a {si [(2k + 1)π (a + 1)] − si [(2k + 1)π a ]} (2k + 1)π

13

−

log(a + 1) + log a (2a + 1)γ − (2k + 1)π (2k + 1)π

+

cos(2k + 1)π a Ci [(2k + 1)π (a + 1)] 2 cos(2k + 1)π a ∞ + (−1) n Ci [(2k + 1)π (a + n )] ∑ (2k + 1)π (2k + 1)π n =1

+

sin(2k + 1)π a si [(2k + 1)π (a + 1)] 2sin(2k + 1)π a ∞ + (−1) n si [(2k + 1)π (a + n )] ∑ (2k + 1)π (2k + 1)π n =1

+

2 log Γ(1 + a) + log(1 + a) + γ (1 + a) (2k + 1)π

which simplifies to 1

∫ log Γ( x + a) sin(2k + 1)π x dx

(1.17)

0

=

cos(2k + 1)π a Ci [(2k + 1)π a] + sin(2k + 1)π a si [(2k + 1)π a] (2k + 1)π

+

2 log Γ(1 + a) − γ a − log a (2k + 1)π

+

2 cos(2k + 1)π a ∞ (−1) n Ci [(2k + 1)π (a + n )] ∑ (2k + 1)π n =1

+

2sin(2k + 1)π a ∞ (−1) n si [(2k + 1)π (a + n )] ∑ (2k + 1)π n =1

With a = 0 we obtain 1

∫ log Γ( x) sin(2k + 1)π x dx 0

= lim a →0

Ci [(2k + 1)π a ] − log a 2 + (2k + 1)π (2k + 1)π

and referring to (1.15) and (1.19) this becomes

14

∞

∑ (−1) Ci [(2k + 1)π n] n

n =1

=

γ + log[(2k + 1)π ] 1 + (2k + 1)π (2k + 1)π

k −1 ⎡ 1 1 ⎤ − − − + + + γ log 2 log 2 k 1 2 ( ) ⎢ ⎥ ∑ 2k + 1 j =1 2 j + 1 ⎦ ⎣

Hence we obtain the known result (see also Section 4) 1

∫ log Γ( x) sin(2k + 1)π x dx =

(1.18)

0

k −1 ⎡ ⎛π ⎞ 1 1 1 ⎤ + 2∑ ⎢log ⎜ ⎟ + ⎥ (2k + 1)π ⎣ ⎝ 2 ⎠ 2k + 1 j =1 2 j + 1 ⎦

Referring to (1.3) we see that ux

Ci (ux) = γ + log u + log x + ∫ 0

cos t − 1 dt t

and we have lim [Ci (ux) − log x ] = γ + log u

(1.19)

x →0

We also see that ux

cos(ux)Ci (ux) = [γ + log u ] + cos(ux) log x + cos(ux) ∫ 0

cos t − 1 dt t ux

= [γ + log u ] + log x + [cos(ux) − 1]log x + cos(ux) ∫ 0

cos t − 1 dt t

and thus lim [ cos(ux)Ci (ux) − log x ] = [γ + log u ]cos(ux) + lim[cos(ux) − 1]log x x →0

x →0

Then using (1.10.1) we obtain (1.20)

lim [ cos(ux)Ci (ux) − log x ] = γ + log u x →0

and in particular we have (as used above) (1.21)

lim a →0

Ci [(2k + 1)π a ] − log a γ + log[(2k + 1)π ] = (2k + 1)π (2k + 1)π □

With a = 1 in (1.17) we obtain

15

1

∫ log Γ( x + 1) sin(2k + 1)π x dx

(1.22)

0

=−

1 (2k + 1)π

∞ ⎛ ⎞ [(2 1) ] 2 (−1) n Ci [(2k + 1)π ( n + 1)] ⎟ + + + Ci k π γ ∑ ⎜ n =1 ⎝ ⎠

=−

1 (2k + 1)π

∞ ⎛ ⎞ + + − [(2 1) ] 2 (−1) m Ci [(2k + 1)π m)] ⎟ Ci k π γ ∑ ⎜ m=2 ⎝ ⎠

=−

1 (2k + 1)π

∞ ⎛ ⎞ − + + − [(2 1) ] 2 (−1) n Ci [(2k + 1)π n)] ⎟ Ci k π γ ∑ ⎜ n =1 ⎝ ⎠

Letting p = 2k + 1 in (1.10.3) gives us 1

∫ log x sin(2k + 1)π x dx = 0

Ci [(2k + 1)π ] − γ − log[(2k + 1)π ] (2k + 1)π

and hence we have 1

1

1

0

0

0

∫ log Γ( x + 1) sin(2k + 1)π x dx = ∫ log x sin(2k + 1)π x dx + ∫ log Γ( x) sin(2k + 1)π x dx We then have −

1 (2k + 1)π

∞ ⎛ ⎞ − + + − [(2 1) ] 2 (−1) n Ci [(2k + 1)π n)] ⎟ = Ci k π γ ∑ ⎜ n =1 ⎝ ⎠

Ci [(2k + 1)π ] − γ − log[(2k + 1)π ] + ∫ log Γ( x) sin(2k + 1)π x dx (2k + 1)π 0 1

or 1

(1.23)

∫ log Γ( x) sin(2k + 1)π x dx = 0

2 (2k + 1)π

∞

∑ (−1) Ci [(2k + 1)π n)] + n

n =1

log[(2k + 1)π ] (2k + 1)π

which corresponds with (1.18). Therefore we have ∞ k −1 1 1 ⎛π ⎞ + 2∑ (1.23.1) 2∑ (−1) n Ci [(2k + 1)π n)] + log[(2k + 1)π ] = log ⎜ ⎟ + ⎝ 2 ⎠ 2k + 1 n =1 j =1 2 j + 1

and with k = 0 this becomes

16

∞

(1.23.2) 2∑ (−1) n Ci (π n) = 1 − log 2 n =1

There is however an unexplained difference between this result and (1.72) ∞

2∑ (−1) n Ci (nπ ) = 1 − γ − log 2 n =1

□ 1

We now consider the integral ∫ log Γ( x + a ) cos 2kπ x dx . 0

Proposition 1.3 1

1

∫ log Γ( x + a) cos 2kπ x dx = − 2kπ [ − sin(2kπ a)Ci(2kπ a) + cos(2kπ a)si(2kπ a)] 0

Proof

We multiply (1.7) by cos pπ x and integrate to obtain 1

1

1

0

0

0

∫ log Γ( x + a) cos pπ x dx = − ∫ log( x + a) cos pπ x dx − γ ∫ ( x + a) cos pπ x dx 1

∞ x+a⎤ ⎡ + ∑ ∫ ⎢log n − log ( n + a + x ) + cos pπ x dx n ⎥⎦ n =1 0 ⎣

These three integrals are dealt with in turn below. As before, integration by parts gives us

∫ log(a +x) cos pπ x dx =

log(a + x) sin pπ x 1 − pπ pπ

∫

sin pπ x dx a+x

and with the substitution t = a + x we get 1 sin pπ x cos pπ a sin pπ t − sin pπ a cos pπ t dx = ∫ dt ∫ pπ a+x pπ t

17

= cos pπ a ∫

=

sin pπ t cos pπ t dt − sin pπ a ∫ dt pπ t pπ t

cos pπ a sin u sin upπ a cos du − du ∫ pπ u pπ ∫ u

and reference to the definitions of the sine and cosine integrals shows that this is equivalent to cos pπ a Si (u ) − sin pπ a Ci (u ) = pπ Hence we obtain

∫ log(a +x) cos pπ x dx =

sin pπ a Ci[ pπ ( a + x )] − cos pπ a Si[ pπ ( a + x )] + log( a + x) sin pπ x pπ

The definite integral becomes 1

(1.24)

∫ log(a +x) cos pπ x dx 0

=

sin pπ a {Ci [ pπ (a + 1)] − Ci [ pπ a]} − cos pπ a {si [ pπ ( a + 1)] − si [ pπ a]} pπ

+

log(a + 1) sin pπ pπ

We also have the definite integral (where k is an integer) 1

(1.25)

∫ log(a +x) cos 2kπ x dx 0

=

sin 2kπ a {Ci [2kπ (a + 1)] − Ci [2kπ a]} − cos 2kπ a {si [2kπ (a + 1)] − si [2kπ a]} 2 kπ

With a = 0 we obtain 1

(1.26)

∫ log x cos 2kπ x dx = − 0

Si [2kπ ] 2 kπ

We also have

18

∫ ( x + a) cos pπ x dx =

cos pπ x ( x + a) sin pπ x + p 2π 2 pπ

and therefore 1

∫ ( x + a) cos pπ x dx = 0

cos pπ − 1 (1 + a ) sin pπ + p 2π 2 pπ

With p = 2k we get 1

∫ ( x + a) cos 2kπ x dx = 0 0

We see by letting a → n + a in (1.25) that 1

∫ log(n + a +x) cos 2kπ x dx =

sin 2kπ (a + n ) {Ci [2kπ ( a + n + 1)] − Ci [2kπ (a + n )]} 2kπ

0

−

=

−

cos 2kπ ( a + n ) {si [2kπ ( a + n + 1)] − si [2kπ (a + n )]} 2 kπ sin 2kπ a {Ci [2kπ (a + n + 1)] − Ci [2kπ (a + n )]} 2 kπ

cos 2kπ a {si [2kπ (a + n + 1)] − si [2kπ (a + n )]} 2 kπ

As before, due to telescoping we see that sin 2kπ a 2 kπ

cos 2kπ a 2 kπ

∞

∑ {Ci [2kπ (a + n + 1)] − Ci [2kπ (a + n )]} = − n =1

∞

∑ {si [2kπ (a + n + 1)] − si [2kπ (a + n )]} = − n =1

1

sin 2kπ a Ci [2kπ (a + 1)] 2 kπ

cos 2kπ a si [2kπ (a + 1)] 2 kπ

Finally, we note that the integral involving ∫ log n cos 2kπ x dx vanishes and we have 0

thereby obtained

19

1

∫ log Γ( x + a) cos 2kπ x dx 0

=−

+

sin 2kπ a {Ci [2kπ (a + 1)] − Ci [2kπ a]} − cos 2kπ a {si [2kπ (a + 1)] − si [2kπ a]} 2 kπ

sin 2kπ a Ci [2kπ (a + 1)] cos 2kπ a si [2kπ (a + 1)] − 2 kπ 2 kπ

which simplifies to 1

(1.27)

1

∫ log Γ( x + a) cos 2kπ x dx = − 2kπ [ − sin(2kπ a)Ci(2kπ a) + cos(2kπ a) si(2kπ a)] 0

This is the companion integral to (1.1) and this corrects the entry reported in [33, p.650, 6.443.3] for a > 0; the sign error referred to in (1.1) is also replicated here. This sign error also features in (1.76) and (1.78). We may also note that (1.27) is also valid for a = 0 because ⎡ sin u ⎤ lim[sin u.log u ] = lim ⎢ u log u ⎥ = 0 u →0 u →0 ⎣ u ⎦

and therefore using (1.3) we see that lim sin(2kπ a)Ci (2kπ a) = 0 . a →0

Since from (1.16) si (0) = −

π 2

, we therefore obtain the well known Fourier coefficients

[33, p.650, 6.443.3] (1.28)

1

1

∫ log Γ( x) cos 2kπ x dx = 4k 0

Further derivations of this integral are contained in (2.28) and (3.9) below. Letting a = 1/ 2 results in 1

(1.28.1)

1⎞ (−1) k +1 ⎛ log Γ + cos 2 = x k π x dx si (kπ ) ⎜ ⎟ ∫0 2⎠ 2 kπ ⎝

With a = 1 in (1.27) we get for k ≥ 1

20

1

∫ log Γ( x + 1) cos 2kπ x dx = −

(1.29)

0

si (2kπ ) 2 kπ

This may be easily confirmed as follows. We have 1

1

1

0

0

0

∫ log Γ( x + 1) cos 2kπ x dx = ∫ log x cos 2kπ x dx + ∫ log Γ( x) cos 2kπ x dx and (1.29) results by using (1.26) and (1.28). Using the duplication formula for the gamma function 1⎞ 1 ⎛ log Γ(2 x) = log Γ( x) + log Γ ⎜ x + ⎟ + ( 2 x − 1) log 2 − log π 2⎠ 2 ⎝ we have 1

1

1

1⎞ ⎛ ∫0 log Γ(2 x) cos 2kπ x dx = ∫0 log Γ( x) cos 2kπ x dx + ∫0 log Γ ⎜⎝ x + 2 ⎟⎠ cos 2kπ x dx 1

+ log 2 ∫ ( 2 x − 1) cos 2kπ x dx 0

We note that 1

∫ ( 2 x − 1) cos 2kπ x dx = 0 0

1

∫ log Γ(2 x) cos 2kπ x dx = 0

1

∫ log Γ(2 x) cos 2kπ x dx = 0

1 (−1) k +1 + si (kπ ) 4k 2 kπ 2

1 log Γ(u ) cos kπ u du 2 ∫0

2

1

2

0

0

1

2

1

1

0

∫ log Γ(u ) cos kπ u du = ∫ log Γ(u ) cos kπ u du + ∫ log Γ(u ) cos kπ u du ∫ log Γ(u ) cos kπ u du = ∫ log Γ(1 + x) cos kπ (1 + x) dx 21

1

= (−1) k ∫ log Γ(1 + x) cos kπ x dx 0

1

∫ log Γ(2 x) cos 2kπ x dx = 0

1

1

1 1 log Γ( x ) cos kπ x dx +(−1) k ∫ log Γ(1 + x) cos kπ x dx ∫ 20 20 1

1

1 1 = [1 + (−1) k ]∫ log Γ( x) cos kπ x dx + (−1) k ∫ log x cos kπ x dx 2 20 0

We note from Appendix B that 1

∫ log x cos kπ x dx = − 0

Si (kπ ) kπ

and unfortunately we simply end up with 1

1 1 (−1) k k [1 + (−1) ]∫ log Γ( x) cos kπ x dx = + 2 4k 4k 0

□ We recall (1.1) and (1.27) 1

1

∫ log Γ( x + a) sin 2nπ x dx = − 2π n [log a − cos(2nπ a)Ci(2nπ a) + sin(2nπ a)si(2nπ a)] 0

1

1

∫ log Γ( x + a) cos 2nπ x dx = − 2nπ [ − sin(2nπ a)Ci(2nπ a) + cos(2nπ a)si(2nπ a)] 0

and multiplying each by sin(2nπ a) and cos(2nπ a) respectively we get 1

sin(2nπ a ) ∫ log Γ( x + a ) sin 2nπ x dx = 0

−

1

⎡⎣ log a sin(2nπ a ) − cos(2nπ a ) sin(2nπ a )Ci (2nπ a ) − sin 2 (2nπ a ) si (2nπ a ) ⎤⎦ 2π n

and 1

cos(2nπ a ) ∫ log Γ( x + a ) cos 2nπ x dx = 0

22

−

1

⎡⎣ − cos(2nπ a) sin(2nπ a)Ci (2nπ a) + cos 2 (2nπ a) si (2nπ a) ⎤⎦ 2π n

Adding these two equations results in 1

∫ log Γ( x + a)[sin(2nπ a) sin 2nπ x + cos(2nπ a) cos 2nπ x]dx 0

=−

1 2π n

[log a sin(2nπ a) − sin(4nπ a)Ci(2nπ a) + cos(4nπ a)si(2nπ a)]

or equivalently 1

∫ log Γ( x + a) cos[2nπ ( x − a)]dx 0

=−

1 2π n

[log a sin(2nπ a) − sin(4nπ a)Ci(2nπ a) + cos(4nπ a)si(2nπ a)]

With a = 0 we immediately obtain (1.28) 1

1

∫ log Γ( x) cos 2nπ x dx = 4n 0

and a = 1/ 2 gives us (1.29) 1

∫ log Γ( x + 1/ 2) cos 2nπ x dx = (−1) 0

n +1

si (nπ ) 2π n

Similarly subtraction gives us 1

1

0

0

sin(2nπ a ) ∫ log Γ( x + a) sin 2nπ x dx − cos(2nπ a) ∫ log Γ( x + a) cos 2nπ x dx

=−

+

1

⎡⎣log a sin(2nπ a ) − cos(2nπ a ) sin(2nπ a )Ci (2nπ a ) − sin 2 (2nπ a ) si (2nπ a ) ⎤⎦ 2π n

1

⎡ − cos(2nπ a ) sin(2nπ a )Ci (2nπ a) + cos 2 (2nπ a ) si (2nπ a) ⎤⎦

2π n ⎣

23

=

1 2π n

[ − log a sin(2nπ a) + si(2nπ a)]

and we have 1

1

∫ log Γ( x + a) cos[2nπ ( x + a)]dx = 2π n [log a sin(2nπ a) − si(2nπ a)]

(1.29.1)

0

The substitution u = x + a gives us a +1

1

∫ log Γ(u ) cos 2nπ u du = 2π n [log a sin(2nπ a) − si(2nπ a)]

(1.29.2)

a

and differentiation with respect to a results in the very obvious identity log Γ(a + 1) cos 2nπ (a + 1) − log Γ(a ) cos 2nπ a

(1.29.3)

=

1 ⎡ 1 1 ⎤ 2nπ log a cos(2nπ a ) + sin(2nπ a ) − sin(2nπ a) ⎥ ⎢ 2π n ⎣ a a ⎦

= log a cos(2nπ a) 1

Finally, we consider the integral ∫ log Γ( x + a ) cos(2k + 1)π x dx . 0

Proposition 1.4 1

∫ logΓ( x) cos(2k + 1)π x dx = 0

∞ ⎤ 2 ⎡ log(2π ) + γ log n + 2 ∑ 2 ⎢ 2 2 2 ⎥ π ⎣ (2k + 1) n =1 4 n − (2 k + 1) ⎦

An attempted proof

As before we have 1

1

1

0

0

0

∫ log Γ( x + a) cos(2k + 1)π x dx = − ∫ log( x + a) cos(2k + 1)π x dx − γ ∫ ( x + a) cos(2k + 1)π x dx 1

∞ x+a⎤ ⎡ + ∑ ∫ ⎢log n − log ( n + a + x ) + cos(2k + 1)π x dx n ⎥⎦ n =1 0 ⎣

Using the definite integral (1.24) we see that

24

1

∫ log(a +x) cos(2k + 1)π x dx

(1.30)

0

=

sin(2k + 1)π a {Ci [(2k + 1)π ( a + 1)] − Ci [(2k + 1)π a]}

−

(2k + 1)π cos(2k + 1)π a {si [(2k + 1)π (a + 1)] − si [(2k + 1)π a]}

1

(2k + 1)π 2

∫ ( x + a) cos(2k + 1)π x dx = − (2k + 1) π 2

2

0

1

∫ log(n + a +x) cos(2k + 1)π x dx 0

=

−

sin[(2k + 1)(n + a)π ]{Ci [(2k + 1)π (a + n + 1)] − Ci [(2k + 1)π (a + n )]} (2k + 1)π

cos[(2k + 1)( n + a)π ]{si [(2k + 1)π ( a + n + 1)] − si [(2k + 1)π ( a + n )]} (2k + 1)π

Since sin[(2k + 1)(n + a )π ] = (−1) n sin(2k + 1)aπ cos[(2k + 1)(n + a )π ] = (−1)n cos(2k + 1)aπ this becomes =

−

(−1) n sin(2k + 1)aπ {Ci [(2k + 1)π (a + n + 1)] − Ci [(2k + 1)π (a + n )]} (2k + 1)π

(−1) n cos(2k + 1)aπ {si [(2k + 1)π (a + n + 1)] − si [(2k + 1)π (a + n )]} (2k + 1)π

We have as in (1.13.1) ∞

∑ (−1) {Ci [(2k + 1)π (a + n + 1)] − Ci [(2k + 1)π (a + n )]} n

n =1

25

∞

= − Ci [(2k + 1)π (a + 1)] − 2∑ (−1) n Ci [(2k + 1)π (a + n )] n =1

and ∞

∑ (−1) {si [(2k + 1)π (a + n + 1)] − si [(2k + 1)π (a + n )]} n

n =1

∞

= − si [(2k + 1)π (a + 1)] − 2∑ (−1) n si [(2k + 1)π (a + n )] n =1

1

Finally, we note that the integral involving ∫ log n cos(2k + 1)π x dx vanishes and we have 0

thereby obtained 1

∫ log Γ( x + a) cos(2k + 1)π x dx 0

=−

+

+

+

−

sin(2k + 1)π a {Ci [(2k + 1)π (a + 1)] − Ci [(2k + 1)π a ]} (2k + 1)π

cos(2k + 1)π a {si [(2k + 1)π (a + 1)] − si [(2k + 1)π a ]} (2k + 1)π 2γ (2k + 1) 2 π 2

sin(2k + 1)aπ Ci [(2k + 1)π (a + 1)] 2sin(2k + 1)aπ + (2k + 1)π (2k + 1)π cos(2k + 1)aπ si [(2k + 1)π (a + 1)] 2 cos(2k + 1)aπ − (2k + 1)π (2k + 1)π

∞

∑ (−1) Ci [(2k + 1)π (a + n )] n

n =1 ∞

∑ (−1) n =1

∞

2 2 2 n =1 (2k + 1) π n

−∑

which simplifies to sin(2k + 1)π a Ci [(2k + 1)π a ] − cos(2k + 1)π a si [(2k + 1)π a] (2k + 1)π 2γ + (2k + 1) 2 π 2

=

26

n

si [(2k + 1)π (a + n )]

2sin(2k + 1)aπ + (2k + 1)π −

2 cos(2k + 1)aπ (2k + 1)π

∞

∑ (−1) Ci [(2k + 1)π (a + n )] n

n =1 ∞

∑ (−1)

n

si [(2k + 1)π (a + n )]

n =1

∞

2 2 2 n =1 (2k + 1) π n

−∑

With a = 0 we have 1

∫ log Γ( x) cos(2k + 1)π x dx 0

1 2γ 2 + − 2 2 2k + 1 (2k + 1) π (2k + 1)π

=

∞

∞

2 2 2 n =1 (2k + 1) π n

∑ (−1)n si [(2k + 1)π n] + ∑ n =1

This is obviously incorrect because of the appearance of the divergent series; the author would appreciate receiving a corrected version of the proof. Using a different approach we note from (4.9) that 1

∫ logΓ( x) cos(2k + 1)π x dx = 0

∞ ⎤ 2 ⎡ log(2π ) + γ log n + 2 ∑ 2 ⎢ 2 2 2 ⎥ π ⎣ (2k + 1) n =1 4 n − (2 k + 1) ⎦

Representations of log Γ( x) in terms of the sine and cosine integrals

The following analysis is extracted from an earlier paper [21]; it indicates how the log Γ( x) function is itself intimately connected with the sine and cosine integrals. Whittaker & Watson [56, p.261] posed the following question: Prove that for all values of a except negative real values we have ∞

(1.40)

1 1⎞ 1 ∞ sin(2nπ x) ⎛ log Γ(a) = log(2π ) + ⎜ a − ⎟ log a − a + ∑ ∫ dx π n =1 0 n( x + a ) 2 2⎠ ⎝

and this result was attributed by Stieltjes to Bourguet. Equation (1.40) may also be derived using the Euler-Maclaurin summation formula (see for example Knopp’s book [39, p.530]). By differentiation we can easily see that

27

d sin(2nπ x) ( cos(2nπ a) Si[2nπ ( x + a)] − sin(2nπ a)Ci[2nπ ( x + a)]) = dx x+a and we therefore have M

sin(2nπ x) ∫0 x + a dx = ( cos(2nπ a)Si[2nπ ( x + a)] − sin(2nπ a)Ci[2nπ ( x + a)]) 0

M

= cos(2nπ a ) {Si[2nπ ( M + a )] − Si[2nπ a ]} − sin(2nπ a ) {Ci[2nπ ( M + a )] − Ci[2nπ a ]}

From (1.4) and (1.5) we see that lim Si[2nπ ( M + a )] =

M →∞

π 2

and from (1.3) we see that

lim Ci[2nπ ( M + a)] = 0

M →∞

Hence we obtain as M → ∞ ∞

sin(2nπ x) ⎧π ⎫ dx = cos(2nπ a) ⎨ − Si (2nπ a) ⎬ + sin(2nπ a)Ci (2nπ a) x+a ⎩2 ⎭ 0

∫

and reference to (1.6) shows that this is equal to

= − cos(2nπ a) si (2nπ a ) + sin(2nπ a)Ci (2nπ a) Therefore using Bourguet’s formula we have (1.41)

log Γ(a ) = 1 1⎞ 1 ∞ 1 ⎛ log(2π ) + ⎜ a − ⎟ log a − a + ∑ [sin(2nπ a)Ci (2nπ a) − cos(2nπ a) si (2nπ a)] π n =1 n 2 2⎠ ⎝

Nielsen [44, p.79] also reports a similar formula which is valid for 0 < a < 1 (1.41.1) log Γ(a) =

1 1 ∞ 1 log(2π ) − 1 − log a + ∑ [sin 2nπ a Ci(2nπ ) − cos 2nπ a si (2nπ )] π n =1 n 2

and a derivation of this is contained in Appendix B. Note that in this case, the sine and cosine integral functions do not contain the variable a in their arguments. 28

Equation (1.41) may be expressed as ∞ 1 1⎞ ⎛ log Γ(a) = log(2π ) + ⎜ a − ⎟ log a − a + ∑ Cn 2 2⎠ ⎝ n =1

and we have with a → 2a ∞ 1 1⎞ ⎛ log Γ(2a ) = log(2π ) + ⎜ 2a − ⎟ log(2a) − 2a + 2∑ C2 n 2 2⎠ ⎝ n =1 ∞

∞

∞

n =1

n =1

n =1

Since 2∑ C2 n = ∑ Cn + ∑ (−1) n Cn we have ∞ 1⎞ 1⎞ ⎛ ⎛ log Γ(2a ) = ⎜ 2a − ⎟ log(2a) − a + log Γ(a) − ⎜ a − ⎟ log a + ∑ (−1) n Cn 2⎠ 2⎠ ⎝ ⎝ n =1

and using Legendre’s duplication formula for the gamma function [52, p.7] ⎛ ⎝

1⎞

π Γ(2a) = 22 a −1 Γ(a)Γ ⎜ a + ⎟ 2 ⎠

this results in (1.42)

log Γ(a + 1/ 2) =

1 1 ∞ (−1) n log(2π ) + a log a − a + ∑ [sin(2nπ a)Ci (2nπ a) − cos(2nπ a ) si(2nπ a )] π n =1 n 2

which is also reported by Nörlund [46, p.114]. Since si (0) = −π / 2 this identity may be easily verified for a = 0 . With a = 1/ 2 we obtain (1.43)

si (nπ ) π π = log π − n 2 2 n =1 ∞

∑

which is contained in [44, p,82]. Letting a = 1 we have (1.44)

∞

∑ (−1) n =1

n

si (2nπ ) 3 = π log 2 − π n 2 □

Elizalde [29] reported in 1985 that for a > 0

29

ς ′(−1, a) =

(1.45)

1 1 1 −ς (−1, a) log a − a 2 + − 2 4 12 2π

∞

1

∑n n =1

2

[cos(2nπ a)Ci(2nπ a) + sin(2nπ a) si (2nπ a)]

where ς ( s, a) is the Hurwitz zeta function. Since si ( x) = Si ( x) −

(1.46)

1 4

ς ′(−1, a) = −ς (−1, a ) log a − a 2 +

−

∞

1 2π

2

1

∑n n =1

2

1 1 + 12 4π

π 2

this may be written as

sin(2nπ a ) n2 n =1 ∞

∑

[cos(2nπ a)Ci (2nπ a) + sin(2nπ a) Si (2nπ a)]

Elizalde [29] gave no indication of the source of the above identity but differentiation of (1.45) sheds more light on the subject: we have ∂ 1⎞ 1⎛ 1⎞1 1 1 ς (2) ⎛ ς ′(−1, a) = ⎜ a − ⎟ log a + ⎜ a 2 − a + ⎟ − a − 2 ∂a 2⎠ 2⎝ 6⎠a 2 2π a ⎝ −

1

∞

1

∑ n [− sin(2nπ a)Ci(2nπ a) + cos(2nπ a)si(2nπ a)] π n =1

since

d cos x d sin x Ci ( x) = and si ( x) = . dx x dx x

This simplifies to ∂ 1⎞ 1 1 ∞ 1 ⎛ ς ′(−1, a) = ⎜ a − ⎟ log a − + ∑ [sin(2nπ a)Ci (2nπ a) − cos(2nπ a ) si (2nπ a )] ∂a 2⎠ 2 π n =1 n ⎝ We have ∂ ∂ ς ( s, a) = −ς ( s + 1, a) − sς ′( s + 1, a) ∂a ∂s and hence ∂ ς ′(−1, a) = −ς (0, a) + ς ′(0, a) ∂a Then using Lerch’s identity

30

1 2

ς ′(0, a) = log Γ(a) − log(2π )

(1.46.1) this becomes

∂ 1 ς ′(−1, a) = −ς (0, a) + log Γ(a) − log(2π ) ∂a 2 We then obtain 1 −ς (0, a) + log Γ(a) − log(2π ) 2 1⎞ 1 1 ∞ 1 ⎛ = ⎜ a − ⎟ log a − + ∑ [sin(2nπ a)Ci (2nπ a ) − cos(2nπ a) si (2nπ a)] 2⎠ 2 π n =1 n ⎝ which, for 0 < a < 1, simplifies to

log Γ(a) =

(1.47)

1 1⎞ 1 ∞ 1 ⎛ log(2π ) + ⎜ a − ⎟ log a − a + ∑ [sin(2nπ a)Ci (2nπ a) − cos(2nπ a) si (2nπ a)] π n =1 n 2 2⎠ ⎝ 1 1⎞ 1 ∞ cos(2nπ a) ⎛ = log(2π ) + ⎜ a − ⎟ log a − a + ∑ 2 2⎠ 2 n =1 n ⎝ +

1

∞

1

∑ n [sin(2nπ a)Ci(2nπ a) − cos(2nπ a)Si(2nπ a)] π n =1

and using the Fourier series [55, p.148] cos(2nπ a) = − log[2sin(π a)] n n =1 ∞

∑

(1.47.1) this becomes (1.48)

1 1⎞ 1 ⎛ log Γ(a) = log(2π ) + ⎜ a − ⎟ log a − a − log[2sin(π a )] 2 2⎠ 2 ⎝ +

1

∞

1

∑ n [sin(2nπ a)Ci(2nπ a) − cos(2nπ a)Si(2nπ a)] π n =1

31

which corresponds with (1.41) above. The formula (1.47) was given by Nörlund in [46, p.114]. When a = 1/ 2 we obtain ∞

∑ (−1)

(1.49)

n

n =1

si (nπ ) π π = log 2 − n 2 2

This is a particular case of Nielsen’s formula [44, p.83] ∞

∑ (−1)

(1.49.1)

n =1

n

si (nx) π x = log 2 − n 2 2

It may be noted that Nielsen indicated that (1.49.1) was only valid for x ∈ ( −π , π ) . With a = 1/ 4 in (1.48) we get 1 1 1 ⎛1⎞ 1 log Γ ⎜ ⎟ = log(2π ) + log 2 − − log 2 2 4 4 ⎝4⎠ 2

(1.50)

+

1

∞

1

∑ n [sin(nπ / 2)Ci(nπ / 2) − cos(nπ / 2)Si(nπ / 2)] π n =1

We also have using (1.47) 1⎞ 1⎞ 1 ⎛ ⎛ log Γ(2a) = ⎜ 2a − ⎟ log 2 + log Γ(a) + a log ⎜ a + ⎟ − a − 2⎠ 2⎠ 2 ⎝ ⎝

(1.51)

+

(−1) n [sin(2nπ a )Ci (2nπ a + nπ ) − cos(2nπ a ) si (2nπ a + nπ )] ∑ π n =1 n 1

∞

and with a = 1/ 2 we obtain (1.52)

si (2nπ ) π = log(2π ) − π n 2 n =1 ∞

∑

and this concurs with Nielsen’s result [44, p.79]. A completely different derivation of (1.54) is given in (1.105) below. Using (1.47.1), which is valid for 0 < a < 1, equation (1.48) may be written as (1.53)

1 1⎞ 1 ⎛ log Γ(a) = log(2π ) + ⎜ a − ⎟ log a − a − log[2sin(π a )] 2 2⎠ 2 ⎝

32

+

∞

1

1

∑ n [sin(2nπ a)Ci(2nπ a) − cos(2nπ a)Si(2nπ a)] π n =1

This may be written more compactly as ⎡ Γ 2 (a ) sin(π a ) ⎤ 2 ∞ 1 log ⎢ ⎥ = ∑ [sin(2nπ a)Ci (2nπ a) − cos(2nπ a ) Si(2nπ a)] π a 2 a −1 ⎣ ⎦ π n =1 n

□ Integration of (1.47) results in x

∫ε log Γ(a) = 1 1 1 1 1 ( x − ε ) log(2π ) + x ⎡⎣ 2 − x + 2 ( x − 1) log x ⎤⎦ − ε ⎡⎣ 2 − ε + 2 ( ε − 1) log ε ⎤⎦ − x 2 + ε 2 2 4 4 2 2 +

−

+

1 4π

sin(2nπ x) 1 − ∑ n2 4π n =1 ∞

2

1 ∑ n [cos(2nπ x)Ci(2nπ x) + sin(2nπ x) Si(2nπ x) − log(2nπ x)] 2

n =1 ∞

1 2π

∑

∞

1 2π

sin(2nπε ) n2 n =1 ∞

2

1 ∑ n [cos(2nπε )Ci(2nπε ) + sin(2nπε )Si(2nπε ) − log(2nπε )] 2

n =1

Therefore as ε → 0 , and using (1.9.1) lim[cos y Ci ( y ) − log y ] = γ , we have y →0

x

∫ log Γ(a) da =

(1.60)

0

1 1 1 1 x log(2π ) + x ⎡⎣ 2 − x + 2 ( x − 1) log x ⎤⎦ − x 2 + 2 4 2 4π −

=

∑

1 γς (2) ∑ n [cos(2nπ x)Ci(2nπ x) + sin(2nπ x) Si(2nπ x) − log(2nπ x)] + 2π ∞

1 2π

sin(2nπ x) n2 n =1 ∞

2

n =1

2

2

1 1 1 1 x log(2π ) + x ⎡⎣ 2 − x + 2 ( x − 1) log x ⎤⎦ − x 2 + 2 4 2 4π

33

sin(2nπ x) 1 + log x n2 12 n =1 ∞

∑

−

1 2π 2

1 ς (2) log(2π ) ς (2) γς (2) − + ∑ n [cos(2nπ x)Ci(2nπ x) + sin(2nπ x) Si(2nπ x)] + 2π 2π 2π ′

∞

2

n =1

2

2

2

and with a little algebra and using the well known formula

ς ′(−1) =

(1.61)

1 1 (1 − γ − log 2π ) + 2 ς ′(2) 12 2π

(which is obtained by differentiating the functional equation for the Riemann zeta function) we obtain x

∫ log Γ(a) da =

(1.62)

0

=

−

1 1 1 1 x log(2π ) + x ⎡⎣ 2 − x + 2 ( x − 1) log x ⎤⎦ − x 2 + 2 4 2 4π 1 2π 2

sin(2nπ x) 1 + log x n2 12 n =1 ∞

∑

∞

1 1 ∑ n [cos(2nπ x)Ci(2nπ x) + sin(2nπ x) Si(2nπ x)] + 12 − ς ′(−1) 2

n =1

Letting x = 1 in (1.62) gives us Ci (2nπ ) 1⎤ 1⎤ ⎡ ⎡ = 2π 2 ⎢log A − ⎥ = −2π 2 ⎢ς ′(−1) + ⎥ 2 n 4⎦ 6⎦ ⎣ ⎣ n =1 ∞

∑

(1.63)

and with x = 1/ 2 we get 1

2

1 1 1 1 ∫0 log Γ(a) da = 4 log(2π ) + 16 + 24 log 2 − 2π 2

(−1) n 1 Ci (nπ ) + − ς ′(−1) ∑ 2 12 n =1 n

and using equation (6.117b) in [21] ∞

1 2π

2

∑ (−1) n =1

n

Ci (nπ ) 1 1 1 = log 2 + + ς ′ ( −1) 2 n 12 48 2

we see that 1

(1.64)

2

5

1

3

∫ log Γ(a) da = 24 log 2 + 4 log π + 2 log A 0

which is reported in [52, p.35]. With x = 1/ 4 we obtain

34

∞

1

4

∫ log Γ(a) da = 0

1 5 G 1 = log(2π ) + + + log 2 8 64 4π 48 −

∞

1 2π 2

1 1 ∑ n [cos(nπ / 2)Ci(nπ / 2) + sin(nπ / 2)Si(nπ / 2)] + 12 − ς ′(−1) n =1

2

From [52, p.35] we have 1

4

1

1

9

G

∫ log Γ(a) da = 8 log 2 + 8 log π + 8 log A + 4π

(1.54.1)

0

and we therefore obtain (1.65)

∞

1 2π

2

1 5 1 1 ∑ n [cos(nπ / 2)Ci(nπ / 2) + sin(nπ / 2)Si(nπ / 2)] = 64 + 48 log 2 − 8 log A 2

n =1

Using (1.46) we may write (1.62) as Gosper’s integral x

1

1

∫ log Γ(a)da = 2 x(1 − x) + 2 x log(2π ) + ς ′(−1, x) − ς ′(−1)

(1.66)

0

In fact, we note that equating (1.62) with (1.66) results in Elizalde’s formula (1.45). □ Integrating Elizalde’s identity (1.45) gives us x

∫ ς ′(−1, a)da = 0

1 1 1 1 x [−4 x 2 + 9 x − 6 + 6( x − 1)(2 x − 1) log x] − x3 + x − 2 72 12 12 8π +

1 8π

ς (3) − 2

∞

1 4π

3

1

∑n n =1

3

cos(2nπ x) n3 n =1 ∞

∑

[sin(2nπ x)Ci(2nπ x) − cos(2nπ x) Si(2nπ x)]

where we have used the integrals (easily derived using integration by parts)

35

∫ Ci( x)dx = x Ci( x) − sin x

∫ Si( x)dx = x Si( x) + cos x

and

In evaluating the integral at a = 0 , we have used the fact that Si (0) = 0 and from (1.3) we have x cos t − 1 sin x Ci ( x) = γ sin x + sin x log x + sin x ∫ dt t 0 sin x cos t − 1 x log x + sin x ∫ dt = γ sin x + x t 0 x

We therefore see that lim sin x Ci ( x) = 0 x →0

Since the Hurwitz zeta function is analytic in the whole complex plane except for s ≠ 1 , its partial derivatives commute in the region where the function is analytic: we therefore have ∂ ∂ ∂ ∂ ∂ ς ( s, t ) = ς ( s, t ) = − [ sς ( s + 1, t )] ∂t ∂s ∂s ∂t ∂s = −ς ( s + 1, t ) − s

∂ ς ( s + 1, t ) ∂s

and upon integrating with respect to t we see that v

∂ ∂ ς ( s, t )dt + ∫ ς ( s + 1, t ) dt ∂t ∂s 0 0 v

v

− s ∫ ς ′( s + 1, t ) dt = ∫ 0

We therefore get v

v

0

0

− s ∫ ς ′( s + 1, t ) dt = ς ′( s, v) − ς ′( s, 0) + ∫ ς ( s + 1, t ) dt

and with s = − n we have v

v

0

0

n ∫ ς ′(1 − n, t )du = ς ′(−n, v) − ς ′(−n, 0) + ∫ ς (1 − n, t ) dt

Then, using the well-known result [6, p.264]

36

ς (1 − n, v) = −

Bn (v) for n ≥ 1 n

we obtain v

n ∫ ς ′(1 − n, t ) dt =

(1.66)

0

Bn +1 − Bn +1 (v) + ς ′(− n, v) − ς ′(− n, 0) n(n + 1)

We have lim[ς ( s, a) − a − s ] = ς ( s ) a →0

lim[ς (− s, a) − a s ] = ς (− s ) a →0

∂ lim[ς (− s, a) − a s ] = −ς ′(− s) ∂s a →0 ∂ lim[ς (− s, a) − a s ] = lim[−ς ′(− s, a ) − a s log a] a →0 ∂s a →0 = −ς ′(− s, 0) Therefore we have

ς ′(− s, 0) = ς ′(− s)

and thus we obtain v

n ∫ ς ′(1 − n, t ) dt =

(1.67)

0

Bn +1 − Bn +1 (v) + ς ′(− n, v) − ς ′(−n) n(n + 1)

This integral was originally derived by Adamchik [2] in a different manner in 1998. We have for n = 2 x

1

1

1

∫ ς ′(−1, a)da = − 12 B ( x) + 2 ς ′(−2, x) − 2 ς ′(−2) 3

0

Therefore we obtain 1 1 1 1 x [−4 x 2 + 9 x − 6 + 6( x − 1)(2 x − 1) log x] − x3 + x − 2 72 12 12 8π +

1 8π

ς (3) − 2

∞

1 4π

3

1

∑n n =1

3

cos(2nπ x) n3 n =1 ∞

∑

[sin(2nπ x)Ci(2nπ x) − cos(2nπ x) Si(2nπ x)]

37

=−

1 1 1 B3 ( x) + ς ′(−2, x) − ς ′(−2) 12 2 2

This is easily simplified to (1.68) 1 5 1 1 x( x − 1)(2 x − 1) log x − x3 + x 2 − 2 12 36 8 8π −

∞

1 4π

3

1

∑n n =1

3

cos(2nπ x) 1 + 2 ς (3) 3 n 8π n =1 ∞

∑

[sin(2nπ x)Ci(2nπ x) − cos(2nπ x) Si(2nπ x)]

=−

1 1 1 B3 ( x) + ς ′(−2, x) − ς ′(−2) 12 2 2

Equation (1.68) could then be integrated to produce an identity involving ς ′(−3, x) and so on. □ Differentiating (1.41) term by term (and boldly assuming that the procedure is valid) easily results in (1.70)

ψ (a) = log a −

∞ 1 + 2∑ [cos(2nπ a )Ci (2nπ a) + sin(2nπ a) si (2nπ a)] 2a n =1

which appears in Nörlund’s book [46, p.108]. Letting a = 1 results in (1.71)

∞ 1 − γ = 2∑ Ci (2nπ ) 2 n =1

and this corrects the corresponding formula given by Nielsen [44, p.80]. It appears that Nielsen’s analysis is incorrect because he effectively used the Fourier series (1.41.1) which does not hold at the point a = 1 . This formula was also used in [20]. With a = 1/ 2 we get ⎛1⎞ ⎝ ⎠

∞

ψ ⎜ ⎟ = − log 2 − 1 + 2∑ (−1) n Ci (nπ ) 2 n =1

and hence we have

38

∞

2∑ (−1) n Ci (nπ ) = 1 − γ − log 2

(1.72)

n =1

With a → a / 2 we see that ∞

1 a

ψ (a / 2) = log(a / 2) − + 2∑ [cos(nπ a)Ci(nπ a) + sin(nπ a) si (nπ a)]

(1.73)

n =1

and we also have

ψ (a + 1) = log(a + 1) −

1 2(a + 1)

∞

+2∑ cos[2nπ (a + 1)]Ci[2nπ (a + 1)] + sin[2nπ (a + 1)]si[2nπ (a + 1)] n =1

Equation (1.70) may be expressed as

ψ (a) = log a −

∞ 1 + 2∑ an 2a n =1

and we have with a → 2a

ψ (2a) = log(2a) −

∞ 1 + 2∑ a2 n 4a n =1

∞

∞

∞

n =1

n =1

n =1

Since 2∑ a2 n = ∑ an + ∑ (−1) n an we have

ψ (2a) = log(2a) −

∞ 1 1 1 1 + ψ (a) − log a + + ∑ (−1) n an 4a 2 2 4a n =1

and using the duplication formula for the gamma function [52, p.15] (which may be obtained by differentiating (2.16.2)) 1

1

⎛ ⎝

1⎞

ψ (2a) = log 2 + ψ (a) + ψ ⎜ a + ⎟ 2 2 2 ⎠

this results in (1.74)

⎛ ⎝

1⎞

∞

ψ ⎜ a + ⎟ = log a + 2∑ (−1) n [cos(nπ a )Ci(nπ a) + sin(nπ a) si (nπ a)] 2 ⎠

n =1

39

Differentiating (1.41.1) results in (which is valid for 0 < a < 1) (1.75)

1 a

∞

ψ (a) = − + 2∑ [cos 2nπ a Ci(2nπ ) + sin 2nπ a si(2nπ )] n =1

or equivalently ∞

ψ (a + 1) = 2∑ [cos 2nπ a Ci (2nπ ) + sin 2nπ a si(2nπ )] n =1

which may be compared with (1.70)

ψ (a) = log a −

∞ 1 + 2∑ [cos(2nπ a )Ci (2nπ a) + sin(2nπ a) si (2nπ a)] 2a n =1

□ Differentiating (1.1) with respect to a 1

1

∫ log Γ( x + a) sin 2nπ x dx = − 2π n [log a − cos(2nπ a)Ci(2nπ a) − sin(2nπ a)si(2nπ a)] 0

we obtain cos(2nπ a) ⎡1 ⎤ + 2nπ sin(2nπ a)Ci (2nπ a ) ⎥ ⎢ a − cos(2nπ a ) a 1 ⎥ 1 ⎢ ⎥ ∫0 ψ ( x + a) sin 2nπ x dx = − 2π n ⎢ ⎢ ⎥ sin(2nπ a ) − 2nπ cos(2nπ a) si (2nπ a) ⎥ ⎢ − sin(2nπ a ) a ⎣ ⎦

and therefore we have 1

(1.76)

∫ψ ( x + a) sin 2nπ x dx = − sin(2nπ a)Ci(2nπ a) + cos(2nπ a)si(2nπ a) 0

which corrects the entry in [33, p.652, 6.467 1]. With a = 1 we get 1

(1.77)

∫ψ ( x + 1) sin 2nπ x dx = si(2nπ ) 0

and with a = 0 we get 40

π

1

∫ψ ( x) sin 2nπ x dx = si(0) = − 2

(1.77.1)

0

which is correctly reported in [33, p.652, 6.465 2]. We see that 1

1

0

0

sin 2nπ x dx x 0

1

∫ψ ( x + 1) sin 2nπ x dx = ∫ψ ( x) sin 2nπ x dx + ∫ 1

= ∫ψ ( x) sin 2nπ x dx + Si (2nπ ) − Si (0) 0

and this concurs with (1.77) and (1.77.1). Similarly, differentiating (1.27) we obtain 1

(1.78)

∫ψ ( x + a) cos 2nπ x dx = sin(2nπ a)si(2nπ a) + cos(2nπ a)Ci(2nπ a) 0

which corrects the entry in [33, p.652, 6.467 2]. With a = 1 we have 1

(1.78.1)

∫ψ ( x + 1) cos 2nπ x dx = Ci(2nπ ) 0

□ Using integration by parts we see that for a ≥ 0 1

1

∫ψ ( x + a) sin 2nπ x dx = sin 2nπ x log Γ( x + a) 0 − 2nπ ∫ Γ( x + a) cos 2nπ x dx 1

0

0

We have for a ≥ 0 sin 2nπ x x log Γ( x + a) = 0 x →0 x

lim sin 2nπ x log Γ( x + a) = lim x →0

and hence we have for a ≥ 0 (1.79)

1

1

0

0

∫ψ ( x + a) sin 2nπ x dx = −2nπ ∫ Γ( x + a) cos 2nπ x dx

41

Using (1.27) 1

1

∫ log Γ( x + a) cos 2nπ x dx = − 2nπ [sin(2nπ a)Ci(2nπ a) + cos(2nπ a)si(2nπ a)] 0

we again obtain 1

∫ψ ( x + a) sin 2nπ x dx = sin(2nπ a)Ci(2nπ a) + cos(2nπ a) si(2nπ a) 0

Similarly we have for a > 0 (1.80)

1

1

0

0

∫ψ ( x + a) cos 2nπ x dx = log Γ(1 + a) − log Γ(a) + 2nπ ∫ Γ( x + a) sin 2nπ x dx

and using (1.1) 1

1

∫ log Γ( x + a) sin 2nπ x dx = − 2nπ [log a − cos(2nπ a) Ci (2nπ a) − sin(2nπ a) si (2nπ a)] 0

we again obtain 1

∫ψ ( x + a) cos 2nπ x dx = cos(2nπ a) Ci (2nπ a) + sin(2nπ a) si (2nπ a) 0

Nielsen [44, p.80] reports these integrals in the case a = 1 as (1.81)

(1.82)

1

1

0

0

1

1

0

0

∫ψ ( x + 1) sin 2nπ x dx = −2nπ ∫ log Γ( x + 1) cos 2nπ x dx

∫ψ ( x + 1) cos 2nπ x dx = 2nπ ∫ log Γ( x + 1) sin 2nπ x dx □

Applying Parseval’s identity with (1.78) and (1.78.1) we obtain 1

∞

0

n =1

2 ∫ψ 2 ( x + 1) dx = ∑ [Ci 2 (2nπ ) + si 2 (2nπ )]

With regard to this, see also Lewin’s monograph [43, p.24] where he showed that

42

∞

∫e

− ux

0

2 Li2 (− x 2 ) dx = − ⎡⎣Ci 2 (u ) + si 2 (u ) ⎤⎦ u

and with u = 2nπ we have ∞

nπ ∫ e −2 nπ x Li2 (− x 2 ) dx = − ⎡⎣Ci 2 (2nπ ) + si 2 (2nπ ) ⎤⎦ 0

We make the summation ∞ ∞

∞

0 n =1

n =1

π ∫ ∑ n e−2 nπ x Li2 (− x 2 ) dx = −∑ [Ci 2 (2nπ ) + si 2 (2nπ )] Using the derivative of the geometric series we have ∞

∑n z n =1

n

=

z (1 − z ) 2

so that ∞ ∞

∞

e −2 π x Li2 (− x 2 ) dx −2 π x 2 (1 − e ) 0

π ∫ ∑ n e−2 nπ x Li2 (− x 2 ) dx = π ∫ 0 n =1

and hence we have the curious equality ∞

e −2π x Li2 (− x 2 ) dx = −2 ∫ψ 2 ( x + 1) dx −2π x 2 (1 − e ) 0 0

π∫

1

A different representation of Ci 2 (u ) + si 2 (u ) as an integral is given in [44, p.32]. Applications of Nielsen’s representation of log Γ( x)

We recall Nielsen’s representation of log Γ( x) in equation (1.41.1) 1 1 ∞ 1 log Γ( x) = log(2π ) − 1 − log x + ∑ [sin 2nπ x Ci(2nπ ) − cos 2nπ x si(2nπ )] π n =1 n 2 and multiplying this by cos pπ x and integrating gives us

43

1

⎤ sin pπ Si ( pπ ) + pπ pπ

⎡1

∫ log Γ( x) cos pπ x dx = ⎢⎣ 2 log(2π ) − 1⎥⎦

(1.90)

0

+

2(1 − cos pπ )

π

2

Ci (2nπ ) p sin pπ + ∑ 2 2 π2 n =1 4n − p ∞

1 si (2nπ ) 2 − p2 n =1 ∞

∑ n 4n

where we have employed the integral 1

∫ log x cos pπ x dx = pπ [log x sin pπ x − Si( pπ x)] 1

∫ log x cos pπ x dx = −

(1.91)

0

Si ( pπ ) pπ

Letting p = 0 in (1.90) results in 1

⎡1

Si ( pπ ) p →0 pπ

⎤

∫ log Γ( x) dx = ⎢⎣ 2 log(2π ) − 1⎥⎦ + lim 0

Si ( pπ ) = 1 giving us the familiar result p →0 pπ

and using L’Hôpital’s rule we see that lim 1

1

∫ log Γ( x) dx = 2 log(2π ) 0

With p = 1 we have 1

(1.92)

∫ log Γ( x) cos π x dx =

Si(π )

0

π

+

Ci (2nπ ) π n =1 4n 2 − 1 4

2

∞

∑

and p = 2k results in 1

∫ log Γ( x) cos 2kπ x dx = 0

Si (2kπ ) si (2kπ ) 1 − = 4k 2 kπ 2 kπ

in agreement with (3.10). The following limits have been used in the above calculation lim

p→2 k

2(1 − cos pπ )

π

2

Ci (2nπ ) 2π sin pπ Ci(2kπ ) = lim =0 2 2 p k → 2 −p π 2 (−2 p) n =1 ∞

∑ 4n

44

lim

p sin pπ

π

p→2 k

2

1 si (2nπ ) ( pπ cos pπ + sin pπ ) si(2kπ ) si (2kπ ) = lim =− 2 2 2 p k → 2 2 kπ −p kπ (−2 p ) n =1 ∞

∑ n 4n

With p = 2k + 1 we have 1

(1.93)

∫ log Γ( x) cos(2k + 1)π x dx = 0

Si[(2k + 1)π ] 4 + 2 (2k + 1)π π

∞

∑ 4n n =1

Ci (2nπ ) − (2k + 1)2

2

We see from the definition (1.3) that

Ci (2nπ x) = γ + log(2nπ x) +

2 nπ x

∫ 0

cos t − 1 dt t

and we have the summation 2 nπ x ∞ ∞ ∞ Ci (2nπ x) 1 log n 1 ⎡ cos t − 1 ⎤ x dt ⎥ [ γ log(2 π )] = + + + ⎢ ∫ ∑ ∑ ∑ ∑ 2 2 2 2 t n =1 4n − 1 n =1 4n − 1 n =1 4n − 1 n =1 4n − 1 ⎣ 0 ⎦ ∞

2 nπ x ∞ ∞ 1 log n 1 ⎡ cos t − 1 ⎤ dt ⎥ = [γ + log(2π x)] + ∑ 2 +∑ 2 ⎢ ∫ t 2 n =1 4n − 1 n =1 4 n − 1 ⎣ 0 ⎦ 2 nπ x

Using

∫ 0

cos t − 1 cos(2nπ y ) − 1 dt = ∫ dy the last series becomes t y 0 x

x ⎡ 2 nπ x cos t − 1 ⎤ ∞ 1 ⎡ cos(2nπ y ) − 1 ⎤ dt dy ⎥ = ⎢ ∫ ⎥ ∑ 2 ⎢∫ ∑ 2 4 1 t n y − n =1 4n − 1 ⎣ 0 n 1 = ⎦ ⎦ ⎣0 ∞

1

⎡ x cos(2nπ y ) − 1 ⎤ dy ⎥ = ∑ 2 lim ⎢ ∫ a →0 y n =1 4n − 1 ⎣0 ⎦ ∞

1

cos(2nπ y ) − 1 dy (4n 2 − 1) y n =1

x ∞

= lim ∫ ∑ a →0

a

x ∞ ⎡ 1 cos(2nπ y ) ⎤ = lim ∫ ⎢ − +∑ dy a →0 2 y n =1 (4n 2 − 1) y ⎥⎦ a ⎣

It is an exercise in Apostol’s book [7, p.337] to show that for 0 < x < π

45

sin x =

(1.94)

∞

2

π

∞

cos(2nx) 2 n =1 4n − 1

∑ π

1 , this holds true for 0 ≤ x < π . It seems likely that (1.94) −1 2 n =1 may also be determined using the differentiation theorems for Fourier series appearing in Tolstov’s book [55, p.142]. and in fact, since

∑ 4n

We have

1

−

4

2

=

∞ 1 cos(2nπ y ) (2 − π sin π y ) = ∑ 4 4n 2 − 1 n =1

so that x ∞ ⎡ 1 cos(2nπ y ) ⎤ 1 π sin π y π +∑ lim ∫ ⎢ − dy = − lim ∫ dy = − Si (π x) ⎥ 2 → 0 a →0 a 2 y n =1 (4n − 1) y ⎦ 4a y 4 a ⎣ x

Hence we see that ∞ Ci (2nπ x) 1 log n π = γ + π x + − Si (π x) [ log(2 )] ∑ ∑ 2 2 2 4 n =1 4n − 1 n =1 4n − 1 ∞

(1.95)

and referring to (1.93) 1

∫ log Γ( x) cos π x dx = 0

Si (π )

π

+

Ci (2nπ ) π n =1 4n 2 − 1 4

2

∞

∑

we obtain 1

∫ logΓ( x) cos π x dx = 0

∞ 2 ⎡ log n ⎤ π γ log(2 ) 2 + + ∑ ⎥ 2 ⎢ 2 π ⎣ n =1 4n − 1 ⎦

which we shall derive in a different manner in (4.5.2). We may write (1.95) as ∞ ∞ Ci (2nπ x) − log(2nπ x) ∞ log(2nπ x) 1 π 1 log n + − log(2 π x ) + Si ( π x ) = γ + ∑ ∑ ∑ ∑ 2 2 2 2 4n − 1 4n − 1 4 2 n =1 4n − 1 n =1 n =1 n =1 4n − 1 ∞

and take the limit as x → 0 . Having regard to (1.3) we see that the limit equates to the right-hand side. Comparing (1.93) with (4.6) results in 46

Si[(2k + 1)π ] 4 = + 2 2 ∑ 2 2 π n =1 4n − (2k + 1) (2k + 1)π π 4

(1.95.1)

∞

γ + log(2π n)

∞

∑ 4n n =1

Ci (2nπ ) − (2k + 1) 2

2

Integrating (1.95) results in Ci (2nπ x) 1 − 2 2π n =1 4n − 1 ∞

x∑

∞ 1 sin(2nπ x) x x log n x = + − + [ γ log(2 π x )] ∑ ∑ 2 2 4n − 1 2 2 n =1 n n =1 4 n − 1 ∞

−

π

1 x Si (π x) − [cos(π x) − 1] 4 4

and hence we obtain 1 sin(2nπ x) π = [2 x + cos(π x) − 1] 4n 2 − 1 2 n =1 ∞

∑n

(1.96)

which concurs with the result directly obtained by integrating (1.94). In passing, we note that it is an exercise in Bromwich’s book [15, p.371] to show that for 0< x < π

π sin a( π2 − x) cos nx = ∑ 2 2 4a cos a2π n = 0 (2n + 1) − a ∞

(1.96.1)

□ Multiplying (1.41.1) by sin pπ x and integrating gives us 1

(1.97)

⎤ 1 − cos pπ Ci ( pπ ) γ + log( pπ ) − + pπ pπ pπ

⎡1

∫ log Γ( x) sin pπ x dx = ⎢⎣ 2 log(2π ) − 1⎥⎦ 0

−

2sin pπ

π2

Ci (2nπ ) p(1 − cos pπ ) ∞ 1 si (2nπ ) + ∑ 2 2 2 π2 − p2 n =1 n =1 n 4n − p ∞

∑ 4n

where we have employed the integral (1.10.3) 1

∫ log x sin pπ x dx = 0

Ci ( pπ ) − γ − log( pπ ) pπ

With p = 1 we have

47

⎡1 ⎤ 2 Ci (π ) γ + log π 2 ∫0 log Γ( x) sin π x dx = ⎢⎣ 2 log(2π ) − 1⎥⎦ π − π + π + π 2 1

(1.98)

1 si (2nπ ) 4n 2 − 1 n =1 ∞

∑n

and comparing this with (5.7) we deduce that 1 si (2nπ ) π = [3 + Ci (π ) − γ − log(4π ) ] 4n 2 − 1 2 n =1 ∞

∑n

(1.99)

Letting p = 2k results in 1

∫ log Γ( x) sin 2kπ x dx 0

=−

Ci (2kπ ) γ + log(2kπ ) 2π cos pπ Ci (2kπ ) (1 + pπ sin pπ − cos pπ ) si (2kπ ) + − lim + lim 2 2 2 p → k p → k π (−2 p ) π 2 k (−2 p) 2 kπ 2 kπ

and hence we obtain another derivation of 1

∫ logΓ( x) sin 2kπ x dx = 0

γ + log(2π k ) 2π k

With p = 2k + 1 we have (1.100) 1

⎡1

⎤

2

∫ log Γ( x) sin(2k + 1)π x dx = ⎢⎣ 2 log(2π ) − 1⎥⎦ (2k + 1)π − 0

+

2(2k + 1)

π

2

∞

1

∑ n 4n n =1

Ci[(2k + 1)π ] γ + log[(2k + 1)π ] + (2k + 1)π (2k + 1)π

si (2nπ ) − (2k + 1) 2

2

which may be compared with (5.6) 1

∫ logΓ( x) sin(2k + 1)π x dx = 0

k −1 ⎡ ⎛π ⎞ 1 1 1 ⎤ + 2∑ ⎢ log ⎜ ⎟ + ⎥ (2k + 1)π ⎣ ⎝ 2 ⎠ 2k + 1 j =0 2 j + 1 ⎦

Further identities may be obtained by multiplying (1.41.1) by x n sin pπ x or x n cos pπ x and integrating as before. Equation (1.99) may also be derived as follows. Referring to (1.91) we have

48

1

∫ log x cos 2nπ x dx = − 0

Si (2nπ ) 2nπ

and we make the summation ∞ 1

∑ ∫ log x n =1 0

cos 2nπ x 1 dx = − 2 4n − 1 2π

=−

1 Si (2nπ ) 4n 2 − 1 n =1 ∞

∑n

1 2π

1 si (2nπ ) 1 ∞ 1 1 − ∑ ∑ 2 4 n =1 n 4n 2 − 1 n =1 n 4n − 1

1 2π

1 si (2nπ ) 1 1 − log 2 + 2 4 4n − 1 2 n =1

∞

and using (2.15) this becomes =−

∞

∑n

Assuming that the interchange of summation and integration is valid ∞ cos 2nπ x cos 2nπ x dx = ∫ log x ∑ dx log x ∑ 2 2 ∫ 4n − 1 n =1 0 n =1 4n − 1 0 ∞ 1

1

Employing (1.94) this becomes 1

=

1 log x(2 − π sin π x) dx 4 ∫0

and using (1.10.3) we have 1 = − [2 + Ci (π ) − γ − log π ] 4 Therefore we obtain (1.101)

1 si (2nπ ) π = [3 + Ci (π ) − γ − log(4π )] 4n 2 − 1 2 n =1 ∞

∑n

Alternatively we consider 1 si (2nπ ) ∞ 1 Si (2nπ ) π ∞ 1 1 =∑ − ∑ ∑ 2 2 2 n =1 n 4n 2 − 1 n =1 n 4n − 1 n =1 n 4n − 1 ∞

49

1 Si (2nπ ) π − (2 log 2 − 1) 2 2 n =1 n 4n − 1 ∞

=∑ where we have used (2.15). 1 Si (2nπ ) = ∑ 2 n =1 n 4n − 1 ∞

2π ∞

1 sin nt dt 2 −1 t n =1

∫ ∑ n 4n 0

1 sin(2nπ t ) dt 4n 2 − 1 t n =1 n

1 ∞

= ∫∑ 0

and using (1.96.1) this becomes =

π 1 cos(π t ) − 1 + 2t 2 ∫0

t

dt

We note from (1.3) that Ci (π x) = γ + log π + log x +

πx

∫ 0

cos t − 1 dt t

cos(π t ) − 1 dt t 0 x

= γ + log π + log x + ∫

Therefore we obtain another derivation of 1 Si (2nπ ) π = [3 + Ci (π ) − γ − log(4π )] 4n 2 − 1 2 n =1 ∞

∑n

□ Equating (1.90) and (4.4.1) gives us ⎡1 ⎤ sin pπ Si ( pπ ) 2(1 − cos pπ ) ∞ Ci (2nπ ) p sin pπ + ∑ 2 2 ⎢⎣ 2 log(2π ) − 1⎥⎦ pπ + pπ + π2 π2 n =1 4n − p

=

sin pπ [γ + log(2π )] sin pπ + 2 pπ 4 pπ

⎡ ⎛ p⎞ ⎛ p ⎞⎤ ⎢ψ ⎜ 2 ⎟ +ψ ⎜ − 2 ⎟ ⎥ ⎝ ⎠⎦ ⎣ ⎝ ⎠

50

1 si (2nπ ) 2 − p2 n =1 ∞

∑ n 4n

+

2(1 − cos pπ )[γ + log(2π )] ⎡ 1 π ⎛ pπ ⎞ ⎤ 2(1 − cos pπ ) ∞ log n cot − ∑ ⎜ ⎟⎥ + ⎢ 2 p2 4 p 2 2 π2 π2 ⎝ 2 ⎠⎦ n =1 4n − p ⎣

Equating (1.97) and (5.5.1) results in ⎡1 ⎤ 1 − cos pπ Ci ( pπ ) γ + log( pπ ) 2sin pπ − + − ⎢⎣ 2 log(2π ) − 1⎥⎦ pπ pπ pπ π2 +

p (1 − cos pπ )

π =

2

Ci (2nπ ) 2 − p2 n =1 ∞

∑ 4n

1 si (2nπ ) 2 − p2 n =1 ∞

∑ n 4n

(1 − cos pπ )[γ + log(2π )] (1 − cos pπ ) ⎡ ⎛ p⎞ p ⎞⎤ ⎛ ψ ⎜1 + ⎟ +ψ ⎜1 − ⎟ ⎥ + ⎢ 2 pπ 4 pπ 2⎠ ⎝ 2 ⎠⎦ ⎣ ⎝ −

2sin pπ [γ + log(2π )] ⎡ 1 π ⎛ pπ ⎞ ⎤ 2sin pπ cot ⎜ − ⎟⎥ − ⎢ 2 2 π π2 ⎝ 2 ⎠⎦ ⎣2p 4p

∞

log n 2 − p2

∑ 4n n =1

Ci (2nπ ) and 2 − p2 n =1 ∞

and hence we have two simultaneous equations involving

∑ 4n

1 si (2nπ ) . 2 − p2 n =1 ∞

∑ n 4n

□ We showed in [21] that b

∫ p( x) cot x dx

(1.102)

a

∞ b

= 2∑ ∫ p ( x) sin 2nx dx n =1 a

which is valid for a wide class of suitably behaved functions. Specifically we require that p ( x) is a twice continuously differentiable function and that either (i) both sin( x / 2) and cos(x / 2) have no zero in [ a, b ] or (ii) if either sin( a / 2) or cos(a / 2) is equal to zero then p (a) must also be zero. Condition (i) is equivalent to the requirement that sin x has no zero in [ a, b ]. Since log Γ(1) = log Γ(2) = 0 , it is clear that p ( x) = log Γ( x + 1) satisfies the necessary conditions on the interval [0,1] . Therefore we have 1

∞ 1

0

n =1 0

∫ log Γ( x + 1) cot π x dx = 2∑ ∫ log Γ( x + 1) sin 2nπ x dx

51

and using (1.11) this becomes Ci (2nπ ) nπ n =1

1

∞

∫ log Γ( x + 1) cot π x dx = ∑

(1.103)

0

This is an interesting result if only for the fact that Mathematica is not able to evaluate 1

even the seemingly simpler integral ∫ log x cot π x dx . 0

We note that integration by parts gives us 1

∫ log Γ( x + 1) cot π x dx = 0

1

π

1

log Γ( x + 1) log sin π x − 0

1

π

1

∫ψ (1 + x) log sin π x dx 0

and using L’Hôpital’s rule we obtain ⎡ log sin π x ⎤ lim[log Γ( x + 1) log sin π x] = lim ⎢ x →1 x →1 1/ log Γ ( x + 1) ⎥ ⎣ ⎦ ⎡ π cot π x log 2 Γ( x + 1) ⎤ = lim ⎢ − ⎥ x →1 ψ ( x + 1) ⎣ ⎦ ⎡ π log 2 Γ( x + 1) ⎤ = lim ⎢ − ⎥ x →1 ⎣ ψ (2) sin π x ⎦

We also have ⎡ log Γ( x + 1) ⎤ ⎡ ψ ( x + 1) ⎤ lim ⎢ log Γ( x + 1) ⎥ = lim ⎢ lim log Γ( x + 1) = 0 x →1 ⎣ sin π x ⎦ x→1 ⎣ π cos π x ⎦⎥ x→1 Similarly we have ⎡ log Γ( x + 1) ⎤ ⎡ π x log Γ( x + 1) ⎤ lim ⎢ log Γ( x + 1) ⎥ = lim ⎢ log Γ( x + 1) ⎥ x →0 πx ⎣ sin π x ⎦ x →0 ⎣ sin π x ⎦ ⎡ log Γ( x + 1) ⎤ ⎡ψ ( x + 1) ⎤ Γ( x + 1) log sin π x] = 0 = lim ⎢ and since lim ⎢ ⎥ ⎥⎦ we obtain lim[log x →0 x →0 πx ⎣ ⎦ x →0 ⎣ π Hence we have

52

1

∫ log Γ( x + 1) cot π x dx = − 0

1

π

1

∫ψ (1 + x) log sin π x dx 0

□ It was also shown in [21] that under the same conditions we have b

∞ b

a

n =0 a

∫ p( x) dx = 2∑ ∫ p( x) cos α nx dx

(1.104)

and we employ this with p ( x) = log Γ( x + 1) and (1.27) to give us 1

1

∞ 1 1 − ∫ log Γ( x + 1) dx = ∑ ∫ log Γ( x + 1) cos 2nπ x dx = − 20 2π n =1 0

si (2nπ ) n n =1 ∞

∑

We therefore get si (2nπ ) π = log(2π ) − π n 2 n =1 ∞

∑

(1.105)

which we have previously seen in (1.54). A number of other closed form Dirichlet series involving the sine and cosine integrals are given in my earlier paper [21]. □ Motivated by a problem posed by Furdui [31] in 2009, we find a more direct derivation of (1.63) which is set out below. We see from (1.3) that nx

Ci (nx) = γ + log nx + ∫ 0

cos t − 1 dt t

and we make the summation nx ∞ ∞ Ci (nx) 1 log n ∞ 1 ⎡ cos t − 1 ⎤ γ x dt ⎥ ( log ) = + + + ∑ ∑ ∑ ∑ 2 2 2 ⎢∫ n2 t n =1 n =1 n n =1 n n =1 n ⎣ 0 ⎦ ∞

nx 1 ⎡ cos t − 1 ⎤ dt ⎥ 2 ⎢∫ t n =1 n ⎣ 0 ⎦ ∞

= ς (2)(γ + log x) − ς ′(2) + ∑

53

We see that cos t − 1 cos ny − 1 ∫0 t dt = ∫0 y dy

nx

x

and thus nx x 1 ⎡ cos t − 1 ⎤ ∞ 1 ⎡ cos ny − 1 ⎤ dt dy ⎥ = ⎥ ∑ 2 ⎢∫ ∑ 2 ⎢∫ t y n =1 n ⎣ 0 ⎦ ⎦ n =1 n ⎣ 0 ∞

⎡ x cos ny − 1 ⎤ 1 lim dy ⎥ 2 a →0 ⎢ ∫ y n =1 n ⎣a ⎦ ∞

=∑

cos ny − 1 dy n2 y n =1

x ∞

= lim ∫ ∑ a →0

a

We have the well known Fourier series [55, p.148] cos ny π 2 π y y 2 = − + ∑ n2 6 2 4 n =1 ∞

so that cos ny π 2 π y = − + ∑ 2 6y 2 4 n =1 n y ∞

and

cos ny − 1 π y =− + 2 n y 2 4 n =1 ∞

∑

We then easily see that (1.106)

Ci (nx) π x x2 ′ ς γ ς = + − − + (2)( log x ) (2) ∑ n2 2 8 n =1 ∞

and with x = 2π we obtain (1.63) again Ci (2nπ x) 1 = ς (2)[γ + log(2π )] − ς ′(2) − π 2 2 n 2 n =1 ∞

(1.107)

∑

More generally we have

54

∞ Ci (nx) cos ny − 1 ′ dy ς γ ς = ( s )( + log x ) − ( s ) + lim ∑ ∑ s ∫ a →0 n ns y n =1 a n =1 ∞

(1.108)

x

Similarly we may consider the sine integral function x

sin t dt t 0

Si ( x) = ∫

nx

Si (nx) =

∫ 0

sin t dt t

nx Si (nx) ∞ 1 ⎡ sin t ⎤ dt ⎥ = ∑ s ⎢∫ ∑ ns t n =1 n =1 n ⎣ 0 ⎦ ∞

∞

∞ Si (nx) sin nt = dt ∑ ∑ s ∫ n nst n =1 0 n =1 x

and we can certainly evaluate this for s = 3 using the Fourier series [55, p.148] ∞

sin nt 1 3 = (t − 3π t 2 + 2π 2t ) 3 n 12 n =1

∑

In this regard, see also the problems proposed by Choulakian [18] in 1998. □ Using (1.10.3) we make the summation ∞ sin(2k + 1)π x Ci [(2k + 1)π ] − γ − log[(2k + 1)π ] dx = ∑ 2k + 1 (2k + 1) 2 π k =0 k =0

1

∞

∫ log x∑ 0

and substituting the Fourier series [51, p.149] sin(2k + 1)π x π = 2k + 1 4 k =0 ∞

∑

we obtain −

π2 4

∞ ∞ Ci [(2k + 1)π ] 1 log(2k + 1) − ( γ + log π ) − ∑ ∑ 2 2 2 (2k + 1) k =0 k = 0 (2 k + 1) k = 0 (2k + 1) ∞

=∑

55

or equivalently

π2

∞ Ci [(2k + 1)π ] π 2 log(2k + 1) − =∑ − ( γ + log π ) − ∑ 2 2 4 k =0 (2k + 1) 8 k = 0 (2 k + 1) ∞

which may be compared with (B.10) 1 2 (γ + log 2π ) − log 2 = 2 π 4

∞

∑

[Ci[2(2n + 1)π ] − log(2n + 1)] (2n + 1) 2

n =0

□ As a matter of interest, Abramowitz and Stegun [1, p.232] define auxiliary functions ∞

f ( x) = − cos x si ( x) + sin x Ci ( x) = ∫ 0

sin y dy y+x

∞

cos y dy y+x 0

g ( x) = − cos x Ci ( x) − sin x si ( x) = ∫ and report that for Re ( x) > 0 ∞

e − xu du 2 1 + u 0

f ( x) = ∫ ∞

ue − xu du 1+ u2 0

g ( x) = ∫

The above results may be derived by considering the double integral ∞∞

I = ∫ ∫ e − ( a + y ) x sin y dxdy 0 0

where integrating with respect to x gives us ∞

∫e

−(a+ y ) x

0

dx =

1 a+ y

and thus we have

56

∞

I =∫ 0

sin y dy a+ y

Similarly, integrating with respect to y gives us ∞

−(a+ y ) x sin y dy = ∫e

0

∞

e − ax 1 − yx iy −iy e − ax − = e ( e e ) dy 2i ∫0 1 + x2

Therefore we see that ∞

∞

e − ax sin y ∫0 1 + x 2 dx = ∫0 a + y dy

and the validity of the operation ∞

∞

∞

∞

0

0

0

0

−(a+ y ) x sin y dy = ∫ dy ∫ e − ( a + y ) x sin y dx ∫ dx ∫ e

is confirmed by [8, p.282]. The formula ∞

∞

xe − ax cos y ∫0 1 + x 2 dx = ∫0 a + y dy

may be derived in a similar fashion. Letting t = xy we see that ∞

∞

sin( xy ) sin t ∫0 x + a dx = ∫0 t + ay dt

In accordance with the above, we have from [33, p.338] ∞

(1.109)

ve − nv ∫0 a 2 + v 2 dv = −[cos(na)Ci(na) + sin(na) si(na)]

and we make the summation ∞ ∞

∞ ve − nv = − [cos(na )Ci (na ) + sin(na ) si (na )] dv ∑ ∑ ∫ 2 2 n =1 0 a + v n =1

The geometric series gives us 57

∞ ∞

∞

ve − nv ve− v = dv ∑ ∫ 2 2 ∫0 (a 2 + v2 )(1 − e−v ) dv n =1 0 a + v and hence we have ∞

∞

v dv v ( + )( − 1) a v e 0

−∑ [Ci (na ) cos(na ) + si (na ) sin(na )] = ∫ n =1

2

2

We now let a → 2π a and v → 2π v to obtain ∞

∞

t dt (a + t )(e 2π t − 1) 0

(1.110) −∑ [cos(2π na )Ci (2π na ) + sin(2π na) si (2π na )] = ∫ n =1

2

2

Using (1.70)

ψ (a) = log a −

∞ 1 + 2∑ [cos(2nπ a)Ci (2nπ a) + sin(2nπ a) si (2nπ a)] 2a n =1

we then see that (1.111) ψ (a ) = log a −

∞

1 t − 2 ∫ 2 2 2π t dt + − 2a ( a t )( e 1) 0

Equation (1.111) is well known and, inter alia, is reported in [56, p.251]. Integrating (1.110) gives us x

∞

∞ av = − dv ∑ ∫ a[cos(2nπ a)Ci(2nπ a) + sin(2nπ a) si(2nπ a)]da (a 2 + v 2 )(e 2π v − 1) n =1 0 0

∫ da ∫ 0

x

We have in a scintilla temporis using the Wolfram Online Integrator

∫ a[cos(2nπ a)Ci(2nπ a) + sin(2nπ a) Si(2nπ a)]da =

[2nπ sin(2nπ a) + cos(2nπ a)]Ci(2nπ a) − [2nπ a cos(2nπ a) − sin(2nπ a)]Si(2nπ a)] − log a (2nπ ) 2

where we note that Mathematica defines SinIntegral [ x] as Si ( x) . This integral may of course be easily obtained using integration by parts.

58

We have x

∫ a[cos(2nπ a)Ci(2nπ a) + sin(2nπ a) Si(2nπ a)]da 0

=

cos(2nπ x)Ci(2nπ x) + sin(2nπ x) Si(2nπ x) − log x − γ + log(2nπ ) (2nπ ) 2

+

x sin(2nπ x)Ci(2nπ x) − x cos(2nπ x) Si (2nπ x) 2nπ

where we have used (1.9.1). We have x

∫ a[cos(2nπ a)Ci(2nπ a) + sin(2nπ a) Si(2nπ a)]da 0

x

= ∫ a[cos(2nπ a )Ci (2nπ a ) + sin(2nπ a ) si (2nπ a )]da + 0

π

x

2 ∫0

a sin(2nπ a ) da

and x

∫ a sin(2nπ a)da = 0

sin(2nπ x) − 2nπ x cos(2nπ x) (2nπ ) 2

x

∫ a[cos(2nπ a)Ci(2nπ a) + sin(2nπ a) si(2nπ a)]da 0

=

=

cos(2nπ x)Ci(2nπ x) + sin(2nπ x) Si(2nπ x) − log x − γ + log(2nπ ) (2nπ ) 2

+

x sin(2nπ x)Ci(2nπ x) − x cos(2nπ x) Si (2nπ x) 2nπ

−

π sin(2nπ x) π x cos(2nπ x) + 2 (2nπ ) 2 2 2nπ

cos(2nπ x)Ci (2nπ x) + sin(2nπ x) si (2nπ x) − log x − γ + log(2nπ ) (2nπ ) 2

+

x sin(2nπ x)Ci(2nπ x) − x cos(2nπ x) si (2nπ x) 2nπ 59

Therefore we have ∞ x

∑ ∫ a[cos(2nπ a)Ci(2nπ a) + sin(2nπ a) si(2nπ a)]da n =1 0

cos(2nπ x)Ci (2nπ x) + sin(2nπ x) si (2nπ x) 1 ς ′(2) − (log x − log(2π ) + γ ) − 2 (2nπ ) 24 4π 2 n =1 ∞

=∑

sin(2nπ x)Ci (2nπ x) − cos(2nπ x) si (2nπ x) 2nπ n =1 ∞

+ x∑

Using (1.41) this becomes cos(2nπ x)Ci(2nπ x) + sin(2nπ x) si(2nπ x) 1 ς ′(2) − (log x + γ ) − 2 (2nπ ) 24 4π 2 n =1 ∞

=∑

1 1⎞ x x⎛ 1 + x log Γ( x) − log(2π ) − ⎜ x − ⎟ log x + x 2 2 4 2⎝ 2⎠ 2 Integrating (1.110) gives us ∞

x

∞

x

av 1 v dv 2a dv = ∫ 2π v ∫ 2 2 da 2 2 2π v (a + v )(e − 1) 2 0 e −1 0 a + v 0

I = ∫ da ∫ 0

∞

∞

1 v log( x 2 + v 2 ) v log v = ∫ dv − ∫ 2π v dv 2π v 20 e −1 e −1 0 We have [24] (1.112)

∞ v log ( v 2 + x 2 ) 1 2⎡ 3⎤ 1 log G (1 + x) = x ⎢log x − ⎥ + x log(2π ) + ς ′(−1) − ∫ dv 2 ⎣ 2⎦ 2 e 2π v − 1 0

where G ( x) is the Barnes double gamma function. This result was originally obtained by Adamchik [3] in 2004. With x = 0 we have ∞

(1.113)

v log v 1 dv = ς ′(−1) 2 v −1 2

∫e π 0

Therefore we obtain

60

I=

1 2⎡ 3⎤ 1 1 x ⎢ log x − ⎥ + x log(2π ) − log G (1 + x) 4 ⎣ 2⎦ 4 2

Hence we obtain cos(2nπ x)Ci (2nπ x) + sin(2nπ x) si (2nπ x) (2nπ ) 2 n =1 ∞

∑

(1.114)

=

1 1 ⎡ 3⎤ 1 1 ς ′(2) (log x − log(2π ) + γ ) + x 2 ⎢log x − ⎥ + x log(2π ) − log G (1 + x) − 24 4 ⎣ 2⎦ 2 2 4π 2 1 x⎛ 1⎞ 1 − x log Γ( x) + ⎜ x − ⎟ log x − x 2 2 2⎝ 2⎠ 2

Alternatively, we multiply (1.111) by a and integrate to obtain x

1

∫ aψ (a)da = 2 x 0

2

1 1 1 ⎡ 3⎤ 1 log x − x 2 − x − x 2 ⎢ log x − ⎥ − x log(2π ) + log G (1 + x) 4 2 2 ⎣ 2⎦ 2

Integration by parts results in x

x

0

0

∫ aψ (a)da = x log Γ( x) − ∫ log Γ(a)da and we obtain Alexeiewsky’s theorem [52, p.32] x

1 1 1 (1.115) ∫ log Γ(a )da = x log Γ( x) − log G (1 + x) − x 2 + x + x log(2π ) 2 2 2 0 □ We also have from [33, p.338] ∞

e− μv 1 ∫0 n2 + v 2 dv = n [sin(nμ )Ci(nμ ) − cos(nμ )si(nμ )]

and we make the summation ∞ ∞

∞ e− μv 1 = dv [sin(nμ )Ci (nμ ) − cos(nμ ) si(nμ )] ∑ ∑ 2 2 ∫ n =1 0 n + v n =1 n

61

Assuming that interchanging the order of summation and integration is valid ∞ ∞

∞

∞ e− μv 1 − μv = dv e dv ∑ ∑ 2 2 2 2 ∫ ∫ n =1 0 n + v n =1 n + v 0

and using (obtained by letting α = iv in (4.7)) (1.116)

1 v

∞

1 2 n =1 n + v

π coth π v = + 2v ∑

2

we have ∞

(1.117)

∞ 1 π v coth π v − 1 − μ v 1 e dv = [sin(nμ )Ci (nμ ) − cos(nμ ) si (nμ )] ∑ 2 ∫ 20 v n =1 n

Mathematica cannot evaluate this integral. With μ = π we have ∞

∞ 1 π v coth π v − 1 −π v (−1) n = e dv si (nπ ) ∑ 2 ∫0 v2 n n =1

and referring to (1.49) we obtain ∞

(1.118)

1 π v coth π v − 1 −π v π π e dv = log 2 − 2 ∫ 2 20 v 2

With μ = 2π we have ∞

∞ 1 π v coth π v − 1 −2π v si (2nπ ) = e dv ∑ 2 ∫ 20 v n n =1

and referring to (1.54) we obtain ∞

(1.119)

1 π v coth π v − 1 −2π v π e dv = log(2π ) − π 2 ∫ 20 v 2 □

We note [33, p.336, 3.341] that

62

π

2

∫ exp(− p tan x)dx = sin p Ci( p) − cos p si( p) 0

so that π

2

∫ exp(−2nπ a tan x)dx = sin 2nπ a Ci(2nπ a) − cos 2nπ a si(2nπ a) 0

Referring to (1.41) 1 1⎞ 1 ∞ 1 ⎛ log Γ(a ) = log(2π ) + ⎜ a − ⎟ log a − a + ∑ [sin(2nπ a)Ci (2nπ a) − cos(2nπ a) si (2nπ a)] π n =1 n 2 2⎠ ⎝ and using ∞

1

∑ n exp(−2nπ a tan x) = − log[1 − exp(−2π a tan x)] n =1

we obtain the integral representation 1 1⎞ 1 ⎛ log Γ(a ) = log(2π ) + ⎜ a − ⎟ log a − a − π 2 2⎠ ⎝

(1.119.1)

π

2

∫ log[1 − exp(−2π a tan x)] dx 0

and we also note that π

2

∫ 0

2π a sec 2 x dx = exp(−2π a tan x) − 1

π

2

∫ log[1 − exp(−2π a tan x)]dx 0

□ As noted by Nielsen [44, p.72] and Bartle [8, p.345] we have the Fourier series (provided x is not an integer) (1.120)

cos tx =

=

∞ ⎤ 2 x sin π x ⎡ 1 (−1) n +1 + ⎢ 2 ∑ 2 − 2 cos nt ⎥ π x n =1 n ⎣ 2x ⎦ ∞ ⎤ sin π x ⎡ 1 (−1) n +1 x + 2 cos nt ⎥ ∑ ⎢ 2 2 π ⎣x n =1 n − x ⎦

and with t = 2π we obtain

63

∞ π 1 (−1) n +1 = + 2 x∑ 2 2 sin π x x n =1 n − x

We then have cos tx =

∞ ∞ ⎤ sin π x ⎡ π (−1) n +1 (−1) n +1 − + 2 x 2 x cos nt ⎥ ∑ ∑ ⎢ 2 2 2 2 π ⎣ sin π x n =1 n − x n =1 n − x ⎦

which we write as cos tx − 1 =

2 x sin π x

π

(−1) n +1 (cos nt − 1) ∑ 2 2 n =1 n − x ∞

We now divide by t and integrate to obtain cos tx − 1 2 x sin π x ∞ (−1) n +1 cos nx − 1 = dt dt ∑ 2 2∫ ∫0 t t π n =1 n − x 0 u

u

It is easily seen that cos tx − 1 cos y − 1 ∫0 t dt = ∫0 y dy u

ux

and hence we have cos tx − 1 dt = Ci (ux) − γ − log(ux) t 0

u

∫

Therefore we obtain Ci (ux) − γ − log(ux) =

=

=

2 x sin π x

π

2 x sin π x

π 2 x sin π x

π

(−1) n +1 [Ci (nu ) − γ − log(nu )] ∑ 2 2 n =1 n − x ∞

(−1) n +1 2 x sin π x ∞ (−1) n +1 γ [ Ci ( nu ) − log n ] − ( + log u ) ∑ ∑ 2 2 2 2 π n =1 n − x n =1 n − x ∞

(−1) n +1 ⎡ sin π x ⎤ [Ci (nu ) − log n] − (γ + log u ) ⎢1 − ∑ 2 2 π x ⎥⎦ ⎣ n =1 n − x ∞

where we have used

64

1−

sin π x 2 x sin π x ∞ (−1) n +1 = ∑ 2 2 πx π n =1 n − x

Hence we obtain (1.121)

Ci (ux) − log x =

2 x sin π x

π

(−1) n +1 (γ + log u ) sin π x [Ci (nu ) − log n] + ∑ 2 2 πx n =1 n − x ∞

which corrects a misprint in Nielsen’s book [44, p.72]. In examining the limit as x → 0 , using (1.19) we see that both sides of the equation are equal. Using L’Hôpital’s rule we easily see that both sides are also equivalent in the case where x = k where k is an integer. In passing, we note that we may write (1.120) as (1.122)

π x cos tx − sin π x ∞ (−1) n +1 =∑ 2 cos nt 2 2 x 2 sin π x n =1 n − x

and then take the limit as x → 0 . Applying L’Hôpital’s rule three times gives us the well known Fourier series [55, p.148] (−1) n +1 cos nt 1 2 = (π − 3t 2 ) ∑ 2 n 12 n =1 ∞

With t = π in (1.122) we obtain ∞ π x cos π x − sin π x 1 = − ∑ 2 2 2 2 x sin π x n =1 n − x

which is the same as (4.7). Differentiating (1.122) with respect to t results in

π sin tx ∞ (−1) n +1 n =∑ sin nt 2sin π x n =1 n 2 − x 2 and applying L’Hôpital’s rule as x → 0 gives us

π sin tx π t cos tx t = lim = x →0 2sin π x x → 0 2π cos π x 2

lim

65

and we end up with the well-known Fourier series [55, p.148] ∞ t (−1) n +1 =∑ sin nt 2 n =1 n

This could also be employed to derive (1.49). Integrating (1.122) with respect to t results in

π sin tx − t sin π x ∞ 1 (−1) n +1 =∑ sin nt 2 2 2 x 2 sin π x n =1 n n − x □ We note that [37a] gives a closed form for the following Fourier series in response to a problem posed by Fettis and Glasser for x < π , ( p, q) = 1, q ≥ 2 ∞

∑n n =1

⎡p ⎤ sin(kπ p / q) (−1) n 1 q −1 sin(1/ 2q)( x + 2k + 1)π ) = ⋅ log sin nt sin ⎢ ( x + kπ ) ⎥ ∑ 2 2 − ( p / q) pq k =1 sin(1/ 2q)( x + 2k − 1)π ) ⎣q ⎦ sin(π p / q ) □

We have [52, p.14] ∞

⎡

k =0

⎣

⎛ ⎝

ψ ( x) = log x + ∑ ⎢log ⎜1 +

1 ⎞ 1 ⎤ ⎟− x + k ⎠ x + k ⎥⎦ 1

which could then be used to evaluate ∫ψ ( x) sin pπ x dx using (1.8), (1.10.3) and the 0

integral

∫

sin pπ x dx = cos(kp) Si[ p ( x + k )] − sin( kp )Ci[ p( x + k )] x+k

2. Another approach to the log Γ( x) integrals

Part of the following analysis is taken from the 1862 book “Exposé de la théorie, propriétés, des formules de transformation, et méthodes d’évaluation des intégrales définies” by David Bierens de Haan [35, p.269], a copy of which is available on the internet courtesy of the University of Michigan Historical Mathematics Collection. The voluminous integrals evaluated by Bierens de Haan were an important source for the table of integrals subsequently compiled by Gradshteyn and Ryzhik [33]. Bierens de Haan (1822-1895) was, inter alia, a mathematician, an actuary and a historian; a short

66

interesting biography of him appears in Talvila’s paper [53] and a more detailed presentation is given in [49]. Let us designate the following integrals as 1

I ( p ) = ∫ log Γ( x) cos pπ x dx

(2.1)

0

1

K ( p ) = ∫ log Γ( x) sin pπ x dx

(2.2)

0

Then we have by letting x → 1 − x (2.3)

1

1

0

0

I ( p ) = cos pπ ∫ log Γ(1 − x) cos pπ x dx + sin pπ ∫ log Γ(1 − x) sin pπ x dx

and we see that 1

(2.4)

I (1/ 2) = ∫ log Γ(1 − x) sin(π x / 2) dx 0

We have from the definition 1

I (1 − p ) = ∫ log Γ( x)[cosπ x cos pπ x + sinπ x sin pπ x] dx 0

The integral 1

J ( p ) = ∫ log Γ(1 − x) cos pπ x dx 0

becomes by letting x → 1 − x 1

= ∫ log Γ( x )[cos pπ cos pπ x + sin pπ sin pπ x] dx 0

Similarly, with L( p) defined by 1

L( p ) = ∫ log Γ(1 − x) sin pπ x dx 0

67

becomes by letting x → 1 − x 1

= ∫ log Γ( x )[sin pπ cos pπ x − cos pπ sin pπ x]dx 0

In the same way as above we find that (2.5)

1

1

0

0

K ( p ) = sin pπ ∫ log Γ(1 − x) cos pπ x dx − cos pπ ∫ log Γ(1 − x) sin pπ x dx

and we see that 1

(2.5.1)

K (1/ 2) = ∫ log Γ(1 − x) cos(π x / 2) dx 0

We have Euler’s reflection formula [10] for all z ∈ C except z = 0, ± n where n ∈ N

Γ( z )Γ(1 − z ) =

π sin π z

and therefore for 0 < x < 1 log Γ( x) + log Γ(1 − x) = log(2π ) − log(2sin π x)

(2.6)

Multiplying this equation by cos pπ x and integrating, we obtain 1

1

0

0

∫ log Γ( x) cos pπ x dx + ∫ log Γ(1 − x) cos pπ x dx 1

1

0

0

= log(2π ) ∫ cos pπ x dx − ∫ log(2sin π x) cos pπ x dx

Then, using the Fourier series [16, p.241] ∞

cos nt n n =1

log ⎡⎣ 2sin ( t / 2 ) ⎤⎦ = −∑

(2.7)

0 < t < 2π

and, assuming that changing the order of integration and summation is valid, we get (2.8)

1

1

0

0

∫ log Γ( x) cos pπ x dx + ∫ log Γ(1 − x) cos pπ x dx 68

=

∞ sin pπ 1 log(2π ) + ∑ ∫ ⎡⎣ cos {(2n + p )π x} + cos {(2n − p )π x}⎤⎦ dx pπ n =1 2 n 0

=

∞ sin pπ 1 ⎡ sin {(2n + p )π } sin {(2n − p )π } ⎤ + log(2π ) + ∑ ⎢ ⎥ pπ (2n + p )π (2n − p )π ⎦ n =1 2n ⎣

1

Hence, provided p ≠ 2n , as originally determined by Bierens de Haan [35, p.270] we have (where I have corrected a typographical error) (2.9) 1

1

∫ log Γ( x) cos pπ x dx + ∫ log Γ(1 − x) cos pπ x dx = 0

0

sin pπ p sin pπ log(2π ) − pπ π

∞

1

∑ n 4n n =1

2

1 − p2

Rather confusingly, Bierens de Haan [35, p.270] reports the left-hand side of (2.9) as being equal to I ( p ) + I (1 − p) . From (2.9) we see that with p = 0 (2.10)

1

1

0

0

∫ log Γ( x) dx + ∫ log Γ(1 − x) dx = log(2π )

and by a simple change of variable we have 1

1

0

0

∫ log Γ(1 − x)dx = ∫ log Γ( x)dx Hence we obtain Raabe’s integral [56, p.261] 1

(2.11)

1

∫ log Γ( x) dx = 2 log(2π ) 0

From (2.9) we see that with p = 1 (2.12)

1

1

0

0

∫ log Γ( x) cos π x dx + ∫ log Γ(1 − x) cos π x dx = 0

which may of course also be obtained by a simple change of variable. With p = 1 we have

69

1

1

0

0

∫ log Γ( x) cos π x dx + ∫ log Γ(1 − x) cos π x dx 1

1

0

0

= log(2π ) ∫ cos π x dx − ∫ log(2sin π x) cos π x dx 1

= − ∫ log(2sin π x) cos π x dx 0

and, with the substitution y = sin π x , we see that the integral on the right-hand side vanishes. This is easily confirmed because, with the substitution u = 1 − x , we see that 1

1

0

0

∫ log Γ(1 − x) cos π x dx = −∫ log Γ(u) cos π u du 1

We are therefore unable to evaluate ∫ log Γ( x) cos π x dx using this methodology. 0

We note in passing that Raabe’s integral (2.11) may also be easily obtained as follows by integrating (2.6), i.e. with p = 0 1

1

1

0

0

0

∫ log Γ( x) dx + ∫ log Γ(1 − x) dx = log(2π ) − ∫ log(2sin π x) dx where we also use the well-known integral due to Euler 1

∫ log sin π x dx = − log 2 0

to determine that 1

1

0

0

∫ log Γ( x) dx + ∫ log Γ(1 − x) dx = log(2π ) and finally we see that 1

1

0

0

∫ log Γ( x) dx = ∫ log Γ(1 − x) dx □

70

In passing, we note that an interesting derivation of Euler’s integral appears in an 1864 paper by Jeffery [38, p.94]. We consider π

2

∫ 0

1 log sin x dx = 2

π

2

∫ log sin

2

x dx

0

and with the substitution t = sin 2 x we have 1

=

1 log t 1 1 dt ∫ 4 0 t 2 (1 − t ) 2

=

1 ∂ B (u , v) 4 ∂u u = v =1/ 2

=

1⎧ ⎛1⎞ ⎛1⎞ ⎫ 2⎛1⎞ ⎨Γ′ ⎜ ⎟ Γ ⎜ ⎟ − Γ ⎜ ⎟ Γ′ (1) ⎬ 4⎩ ⎝2⎠ ⎝2⎠ ⎝2⎠ ⎭

=−

π 2

log 2 □

Letting p = 1/ 2 in (2.9) we see that 1

1

∫ log Γ( x) cos(π x / 2) dx + ∫ log Γ(1 − x) cos(π x / 2) dx = 0

0

=

2

π 2

π

1 2π

log(2π ) −

log(2π ) −

∞

n =1

∞

2

1

∑ n 4n 1

1

0

0

n =1

2

∫ log Γ(1 − x) cos(π x / 2) dx = ∫ log Γ( x) sin(π x / 2) dx Hence we have 1

(2.13)

∫ log Γ( x)[cos(π x / 2) + sin(π x / 2)] dx = 0

2

π

log(2π ) −

2

∞

1

1

∑ n 16n π n =1

2

With regard to the above infinite series, Ramanujan [11, p.29] showed that

71

1 − (1/ 4)

1

∑ n 16n π

With the obvious substitution we see that 1

2

−1

−1

∞

1

1

∑ n 16n

(2.14)

n =1

2

−1

= 3log 2 − 2

In fact, as reported in [11, p.29], this summation was the first problem submitted by Ramanujan in 1911 to the Journal of the Indian Mathematical Society. Ramanujan [11, p.26] also showed that ∞

1

∑ n 4n

(2.15)

n =1

1 2

−1

= 2 log 2 − 1

This series is contained in [33, p.9,] where it is attributed to Bromwich’s text [15, p.51]. A proof is also given in Ramanujan’s Notebooks [11, p.26]. In [15] the proof is based on the definition of Euler’s constant γ and it is therefore quite apt that the series should be connected with the gamma function. Therefore we obtain from (2.13) and (2.14) 1

(2.16)

2

∫ log Γ( x)[cos(π x / 2) + sin(π x / 2)] dx = π [log π − 2 log 2 + 2] 0

or equivalently ⎛π

1

(2.16.1)

⎞

∫ log Γ( x) cos ⎜⎝ 4 [2 x − 1] ⎟⎠ dx = 0

2

π

[log π − 2 log 2 + 2] □

By definition we have 1

⎛1⎞ I ⎜ ⎟ = ∫ log Γ( x) cos(π x / 2) dx ⎝2⎠ 0 1

2

= 2 ∫ log Γ(2t ) cos(π t ) dt 0

We have Legendre’s duplication formula for the gamma function (see [56, p.240] and [25] for example) 1⎞ ⎛ (2.16.2) 22 x −1 Γ( x)Γ ⎜ x + ⎟ = π Γ(2 x) 2⎠ ⎝

72

which may be expressed as 1⎞ 1 ⎛ log Γ(2 x) = log Γ( x) + log Γ ⎜ x + ⎟ + ( 2 x − 1) log 2 − log π 2⎠ 2 ⎝ Hence we have 1 I (1/ 2) = 2

1

1

2

∫ log Γ(t ) cos(π t ) dt + 0

2

⎛

1⎞

∫ log Γ ⎜⎝ t + 2 ⎟⎠ cos(π t ) dt 0

1

2 1 ⎡ ⎤ + ∫ ⎢( 2t − 1) log 2 − log π ⎥ cos(π t ) dt 2 ⎦ 0 ⎣

Using the obvious change of variables we have 1

2

⎛ 1⎞ ∫0 log Γ ⎝⎜ t + 2 ⎠⎟ cos(π t ) dt =

1

⎛1

⎞

∫ log Γ(u) cos π ⎜⎝ 2 − u ⎟⎠ du

1

2

1

=

∫ log Γ(u) sin π u du

1

2 1

1

2

= ∫ log Γ(u ) sin π u du − ∫ log Γ(u ) sin π u du 0

0

1

1⎡ π ⎤ 2 = ⎢log + 1⎥ − ∫ log Γ(u ) sin π u du π⎣ 2 ⎦ 0 where we used (2.26) below. Therefore we have 1

(2.17)

2

⎡1

2 ⎤ 1 log 2 − 2⎥ π ⎦

∫ log Γ(t )[cosπ t − sin π t ] dt + ⎢⎣ π − π 0

1 ⎡ ⎤ 1 ⎢⎣log 2 + 2 log π ⎥⎦ + π

⎡ π ⎤ ⎢⎣log 2 + 1⎥⎦ □

We define

73

1 1 2 n =1 an ( an) − 1 k

φ ( a , k ) = 1 + 2∑ and

∞

1 1 2 k →∞ n =1 an ( an) − 1 Sitaramachandrarao [50] has shown that if p, q are positive integers and if φ ( p, q ) exists, then

φ (a) = lim φ (a, k ) = 1 + 2∑

(2.18)

⎣⎢ q / p ⎦⎥ ⎛ p ⎞ 2q ⎡ ⎛ 2π qj ⎞ ⎛ π j ⎞⎤ 1 q φ ⎜ ⎟ = ⎢log 2 p − 2∑ cos ⎜ + log sin 2 ∑ ⎟ ⎜ ⎟⎥ j =1 pj − q ⎝q⎠ p ⎣ ⎝ p ⎠ ⎝ p ⎠⎦

(2.19)

where the (corrected) first summation runs over 0 < j < p / 2 , and integer ≤ x .

⎢⎣ x ⎥⎦

denotes the largest

Hence we have from (2.9) ⎛ p⎞ ⎛ p ⎞ q sin( pπ / q ) p sin( pπ / q) ∞ 1 1 I ⎜ ⎟ + J ⎜1 − ⎟ = log(2π ) − ∑ 2 2 pπ qπ n =1 n 4n − ( p / q ) ⎝q⎠ ⎝ q⎠ =

q sin( pπ / q ) q 2 sin( pπ / q ) ∞ 2 1 log(2π ) − ∑ 2 2 pπ pπ n =1 2 qn / p (2qn / p ) − 1

and thus ⎛ p⎞ ⎛ p ⎞ q sin( pπ / q ) q 2 sin( pπ / q ) ⎡ ⎛ 2q ⎞ ⎤ (2.20) I ⎜ ⎟ + J ⎜ 1 − ⎟ = log(2π ) − ⎢φ ⎜ ⎟ − 1⎥ pπ p 2π ⎝q⎠ ⎝ q⎠ ⎣ ⎝ p ⎠ ⎦

As mentioned by Sitaramachandrarao, (2.19) is clearly related to Gauss’s formula for the digamma function (see [15, p.522] and [52, p.18]. q −1 ⎡ ⎛ p⎞ ⎛ pπ ⎞ ⎛ 2π pj ⎞ ⎛ π j ⎞⎤ π q ψ ⎜ ⎟ = −γ − cot ⎜ − + log cos ⎜ ∑ ⎟ ⎟ log ⎢ 2sin ⎜ ⎟⎥ 2 j =1 ⎝q⎠ ⎝ q ⎠ ⎝ q ⎠ ⎝ q ⎠⎦ ⎣

With p = 2 and q = 2k +1 in (2.19) we obtain k 1 ⎛ 2 ⎞ k k 2(2 1) log 2 2(2 1) = + + + ∑ ⎟ ⎝ 2k + 1 ⎠ j =1 2 j − (2 k + 1)

φ⎜

or equivalently

74

k 1 ⎛ 2 ⎞ k k 2(2 1) log 2 2(2 1) = + − + ∑ ⎟ ⎝ 2k + 1 ⎠ j =1 2 j − 1

φ⎜

By definition we have ∞ 1 1 ⎛ 2 ⎞ 3 ⎟ = 1 + (2k + 1) ∑ 2 2 ⎝ 2k + 1 ⎠ n =1 n 4 n − (2 k + 1)

φ⎜

and hence we have ∞

1 1 1 = ∑ 2 2 (2k + 1) 2 n =1 n 4n − (2k + 1)

(2.21)

k −1 ⎡ 1 1 ⎤ − − 2 log 2 2 ⎢ ∑ ⎥ 2k + 1 j =0 2 j + 1 ⎦ ⎣

This identity is employed in (5.6). For example, with k = 0 in (2.21) we recover Ramanujan’s formula (2.15). We immediately notice the strong similarity with the following integral which is evaluated in Section 6 1

∫ log sin π x sin(2k + 1)π x dx = 0

2 (2k + 1)π

k −1 ⎡ 1 1 ⎤ − − log 2 2 ⎢ ⎥ ∑ 2k + 1 j =0 2 j + 1 ⎦ ⎣

and, with a minor adjustment, we see that 1

∫ log(2sin π x) sin(2k + 1)π x dx = 0

2 (2k + 1)π

k −1 ⎡ 1 1 ⎤ − − 2 log 2 2 ⎢ ⎥ ∑ 2k + 1 j =0 2 j + 1 ⎦ ⎣

The very fact that Ramanujan deals with φ (a) in Chapter 8 of [11], which also involves connections with Kummer’s formula (3.11) and related matters, strongly suggests that he was well aware of equation (2.9). We note from equation (A.3) in the Appendix that ∞

(2.22)

1 1 2 2 n =1 n n − x

ψ (1 + x) +ψ (1 − x) + 2γ = −2 x 2 ∑

which concurs with Prudnikov et al. [48, 5.1.25-13]. In another paper [27] we have derived rapidly converging series for Catalan’s constant G = β (2) and for Apéry’s constant ς (3) by employing (2.22). These are set out below:

75

∞

1 1 2 2 n =1 n (16n − 1)

G = 3(1 − log 2) − ∑

∞

1 1 2 3 n =1 n (16n − 1)

7ς (3) = 12(5 − 4 log 2) − 20G + 16∑

It may be noted that Mathematica confirms these identities both numerically and analytically. The method may be easily generalised to produce new series representations for values of the Riemann zeta function ς (2n + 1) and the Dirichlet beta function β (2n) . With x → 1/ x in (2.22) we get ⎛1⎞ ⎝ ⎠

⎛ ⎝

1⎞

⎡

∞

⎠

⎣

n =1

1

1

⎤

ψ ⎜ ⎟ +ψ ⎜1 − ⎟ = −2γ − x ⎢1 + 2∑ x x nx (nx) 2 − 1 ⎥ ⎦

as reported by Ramanujan [11, p.186]. With x = p / 2 in (2.22) we have (2.23)

⎛ ⎝

ψ ⎜1 +

∞ p⎞ p⎞ 1 1 ⎛ 2 ⎟ + ψ ⎜1 − ⎟ + 2γ = −2 p ∑ 2 2 2⎠ ⎝ 2⎠ n =1 n 4n − p

with the result that we may write (2.9) as (2.24)

1

1

0

0

∫ log Γ( x) cos pπ x dx + ∫ log Γ(1 − x) cos pπ x dx =

sin pπ pπ

⎡ 1 ⎛ p⎞ 1 ⎛ p ⎞⎤ ⎢γ + log(2π ) + 2 ψ ⎜1 + 2 ⎟ + 2 ψ ⎜1 − 2 ⎟ ⎥ ⎝ ⎠ ⎝ ⎠⎦ ⎣

For example, with p = 1/ 2 in (2.23) we have =

1⎡ 1 ⎛ 5 ⎞ 1 ⎛ 3 ⎞⎤ γ + log(2π ) + ψ ⎜ ⎟ + ψ ⎜ ⎟ ⎥ ⎢ π⎣ 2 ⎝ 4 ⎠ 2 ⎝ 4 ⎠⎦

and using [52, p.20] ⎛3⎞

1

ψ ⎜ ⎟ = −γ + π − 3log 2 2 ⎝4⎠

⎛5⎞

1

ψ ⎜ ⎟ = −γ − π − 3log 2 + 4 2 ⎝4⎠

76

we therefore have another derivation of (2.16). □ 1

We now consider integrals of the form ∫ log Γ( x) sin pπ x dx . Multiplying (2.6) by 0

sin pπ x and integrating, we obtain 1

1

0

0

∫ log Γ( x)sin pπ x dx + ∫ log Γ(1 − x)sin pπ x dx =

∞ 1 − cos pπ 1 log(2π ) + ∑ ∫ ⎡⎣sin {(2n + p )π x} − sin {(2n − p )π x}⎤⎦ dx pπ n =1 2n 0

=

∞ 1 − cos pπ 1 ⎡1 − cos {(2n + p )π } 1 − cos {(2n − p)π } ⎤ − log(2π ) + ∑ ⎢ ⎥ pπ (2n + p )π (2n − p )π n =1 2n ⎣ ⎦

1

Hence, provided p ≠ 2n , we have (2.25)

1

1

0

0

∫ log Γ( x) sin pπ x dx + ∫ log Γ(1 − x) sin pπ x dx =

1 − cos pπ p(1 − cos pπ ) ∞ 1 1 log(2π ) − ∑ 2 2 pπ π n =1 n 4n − p

With p = 1 in (2.25) we obtain 1

1

0

0

∫ log Γ( x) sin π x dx + ∫ log Γ(1 − x) sin π x dx =

2

π

log(2π ) −

∞

2

1

∑ n 4n π n =1

1 2

−1

and substituting (2.15) we have =

= It is easily seen that 1

1

0

0

∫ log Γ(1 − x) sin π x dx = ∫ log Γ( x) sin π x dx 77

2

π

log(2π ) −

2

π

[2 log 2 − 1]

2 ⎡ ⎛π ⎞ ⎤ +1 log π ⎢⎣ ⎜⎝ 2 ⎟⎠ ⎥⎦

and we therefore obtain 1

(2.26) ∫ log Γ( x) sin π x dx = 0

1 ⎡ ⎛π ⎞ ⎤ +1 log π ⎢⎣ ⎜⎝ 2 ⎟⎠ ⎦⎥

which we shall also obtain in (5.7). Another approach is set out below. We have using Euler’s reflection formula 1

1

0

0

∫ [log Γ( x) + log Γ(1 − x)]sin π x dx = ∫ [log π − log sin π x]sin π x dx =

2

π

1

log π − ∫ log sin π x sin π x dx 0

so that 1

2 ∫ log Γ( x) sin π x dx = 0

2

π

1

log π − ∫ log sin π x sin π x dx 0

Dwilewicz and Mináč [28] have shown that (see also equation (6.4) in Section 6 below) 2 pπ 1 log(sin π x) sin 2 pπ x dx = − [ψ (1 + p ) + ψ (1 − p ) + 2γ + 2 log 2] ∫ 2 1 − cos 2 pπ 0 1

(2.27)

and with p = 1/ 2 this becomes

π

1

1

log(sin π x) sinπ x dx = − [ψ (3 / 2) + ψ (1/ 2) + 2γ + 2 log 2] 2∫ 2 0

= log 2 − 1 and this also results in 1

1⎡

⎛π ⎞

⎤

∫ logΓ( x) sin π x dx = π ⎢⎣log ⎜⎝ 2 ⎟⎠ + 1⎥⎦ 0

which we will also see in (5.7). □ With p = 1/ 2 in (2.25) we obtain

78

1

1

∫ log Γ( x) sin(π x / 2) dx + ∫ log Γ(1 − x) sin(π x / 2) dx = 0

0

2

π

log(2π ) −

2

∞

1

1

∑ n 16n π n =1

2

−1

It is easily shown that 1

1

0

0

∫ log Γ(1 − x) sin(π x / 2) dx = ∫ log Γ( x) cos(π x / 2) dx and we simply obtain (2.13) again. □ Letting p = 2k in (2.8) where k is an integer we get 1

1

0

0

∫ log Γ( x) cos 2kπ x dx + ∫ log Γ(1 − x) cos 2kπ x dx =

∞ sin 2kπ 1 ⎡ sin {(n + k )2π } ⎤ k −1 sin {(k − n)2π } log(2π ) + ∑ ⎢ ⎥+∑ 2 kπ (n + k )2π ⎦ n =1 (k − n)2π n =1 2 n ⎣

∞ sin ( k − n)2π { } 1 dx + cos(0) + ∑ ∫ 2k 0 (k − n)2π n = k +1

1

We have by a simple substitution 1

1

0

0

∫ log Γ(1 − x) cos 2kπ x dx = ∫ log Γ( x) cos 2kπ x dx so that 1

(2.28)

1

∫ log Γ( x) cos 2kπ x dx = 4k 0

Letting p = 2k + 1 in (2.8) does not provide any information because both sides of the equation vanish. This result (2.28) may also be obtained more directly from (2.9) by taking the limit as p → 2k . It may be seen from

79

p sin pπ

π

⎡ ⎤ ∞ 1 1 p sin pπ ⎢ 1 1 1 1 ⎥ = +∑ ∑ 2 2 2 2 2 2⎥ ⎢ π k 4k − p n =1 n 4n − p n =1 n 4n − p ⎢ ⎥ n≠ k ⎣ ⎦ ∞

that all, bar one, of the terms automatically vanish and we have the limit via L’Hôpital’s rule 1 p sin pπ 1 =− lim 2 2 p → 2 k π k 4k − p 2k □ We may also write 1

∫0 [log Γ( x) + log Γ(1 − x)]cos pπ x dx 1

∫

= [log π − log sin π x]cos pπ x dx 0

=

sin pπ pπ

1

∫

log π − log sin π x cos pπ x dx 0

and similarly we have 1

∫ [log Γ( x) + log Γ(1 − x)]sin pπ x dx = 0

cos pπ pπ

1

∫

log π − log sin π x sin pπ x dx 0

We note that the Fourier coefficients of log sin π x are given by [33, p.577, 4.384] 1

(2.29)

1

∫ log sin π x cos 2kπ x dx = − 2k

k>0

0

1

(2.30)

∫ log sin π x cos(2k + 1)π x dx = 0 0

1

(2.31)

∫ log sin π x dx = − log 2

k =0

0

1

(2.32)

∫ log sin π x sin 2kπ x dx = 0 0

80

k ≥0

1

∫ log sin π x sin(2k + 1)π x dx =

(2.33)

0

2 (2k + 1)π

k −1 ⎡ 1 1 ⎤ − − log 2 2 ⎢ ⎥ ∑ 2k + 1 j =0 2 j + 1 ⎦ ⎣

It should be noted that the above integral (2.33) is incorrectly reported in Nielsen’s book [45, p.202]. Derivations of these integrals are given in Section 6. We could accordingly use these Fourier coefficients of log sin π x to evaluate the corresponding integrals involving the log gamma function □ Denoting f ( p) by f ( p) =

(2.34)

sin pπ pπ

we have pf ( p) =

sin pπ

π

and thus we get (2.35)

pf ′( p) + f ( p) = cos pπ

or equivalently f ′( p ) =

cos pπ − f ( p ) p

We see that f (0) = 1 and applying L’Hôpital’s rule we find that f ′(0) = lim p →0

cos pπ − f ( p ) = lim[−π sin pπ − f ′( p )] = − f ′(0) p →0 p

and thus f ′(0) = 0 . This could of course have been more readily obtained by using the elementary Maclaurin series for f ( p ) . Differentiating (2.9) results in (2.36)

I ′( p) + J ′( p) = f ′( p) log(2π ) −

81

pπ cos pπ + sin pπ

π

∞

1

∑ n 4n n =1

2

1 − p2

−

2 p 2 sin pπ

π

∞

1

∑ n (4n n =1

2

1 − p 2 )2

where we have 1

I ( p ) = ∫ log Γ( x) cos pπ x dx 0

1

J ( p ) = ∫ log Γ(1 − x) cos pπ x dx 0

1

I ′( p ) + J ′( p ) = −π ∫ x[log Γ( x) + log Γ(1 − x)]sin pπ x dx 0

We have

π 2 p cos pπ − π sin pπ f ′( p ) = π 2 p2 and thus

f ′(1) = −1

With p = 1 we have 1

1

0

0

I ′(1) + J ′(1) = −π ∫ x log Γ( x) sin π x dx − π ∫ x log Γ(1 − x) sin π x dx 1

1

0

0

∫ x log Γ(1 − x) sin π x dx = ∫ (1 − u ) log Γ(u ) sin[π (1 − u )] du 1

= ∫ (1 − u ) log Γ(u ) sin π u du 0

1

1

0

0

= ∫ log Γ(u ) sin π u du − ∫ u log Γ(u ) sin π u du 1

Unfortunately, the terms involving

∫ x log Γ( x) sin π x dx cancel out and hence we have 0

1

I ′(1) + J ′(1) = −π ∫ log Γ( x) sin π x dx 0

82

The right-hand side of (2.36) at p = 1 is ∞

1 1 2 n =1 n 4 n − 1

= − log(2π ) + ∑

= − log(2π ) + 2 log 2 − 1 where we have used (2.15). Therefore we obtain another derivation of 1

π

1⎡

⎤

∫ log Γ( x) sin π x dx = π ⎢⎣log 2 + 1⎥⎦ 0

We also have 1

1

0

0

I ′(1) + J ′(1) = −π log π ∫ x sin π x dx + π ∫ x log sin π x.sin π x dx

and therefore we obtain 1

(2.36.1)

1

∫ x log sin π x.sin π x dx = π [log 2 − 1] 0

□ Differentiating (2.35) gives us pf ′′( p) + 2 f ′( p ) = −π sin pπ or equivalently f ′′( p) = −

π sin pπ + 2 f ′( p ) p

Using L’Hôpital’s rule then gives us f ′′(0) = − lim

π sin pπ + 2 f ′( p )

p →0

= − lim[π 2 cos pπ + 2 f ′′( p )] p →0

p

1 so that we have f ′′(0) = − π 2 . 3 Differentiating (2.36) results in

83

I ′′( p) + J ′′( p ) = f ′′( p) log(2π ) −

−

pπ 2 sin pπ + 2π cos pπ

6 p ( pπ cos pπ + sin pπ )

π

1

∑ n (4n n =1

1 1 8 p 3 sin pπ − ∑ 2 2 π n =1 n 4n − p ∞

π

∞

∞

1

∑ n (4n n =1

2

Therefore we obtain with p = 0 1 1 I ′′(0) + J ′′(0) = − π 2 log(2π ) − ς (3) 3 2 We have from the definitions of I ( p ) and J ( p ) I ′′( p ) + J ′′( p ) = −π

1

2

∫ x [log Γ( x) + log Γ(1 − x)]cos pπ x dx 2

0

I ′′(0) + J ′′(0) = −π

1

2

∫ x [log Γ( x) + log Γ(1 − x)] dx 2

0

= −π

1

2

∫0 x [log π − log sin π x] dx 2

1

1 = − π 2 log π + π 2 ∫ x 2 log sin π x dx 3 0

Hence we deduce that 1

(2.37)

∫0 x

2

1 3

log sin π x dx = − log 2 −

ς (3) 2π 2

as previously reported in [30]. Alternatively, we have 1

1

0

0

I ′′(0) + J ′′(0) = −π 2 ∫ x 2 log Γ( x) dx − π 2 ∫ x 2 log Γ(1 − x) dx

We see that

84

2

1 − p 2 )2

1 − p 2 )3

1

1

0

0

2 2 ∫ x log Γ(1 − x) dx = ∫ (1 − u) log Γ(u ) du

so that we have I ′′(0) + J ′′(0) = −2π

1

2

∫x

2

log Γ( x) dx + 2π

0

1

2

1

∫ x log Γ( x) dx − π ∫ log Γ( x) dx 2

0

0

We have the well known integral (see for example (3.25) and Appendix C) 1

1

∫ x log Γ( x) dx = 4 log(2π ) − log A

(2.38)

0

so that 1

1 1 ⎡1 ⎤ 1 −2π 2 ∫ x 2 log Γ( x)dx + 2π 2 ⎢ log(2π ) − log A⎥ − π 2 log(2π ) = − π 2 log(2π ) − ς (3) 3 2 ⎣4 ⎦ 2 0

Hence we deduce the known integral 1

∫x

(2.39)

0

2

1 ς (3) log Γ( x) dx = log(2π ) − log A + 2 6 4π

which is also shown in (C.6) in Appendix C. □ Alternatively, we now rewrite (2.9) as ∞

1

∑ n 4n n =1

2

1 log(2π ) sin pπ − π p[ I ( p ) + J ( p )] = 2 −p p 2 sin pπ

and we have the limit as p → 0 1 log(2π ) sin pπ − π p[ I ( p) + J ( p)] ς (3) = lim p →0 p 2 sin pπ 4 Applying L’Hôpital’s rule three times we have

π log(2π ) cos pπ − π p[ I ′( p ) + J ′( p )] − π [ I ( p) + J ( p )] p →0 π p 2 cos pπ + 2 p sin pπ

= lim

85

−π 2 log(2π ) sin pπ − π p[ I ′′( p ) + J ′′( p )] − 2π [ I ′( p ) + J ′( p )] p →0 −π 2 p 2 sin pπ + 4 pπ cos pπ + 2sin pπ

= lim

−π 3 log(2π ) cos pπ − π p[ I ′′′( p) + J ′′′( p )] − 3π [ I ′′( p) + J ′′( p)] p →0 −π 3 p 2 cos pπ − 6π 2 p sin pπ + 6π cos pπ

= lim

so that 1 1 1 ς (3) = − π 2 log(2π ) − [ I ′′(0) + J ′′(0)] 4 6 2 and this gives us another derivation of (2.39). See also Section 6. □ We recall (2.25) 1

1

0

0

∫ log Γ( x) sin pπ x dx + ∫ log Γ(1 − x) sin pπ x dx =

1 − cos pπ p(1 − cos pπ ) ∞ 1 1 log(2π ) − ∑ 2 2 pπ π n =1 n 4n − p

and differentiating gives us (2.40)

1

1

0

0

π ∫ x log Γ( x) cos pπ x dx + π ∫ x log Γ(1 − x) cos pπ x dx =

pπ 2 sin pπ − π (1 − cos pπ ) 2 p 2 (1 − cos pπ ) ∞ 1 1 log(2 π ) − ∑ 2 2 2 2 ( pπ ) π n =1 n (4n − p ) −

pπ sin pπ + 1 − cos pπ

π

∞

1

∑ n 4n n =1

2

1 − p2

π

log(2π ) , 2 which, using L’Hôpital’s rule, is also seen to be the only surviving term on the right-hand side. Letting p = 0 in (2.40) we easily determine that the left-hand side becomes

With p = 1 in (2.40) we have

86

1

1

0

0

π ∫ x log Γ( x) cos π x dx + π ∫ x log Γ(1 − x) cos π x dx

(2.41)

2

4

= − log(2π ) − π π

∞

1 1 2 ∞ 1 1 − ∑ ∑ 2 2 π n =1 n 4n 2 − 1 n =1 n (4n − 1)

Differentiating (2.23) results in ∞ ∞ 1 ⎛ p⎞ 1 ⎛ p⎞ 1 1 1 1 ψ ′ ⎜1 + ⎟ − ψ ′ ⎜1 − ⎟ = −4 p 3 ∑ − 4 p ∑ 2 2 2 2 2 2 ⎝ 2⎠ 2 ⎝ 2⎠ n =1 n (4n − p ) n =1 n 4n − p

(2.42)

and with p = 1 we obtain ∞ ∞ 1 ⎛ 1⎞ 1 ⎛1⎞ 1 1 1 1 ψ ′ ⎜1 + ⎟ − ψ ′ ⎜ ⎟ = −4∑ − 4 ∑ 2 2 2 2 ⎝ 2⎠ 2 ⎝2⎠ n =1 n (4 n − 1) n =1 n 4n − 1

Since ψ (1 + x) −ψ ( x) =

1 1 we have ψ ′(1 + x) −ψ ′( x) = − 2 and thus x x

1 ⎛ 1⎞ 1 ⎛1⎞ ψ ′ ⎜1 + ⎟ − ψ ′ ⎜ ⎟ = −2 2 ⎝ 2⎠ 2 ⎝2⎠ Hence using (2.15) we obtain ∞

1

∑ n (4n

(2.43)

n =1

1 3 = − 2 log 2 2 − 1) 2

2

which is reported in [33, p.9]. The right-hand side of (2.41) becomes using (2.15) and (2.43) 2

4 ⎡3

⎤

2

= − log(2π ) − ⎢ − 2 log 2 ⎥ − (2 log 2 − 1) π π ⎣2 ⎦ π =−

2⎡ π ⎤ log + 2 ⎥ ⎢ π⎣ 2 ⎦

Having regard to (2.40), we see that with p = 1 1

1

0

0

∫ x log Γ(1 − x) cos π x dx = ∫ (1 − u ) log Γ(u ) cos π (1 − u ) du 87

1

= ∫ (u − 1) log Γ(u ) cos π u du 0

which gives us 1

1

1

1⎞ ⎛ 2π ∫ x log Γ( x) cos π x dx − π ∫ log Γ( x) cos π x dx = 2π ∫ ⎜ x − ⎟ log Γ( x) cos π x dx 2⎠ 0 0 0⎝

and we have the result 1

⎛

1 ⎡ π ⎤ log + 2 ⎥ 2 ⎢ 2 ⎣ ⎦

1⎞

∫ ⎝⎜ x − 2 ⎠⎟ log Γ( x) cos π x dx = − π

(2.44)

0

We note from (4.5.2) that 1

∞

4

∫ log Γ( x) cos π x dx = π ∑ 2

0

γ + log(2π n) 4n 2 − 1

n =1

so that we have 1

∫ x log Γ( x) cos π x dx =

(2.45)

0

2

π

2

∞

∑

γ + log(2π n)

n =1

4n − 1 2

−

1 ⎡ π ⎤ log + 2 ⎥ 2 ⎢ π ⎣ 2 ⎦

This may also be obtained by differentiating (5.5) with respect to p . This gives us 1

∫ x log Γ( x) cos π x dx 0

=−

=−

=

1

π

2

π

2

1

π

2

log(2π ) −

2

log(2π ) −

∞

∑ n =1

2

1 1 1 − 2 2 ∑ 2 2 π n =1 n (4n − 1) π

∞

1 1 2 + 2 ∑ 2 π n =1 n 4n − 1

2 ⎡3 2 ⎤ 1 − 2 log 2 ⎥ − 2 (2 log 2 − 1) + 2 2 ⎢ π ⎣2 π ⎦ π

γ + log(2π n) 4n − 1 2

∞

−

∞

∑ n =1

∞

∑ n =1

γ + log(2π n) 4n 2 − 1

γ + log(2π n) 4n 2 − 1

1 ⎡ π ⎤ log + 2 ⎥ 2 ⎢ π ⎣ 2 ⎦ □

88

As noted in Section 4 we have 1

1

∂ 1 ς ( s, x)ς (0, x) cos π x dx = log(2π ) ∫ B1 ( x) cos π x dx ∫ ∂s 0 2 0 s =0 1

− ∫ B1 ( x) log Γ( x) cos π x dx 0

where Bn ( x) are the Bernoulli polynomials and in particular we have B1 ( x) = x −

1 2 □

Differentiating (2.42) results in ∞ ∞ 1 ⎛ p⎞ 1 ⎛ p⎞ 1 1 1 1 2 ψ ′′ ⎜1 + ⎟ + ψ ′′ ⎜1 − ⎟ = −16 p 4 ∑ − 20 p ∑ 2 2 3 2 2 2 4 ⎝ 2⎠ 4 ⎝ 2⎠ n =1 n (4n − p ) n =1 n (4n − p )

∞

1 1 2 2 n =1 n 4n − p

− 4∑ and with p = 1 we obtain ⎛ ⎝

1⎞

⎛1⎞ ⎝ ⎠

∞

1

∞

1

1

1

∞

1

1

ψ ′′ ⎜1 + ⎟ +ψ ′′ ⎜ ⎟ = −64∑ − 80∑ − 16∑ n (4n 2 − 1)3 n (4n 2 − 1) 2 n 4n 2 − 1 2 2 ⎠

n =1

n =1

It is well known that (see [1, 6.4.4] and [41]) ⎛1⎞ ⎝ ⎠

ψ ( k ) ⎜ ⎟ = (−1) k +1 k !(2k +1 − 1)ς (k + 1) 2 ⎛1⎞ ⎝ ⎠

ψ (2) ⎜ ⎟ = −14ς (3) 2 ψ ′′(1 + x) −ψ ′′( x) = ⎛ ⎝

1⎞

2 x3

⎛1⎞ ⎝ ⎠

ψ ′′ ⎜1 + ⎟ +ψ ′′ ⎜ ⎟ = 16 − 28ς (3) 2 2 ⎠

89

n =1

∞

∞ ∞ 1 1 1 1 1 1 + 80∑ + 16∑ 28ς (3) − 16 = 64∑ 2 3 2 2 2 n =1 n (4 n − 1) n =1 n (4 n − 1) n =1 n 4n − 1

and using (2.15) and (2.43) this becomes ∞

1 1 ⎡3 ⎤ + 80 ⎢ − 2 log 2 ⎥ + 16 [ 2 log 2 − 1] 2 3 ⎣2 ⎦ n =1 n (4n − 1)

28ς (3) − 16 = 64∑ which gives us ∞

1 1 = 7ς (3) + 32 log 2 − 30 2 3 n =1 n (4 n − 1)

16∑

(2.46)

□ We recall (2.24) 1

1

0

0

∫ log Γ( x) cos pπ x dx + ∫ log Γ(1 − x) cos pπ x dx =

sin pπ pπ

⎡ 1 ⎛ p⎞ 1 ⎛ p ⎞⎤ ⎢γ + log(2π ) + 2 ψ ⎜ 1 + 2 ⎟ + 2 ψ ⎜1 − 2 ⎟ ⎥ ⎝ ⎠ ⎝ ⎠⎦ ⎣

and differentiation results in 1

1

0

0

−π ∫ x log Γ( x) sin pπ x dx − π ∫ x log Γ(1 − x) sin pπ x dx

=

sin pπ pπ

+

π 2 p cos pπ − π sin pπ π 2 p2

⎡1 ⎛ p⎞ 1 ⎛ p ⎞⎤ ⎢ 22 ψ ′ ⎜1 + 2 ⎟ − 22 ψ ′ ⎜1 − 2 ⎟ ⎥ ⎝ ⎠ ⎝ ⎠⎦ ⎣ ⎡ 1 ⎛ p⎞ 1 ⎛ p ⎞⎤ ⎢γ + log(2π ) + 2 ψ ⎜ 1 + 2 ⎟ + 2 ψ ⎜1 − 2 ⎟ ⎥ ⎝ ⎠ ⎝ ⎠⎦ ⎣

With p = 1 this becomes 1

1

0

0

−π ∫ x log Γ( x) sin π x dx − π ∫ x log Γ(1 − x) sin π x dx

⎡ 1 ⎛ 1 ⎞ 1 ⎛ 1 ⎞⎤ = − ⎢γ + log(2π ) + ψ ⎜1 + ⎟ + ψ ⎜ ⎟ ⎥ 2 ⎝ 2 ⎠ 2 ⎝ 2 ⎠⎦ ⎣

90

⎡ π ⎤ = − ⎢ log + 1⎥ 2 ⎦ ⎣ We see that 1

1

0

0

∫ x log Γ(1 − x) sin π x dx = ∫ (1 − x) log Γ( x) sin π x dx 1

1

1

0

0

0

−π ∫ x log Γ( x) sin π x dx − π ∫ x log Γ(1 − x) sin π x dx = −π ∫ log Γ( x) sin π x dx

and hence we obtain 1

π

1⎡

⎤

∫ log Γ( x) sin π x dx = π ⎢⎣log 2 + 1⎥⎦ 0

An observation by Glasser

Part of the following is based on an observation made by Glasser [32] in 1966. Let us consider the integral 1

I = ∫ f ( x)ψ ( x)dx 0

where f ( x) = − f (1 − x) and f ( x) is selected so that the integral converges. We then have 1

1

0

0

I = ∫ f (1 − t )ψ (1 − t )dt = − ∫ f (t )ψ (1 − t )dt

and hence we see that 1

2 I = ∫ f ( x)[ψ ( x) −ψ (1 − x)]dx 0

Therefore, since ψ ( x) −ψ (1 − x) = −π cot π x , we have 1

(2.47)

∫ 0

f ( x)ψ ( x)dx = −

π

1

2 ∫0

1

f ( x) cot π x dx = −π

2

∫ 0

91

f ( x) cot π x dx

Glasser [32] used the function f ( x) = x(1 − x) cos π x to show that 1

1 ⎡ 7 ⎤ 2 − ς (3) ⎥ 2 ⎢ 2 ⎣ ⎦

∫ ψ ( x) x(1 − x) cos π x dx = π

(2.48)

0

Integration by parts gives us 1

∫ ψ ( x) x(1 − x) cos π x dx = log Γ( x) x(1 − x) cos π x

1 0

0

1

+ ∫ log Γ( x)[ x(1 − x)π sin π x + (2 x − 1) cos π x] dx 0

Since x log Γ( x) = x log Γ(1 + x) − x log x we see that the integrated part vanishes and we obtain 1

1 ⎡ 7 ⎤ 2 − ς (3) ⎥ 2 ⎢ 2 ⎣ ⎦

∫ log Γ( x)[ x(1 − x)π sin π x + (2 x − 1) cos π x] dx = π 0

Unfortunately, if we substitute the relevant component integrals (which are listed in Appendix D) we find that the interesting summations cancel out. Some functions which satisfy the condition f ( x) = − f (1 − x) are listed below: B2 n +1 ( x) , B2 n ( x) cos(2m + 1)π x , B2 n +1 ( x) sin 2mπ x and h ( x(1 − x) ) cos(2m + 1)π x since Bn ( x) = (−1) n Bn (1 − x) where n and m are positive integers. Similarly, if g ( x) = g (1 − x) , we obtain 1

1

0

0

∫ g ( x)[ψ ( x) −ψ (1 − x)] dx = ∫ g ( x) cot π x dx = 0 1

1

I = ∫ f ( x)ψ ( x)dx = f ( x) log Γ( x) 0 − ∫ f ′( x) log Γ( x) dx 0

1

0

We have for suitably behaved functions

92

1

∫ 0

1

f ( x)ψ ( x)dx = − ∫ f ′( x) log Γ( x) dx 0

and an example of this is 1

∫B

2 n +1

0

1

( x)ψ ( x) dx = B2 n +1 ( x) log Γ( x) 0 − (2n + 1) ∫ B2 n ( x) log Γ( x) dx 1

0

1

= −(2n + 1) ∫ B2 n ( x) log Γ( x) dx 0

Glasser’s formula gives us 1

(2.49)

∫B

2 n +1

( x)ψ ( x) dx = −

0

π

1

2 ∫0

B2 n +1 ( x) cot π x dx

and hence we have 1

(2.50)

∫B

2n

( x) log Γ( x) dx =

0

π

1

2(2n + 1) ∫0

B2 n +1 ( x) cot π x dx

The following integral appears in Abramowitz and Stegun [1, p.807] 1

(2.51)

∫ B2n+1 ( x) cot π x dx = 0

(−1) n +1 2(2n + 1)!ς (2n + 1) (2π ) 2 n +1

and there are many derivations; for example, see the recent one by Dwilewicz and Mináč [28]. A similar identity was also derived by Espinosa and Moll [30] in the form 1

(2.52)

∫B

2n

( x) log sin π x dx = (−1) n

0

(2n)!ς (2n + 1) (2π ) 2 n

and it is easily shown that equation (2.52) above is equivalent to (2.51) following a simple integration by parts. Hence we obtain 1

(2.53)

∫ B2n ( x) log Γ( x) dx = 0

(−1) n +1 (2n)!ς (2n + 1) 2(2π ) 2 n +1

93

in agreement with [30]. □ Some of the above analysis could be replicated by employing the well-known relation 1⎞ ⎛1 π ⎛ ⎞ Γ⎜ x + ⎟Γ⎜ − x⎟ = 2⎠ ⎝2 ⎝ ⎠ cos π x which results from Euler’s reflection formula with the substitution x → x + 1/ 2 . □ Using the Maclaurin series for cosπ x we have 1

∞

∫ log Γ( x) cosπ x dx = ∑ (−1)

n

n =0

0

π 2n

1

x (2n)!∫

2n

log Γ( x) dx

0

From [30] we have the inversion formula for the Bernoulli polynomials xj =

1 j ⎛ j + 1⎞ ∑ ⎜ ⎟ Bk ( x) j + 1 k =0 ⎝ k ⎠

1 1 2 n ⎛ 2n + 1 ⎞ ⎜ ⎟ Bk ( x) log Γ( x) dx ∫0 x log Γ( x) dx = 2n + 1 ∑ k ⎠ ∫0 k =0 ⎝ 1

2n

We have from equation (6.10) of [30] 1

∫ B ( x) log Γ( x) dx = H ς (−k ) + (−1) k

k

k +1

ς ′(−k )

0

and thus we have 1

∞

∫ log Γ( x) cosπ x dx = ∑ (−1) 0

n =0

n

π 2n

⎛ 2n + 1 ⎞ k +1 ⎟ ⎡⎣ H k ς (− k ) + (−1) ς ′(−k ) ⎤⎦ k ⎠ k =0 ⎝ 2n

∑⎜ (2n + 1)!

We note that via integration by parts this links up with equation (12.9) of [30] 2 n 2n + 1 ⎛ ⎞ 2 n +1 ⎡⎣ H k ς (− k ) + (−1) k +1ς ′(− k ) ⎤⎦ ( ) = x ψ x dx ∑ ⎜ ⎟ ∫0 k ⎠ k =0 ⎝ 1

94

3. Some applications of the Fourier series for the Hurwitz zeta function

In Titchmarsh’s treatise [54, p.37] we have Hurwitz’s formula for the Fourier series of the Hurwitz zeta function ς ( s, x) (3.1)

⎡

⎛ π s ⎞ ∞ cos 2kπ x ⎛ π s ⎞ ∞ sin 2kπ x ⎤ cos + ⎟∑ ⎜ ⎟∑ 1− s 1− s ⎥ ⎝ 2 ⎠ k =1 (2π k ) ⎝ 2 ⎠ n =1 (2π k ) ⎦

ς ( s, x) = 2Γ(1 − s) ⎢sin ⎜ ⎣

where Re ( s) < 0 and 0 < x ≤ 1 . In 2000, Boudjelkha [14] showed that this formula also applies in the region Re ( s ) < 1. It may be noted that when x = 1 this reduces to Riemann’s functional equation for ς ( s ) . From the above are immediately found the corresponding Fourier coefficients 1

(3.2)

Γ(1 − s )

∫ ς (s, x) sin 2nπ x dx = (2π n)

1− s

0

1

(3.3)

Γ(1 − s )

∫ ς (s, x) cos 2nπ x dx = (2π n)

1− s

0

⎛πs ⎞ cos ⎜ ⎟ ⎝ 2 ⎠ ⎛πs ⎞ sin ⎜ ⎟ ⎝ 2 ⎠

Using Euler’s reflection formula (2.6) these may be written as (see [30]) (3.4)

(2π ) s n s −1 ⎛πs ⎞ ∫0 ς (s, x) sin 2nπ x dx = 4Γ(s) csc ⎜⎝ 2 ⎟⎠

(3.5)

(2π ) s n s −1 ⎛πs ⎞ ∫0 ς (s, x) cos 2nπ x dx = 4Γ(s) sec ⎜⎝ 2 ⎟⎠

1

1

Having regard to (3.2) we consider f ( s) =

Γ(1 − s ) ⎛πs ⎞ cos ⎜ ⎟ 1− s (2π n) ⎝ 2 ⎠

and logarithmic differentiation gives us f ′( s ) π ⎛πs ⎞ = log(2π n) −ψ (1 − s ) − tan ⎜ ⎟ f (s) 2 ⎝ 2 ⎠ Therefore differentiating (3.2) results in (this idea was inspired by a paper by Espinosa and Moll [30])

95

1

π

⎡

⎛ π s ⎞ ⎤ Γ(1 − s ) ⎛πs ⎞ cos ⎜ ⎟ ⎟⎥ 1− s 2 ⎠ ⎦ (2π n) ⎝ 2 ⎠

∫ ς ′(s, x) sin 2nπ x dx = ⎢⎣log(2π n) −ψ (1 − s) − 2 tan ⎜⎝

(3.6)

0

Applying Lerch’s identity [10] 1 2

ς ′(0, x) = log Γ( x) − log(2π )

(3.7)

we then have as s → 0 1

⎡

1

⎤

∫ ⎢⎣log Γ( x) − 2 log(2π ) ⎥⎦ sin 2nπ x dx = 0

log(2π n) + γ 2π n

and this becomes the well-known result which we derived in a different manner in (1.9.2) 1

(3.8)

∫ log Γ( x) sin 2nπ x dx = 0

log(2π n) + γ 2π n

Similarly differentiating (3.3) gives us 1

(3.9)

π

⎡

⎛ π s ⎞ ⎤ Γ(1 − s ) ⎛πs ⎞ sin ⎜ ⎟ ⎟⎥ 1− s 2 ⎠ ⎦ (2π n) ⎝ 2 ⎠

∫ ς ′(s, x) cos 2nπ x dx = ⎢⎣log(2π n) −ψ (1 − s) + 2 cot ⎜⎝ 0

and as s → 0 we obtain 1

⎡

1

⎤

1

∫ ⎢⎣log Γ( x) − 2 log(2π ) ⎥⎦ cos 2nπ x dx = 4n 0

This then gives us (1.9) 1

(3.10)

1

∫ log Γ( x) cos 2nπ x dx = 4n 0

We have thereby obtained the Fourier coefficients for log Γ( x ) and, in the process, have thus rediscovered Kummer’s Fourier series for log Γ( x) (3.11)

log Γ( x) =

∞ 1 1 (γ + log 2π n) sin 2π nx (0 < x < 1) log π − log sin π x + ∑ 2 2 πn n =1

(this formula was originally derived by Kummer in 1847 [42]). Reference to (2.7) confirms that (3.11) is properly described as a Fourier series expansion for log Γ( x) . 96

Kummer’s formula may also be written as (cf. Nielsen [45, p.201]) log Γ( x) =

(3.12)

1 π 1 ∞ log n ⎛1 ⎞ + ⎜ − x ⎟ (γ + log 2π ) + ∑ log sin 2π nx π n =1 n 2 sin π x ⎝ 2 ⎠

Nörlund [46, p.118] reports that for 0 < x < 1 (3.13) 1 ⎛x⎞ ⎛ 1+ x ⎞ 1 ⎛ π x ⎞ 2 ∞ log(2n + 1) γ π log Γ ⎜ ⎟ − log Γ ⎜ = ( + log 2 ) + log cot sin(2n + 1)π x ⎟ ⎜ ⎟+ ∑ 2 ⎝2⎠ ⎝ 2 ⎠ 2 ⎝ 2 ⎠ π n = 0 2n + 1 (this was also derived by Williams and Zhang [57] using Kummer’s formula (3.12)). Integrating this results in 1 1 4 ⎛ x⎞ ⎛ 1+ x ⎞ ⎛πx ⎞ ∫0 log Γ ⎜⎝ 2 ⎟⎠ dx − ∫0 log Γ ⎜⎝ 2 ⎟⎠ dx = 2 (γ + log 2π ) + 2 ∫0 log cot ⎜⎝ 2 ⎟⎠ dx + π 2 1

1

1

We see that ⎛πx⎞ ⎛πx⎞ ⎛πx ⎞ ∫0 log cot ⎜⎝ 2 ⎟⎠ dx = ∫0 log cos ⎜⎝ 2 ⎟⎠ dx − ∫0 log sin ⎜⎝ 2 ⎟⎠ dx 1

1

1

and we have the integrals [33, p.526] 2 ⎛πx ⎞ ∫0 log cos ⎜⎝ 2 ⎟⎠ dx = π 1

2 ⎛πx ⎞ ∫0 log sin ⎜⎝ 2 ⎟⎠ dx = π 1

π

2

∫ log cos u du = − log 2 0

π

2

∫ log sin u du = − log 2 0

We have 1

1

2

⎛x⎞ ∫0 log Γ ⎜⎝ 2 ⎟⎠ dx = 2 ∫0 log Γ(u) du and using (1.64) this becomes

97

log(2n + 1) 2 n = 0 (2n + 1) ∞

∑

1

⎛x⎞

5

1

∫ log Γ ⎜⎝ 2 ⎟⎠ dx = 12 log 2 + 2 log π + 3log A 0

Hence we obtain 1

(3.13.1)

1 1 4 ⎛ 1+ x ⎞ ∫0 log Γ ⎜⎝ 2 ⎟⎠ dx = 3log A − 12 log 2 − 2 γ − π 2

log(2n + 1) 2 n = 0 (2 n + 1) ∞

∑

We have for Re (s ) > 1 ∞

1 s n = 0 (2n + 1)

(1 − 2− s )ς ( s ) = ∑

and differentiation results in log(2n + 1) s n = 0 (2 n + 1) ∞

(1 − 2− s )ς ′( s ) + 2− s ς ( s ) log 2 = −∑ We may therefore write (3.13.1) as 1

(3.14)

1 1 1 ⎛ 1+ x ⎞ ⎟ dx = 3log A − log 2 − γ − 2 [3ς ′(2) + ς (2) log 2] 2 ⎠ 12 2 π

∫ log Γ ⎜⎝ 0

Another proof of Hurwitz’s formula for the Fourier series expansion of the Hurwitz zeta function

The following proof of Hurwitz’s formula for the Fourier series expansion of the Hurwitz zeta function is based on Kummer’s method which is outlined in [5, p.29]. Since log Γ( x) is differentiable in (0,1) it has a Fourier expansion ∞

∞

k =1

k =1

log Γ( x) = C0 + 2∑ Ck cos 2kπ x + 2∑ Dk sin 2kπ x 1

1

0

0

where Ck = ∫ log Γ( x) cos 2kπ x dx and Dk = ∫ log Γ( x) sin 2kπ x dx The Ck coefficients are relatively easy to determine using Euler’s reflection formula for the gamma function log Γ( x) + log Γ(1 − x) = log 2π − log [ 2sin π x ]

98

cos 2nπ x n n =1 ∞

= log 2π + ∑ The corresponding Fourier series is ∞

log Γ( x) + log Γ(1 − x) = 2C0 + 4∑ Ck cos 2kπ x k =1

and equating coefficients in the last two equations gives C0 =

(3.15)

1 1 for k ≥ 1 log 2π and Ck = 4k 2

We now refer to Malmstén’s formula [52, p.16] (which is also derived in equation (E.22g) of [22]) ∞ ⎡ e −α − e −α x ⎤ dα log Γ( x) = ∫ ⎢( x − 1)e −α − ⎥ 1 − e −α ⎦ α 0 ⎣

(3.16)

and a change of variables u = e −α gives us 1

log Γ( x) = ∫

(3.17)

0

⎡1 − u x −1 ⎤ du − x + 1⎥ ⎢ ⎣ 1− u ⎦ log u

We therefore obtain 1

Dk = ∫ log Γ( x) sin 2kπ x dx 0

1 1 ⎡1 − u x −1 ⎤ sin 2kπ x dudx = ∫∫ ⎢ − x + 1⎥ 1− u log u ⎦ 0 0 ⎣

We have 1

∫ sin 2kπ x dx = 0 0

1

1

∫ x sin 2kπ x dx = − 2kπ 0

99

1

x −1 ∫ u sin 2kπ x dx = 0

=

1

1 Im ∫ exp [ x(log u + 2kπ i ) ] dx u 0

1 u −1 Im u log u + 2kπ i

Therefore we get

⎡ 1 ⎤ du −2 k π ⎥ Dk = ∫ ⎢ + 2 2 2 2 π k log 4 π u u k + ⎢ ⎥⎦ log u 0 ⎣ 1

(3.18)

(

)

With u = e −2kπ t we obtain Dk =

1 2π k

∞

⎡ 1 ⎤ −2 kπ t dt ⎢ ⎥ e − ⎢⎣ l+ t 2 ⎥⎦ t

∫ ( 0

)

Taking k = 1 we have 1 D1 = 2π

∞

⎡ 1 ⎤ −2π t dt ⎢ ⎥ e − ⎢⎣ l+ t 2 ⎥⎦ t

∫ ( 0

)

With x = 1 in Malmstén’s formula (3.16) we get

γ 1 − = 2π 2π

∞

⎡

∫ ⎢⎣e

−t

−

0

1 ⎤ dt 1 + t ⎥⎦ t

Hence we have (3.19)

∞ γ 1 e −t − e −2π t 1 = D1 − dt + ∫ 2π 2π 0 2π t

∞

⎡ 1 1 ⎤ dt ⎢ ⎥ − ( l+ t ) ⎥⎦ t ⎢⎣ l+ t 2

∫ ( 0

∞

)

We have 1/ r = ∫ e − rx dx ( r > 0) and, integrating that expression, we obtain Frullani’s 0

integral b

n

∞

∞

b

dr − rx − rx ∫1 r = ∫1 dr ∫0 e dx = ∫0 dx ∫1 e dr

100

which implies that ∞

e − x − e −bx dx x 0

log b = ∫

(3.20)

By (3.20) the first integral in (3.19) is equal to log 2π and a change of variables t → 1/ t shows that the second integral vanishes. Therefore we have D1 =

γ 1 + log 2π 2π 2π

Dk is found by noting that kDk − D1 =

∞

e −2π t − e −2 kπ t 1 dt = log k ∫0 t 2π

1 2π

where the integral is also evaluated with (3.20). Thus we have

γ + log 2kπ 1 Dk = = ∫ log Γ( x) sin 2kπ x dx 2π k 0

(3.21)

and Kummer’s formula thereby follows. □ We see from (3.18) and (3.21) that ⎡ 1 1 ∫0 ⎢⎢ (2π k )2 − u log 2 u + 4k 2π 2 ⎣ 1

(

)

⎤ du γ + log(2kπ ) ⎥ = (2π k ) 2 ⎥⎦ log u

and we make the summation ⎡ 1 1 ⎢ − ∑ 2 ∫ u log 2 u + 4k 2π 2 k =1 0 ⎢ (2π k ) ⎣ ∞ 1

(

)

∞ ⎤ du γ + log(2kπ ) ⎥ =∑ 2 ⎥⎦ log u k =1 (2π k )

Assuming that interchanging of the order of integration and summation is valid we have ∞ ⎡ ς (2) 1 ∞ ⎤ du 1 γ + log(2kπ ) − = ∑ 2 2 2⎥ ∫0 ⎢⎣ (2π )2 u ∑ (2π k ) 2 k =1 log u + 4 k π ⎦ log u k =1

1

101

Letting α → iα in (4.7) gives us

π coth απ =

1

α

∞

1 2 k =1 k + α

+ 2α ∑

2

and we obtain 1 log u coth[(log u ) / 2] − 1 1 1 1 2 = ∑ (2π ) 2 k =1 k 2 + (log u / 2π ) 2 4π log u ∞

=

log u coth[(log u ) / 2] − 2 8π log u

This gives us 1 ⎡ ς (2) 1 ∞ ⎤ du ⎡ ς (2) log u coth[(log u ) / 2] − 2 ⎤ du 1 − = ⎢ (2π ) 2 − ⎥ log u 2 2 2⎥ ∫0 ⎢⎣ (2π )2 u ∑ ∫ 8π u log u k =1 log u + 4k π ⎦ log u ⎦ 0 ⎣ 1

and thus ∞ ⎡ ς (2) log u coth[(log u ) / 2] − 2 ⎤ du γ + log(2kπ ) − = ∑ ⎥ 2 ∫0 ⎢⎣ (2π )2 8π u log u ⎦ log u k =1 (2π k ) 1

Since coth[(log u ) / 2] =

u +1 we may write this as u −1

1 ∞ 1 ⎡π 1 1+ u 1 ⎤ du γ + log(2kπ ) + + = ∑ ⎢ ⎥ ∫ 4π 0 ⎣ 6 2 u (1 − u ) u log u ⎦ log u k =1 (2π k ) 2

It may be noted that this bears a structural similarity to the integral reported in Gradshteyn and Ryzhik [33, p.550, 4.283 6] 1

⎡ 1 1+ u

1

⎤ du

1

∫ ⎢⎣ 2 1 − u + log u − log u ⎥⎦ log u = 2 log(2π )

(3.23)

0

□ We also have Dk =

∞

e −2π t − e −2 kπ t γ + log(2π k ) dt = ∫ 2π k 0 t 2π k 1

102

and we have the summation ∞

Dk 1 = ∑ 2π k =1 k

∞

1 e −2π t − e −2 kπ t 1 dt = ∑ 2∫ t 2π k =1 k 0 ∞

∞

∑

γ + log(2kπ ) k2

k =1

We see that ∞

∞

1 e −2π t − e −2 kπ t ς (2)e−2π t − Li2 (e −2π t ) dt = dt ∑ 2∫ ∫0 t t k =1 k 0 ∞

in terms of the dilogarithm function. With the substitution x = e −2π t the integral becomes 1

= 4π 2 ∫ [ Li2 ( x) − ς (2) x] 0

x dx log x

so that we have 1

(3.24)

∫ [ Li2 ( x) − ς (2) x] 0

∞ x γ + log(2kπ ) dx = ∑ log x (2kπ ) 2 k =1

□ Kummer’s Fourier series expansion (3.12) may also be written as 1 1 sin π x ∞ (γ + log 2π n) sin 2π nx (0 < x < 1) log Γ( x) = − log x − log +∑ πx πn 2 2 n =1 and hence we have 1 Γ( x) log = 2 Γ(1 − x) 1

(γ + log 2π n) sin 2π nx πn n =1 ∞

∑

We have Dk = ∫ log Γ( x) sin 2kπ x dx = 0

1 2π k

(γ + log 2kπ )

∞

∞ Dk 1 = γ + log 2kπ ) ∑ ∑ 2 ( k =1 k k =1 2π k

=

(γ + log 2π ) ς (2) − 2π

1 ς ′(2) 2π

103

for k ≥ 1 and hence we get

We also have ∞ ∞ Dk 1 sin 2kπ x log ( ) sin 2 π log ( ) = Γ x k x dx = Γ x dx ∑ ∑ ∑ ∫ ∫ k k =1 k k =1 k 0 k =1 0 1

∞

1

Now using the Fourier series which is valid for x ∈ (0,1) sin 2kπ x π = (1 − 2 x) k 2 k =1 ∞

∑ we have

Dk π = ∫ log Γ( x)(1 − 2 x) dx ∑ 20 k =1 k 1

∞

1

Using ∫ log Γ( x) dx = 0

π 4

1 log 2π we obtain 2 1

log 2π − π ∫ x log Γ( x) dx =

(γ + log 2π ) ς (2) 2π

0

−

1 ς ′(2) 2π

Therefore we obtain 1

γ

1

1

∫ x log Γ( x) dx = 6 log 2π − 12 + 2π

(3.25)

2

ς ′(2)

0

which we also derived in (2.38). ∞ Dk D and (−1) k pk we may easily obtain further ∑ ∑ p k k =1 k =1 k identities using the Fourier series reported, for example, in [55, p.148].

With reference to series of the form

1

∞

We have Ck = ∫ log Γ( x) cos 2kπ x dx = 0

∞

∑ (−1) k =1

k +1

1 for k ≥ 1 and hence we get 4k

Ck 1 = ς a (2) k 4

We also have

104

∞

∑ (−1)k +1 k =1

∞ ∞ Ck 1 cos 2kπ x = ∑ ∫ log Γ( x)(−1) k +1 cos 2kπ x dx = ∫ log Γ( x)∑ (−1) k +1 dx k k k =1 k 0 k =1 0 1

1

Now using [55, p.148] we have ∞

∑ (−1)

k +1

k =1

cos 2kπ x = log[2 cos π x] k

and accordingly ∞

∑ (−1) k =1

k +1

1

Ck 1 = ∫ log Γ( x) log[2 cos π x] dx = ς a (2) k 4 0

We therefore have 1 π2 ∫0 log Γ( x) log cos π x dx = − 2 log 2 log 2π + 48 1

(3.26)

It is noted in [30, Eq. (7.4)] that 1 π2 log ( x ) log sin x dx log 2 log 2 Γ π = − π − ∫0 2 24 1

(3.27)

which was derived in a different manner in [26]. Addition of the two integrals results in 1 π2 log ( x ) log sin 2 x dx log 2 log 2 Γ π = − π − ∫0 2 48 1

(3.28)

From equation (6.92a) of [21] we have 1

∫ log Γ( x + 1) log [ 2sin(π x)] dx = 0

1 2π

si (2nπ ) n2 n =1 ∞

∑

and we see that 1

1

0

0

∫ log Γ( x + 1) log [ 2sin(π x)] dx = ∫ [log x + log Γ( x)][log 2 + log sin(π x)] dx

105

1

1

1

1

0

0

0

0

= log 2 ∫ log x dx + ∫ log x log sin(π x)dx + log 2 ∫ log Γ( x) dx + ∫ log Γ( x) log sin(π x) dx 1 1 π2 = − log 2 + ∫ log x log sin(π x) dx + log 2 log 2π − log 2 log 2π − 2 2 24 0 Hence we have 1

1

∫ log x log sin(π x)dx =

(3.29)

0

1 2π

si (2nπ ) π2 log 2 + + ∑ n2 24 n =1 ∞

si (2nπ ) may be determined from (6.117j) in [21] n2 n =1 ∞

and

∑

Si (2nπ ) 1 = log A − 2 2π n =1 n 4 ∞

1

2

∑

so that we have

π

1

∫ log x log sin(π x)dx = π log(2 A) − 4

(3.29.1)

0

□ Using Euler’s reflection formula we see that 1

1

0

0

∫ log Γ( x) log sin π x dx + ∫ log Γ(1 − x) log sin π x dx 1

1

0

0

= log π ∫ log sin π x dx − ∫ log 2 sin π x dx

and we have 1

1

0

0

∫ log Γ(1 − x) log sin π x dx = ∫ log Γ(u ) log sin π u du Therefore using (3.27) we obtain 1

(3.30)

2 ∫ log sin π x dx = log 2 log 2π + 0

π2 12

− log 2 log π = log 2 2 +

106

π2 12

which is derived in a different manner in (6.3). □ In a similar manner we multiply by log Γ( x ) and integrate to obtain 1

∫ log

1

2

0

Γ( x) dx + ∫ log Γ(1 − x) log Γ( x) dx 0

1

1

0

0

= log π ∫ log Γ( x) dx − ∫ log Γ( x) log sin π x dx

Espinosa and Moll [30] proved that 1

(3.32)

∫ log

2

Γ( x) dx =

0

γ2 12

+

1 1 ς ′(2) ς ′′(2) + γ log(2π ) + log 2 (2π ) − [γ + log(2π )] 2 + 48 6 3 2π 2 π

π2

and a different derivation is given in [26].Therefore we obtain 1 1 π2 ∫0 log Γ(1 − x) log Γ( x) dx = 2 log π log(2π ) + 2 log 2 log 2π + 24 1

⎡γ 2 π 2 1 1 ς ′(2) ς ′′(2) ⎤ −⎢ + + γ log(2π ) + log 2 (2π ) − [γ + log(2π )] 2 + ⎥ 3 π 2π 2 ⎦ ⎣ 12 48 6

which simplifies to (3.33) ⎡γ 2 π 2 1 1 ς ′(2) ς ′′(2) ⎤ log (1 x ) log ( x ) dx Γ − Γ = − + γ log(2π ) − log 2 (2π ) − [γ + log(2π )] 2 + ⎢ − ⎥ ∫0 6 π 2π 2 ⎦ ⎣ 12 48 6 1

1

We may also use Parseval’s theorem to evaluate ∫ log 2 Γ( x) dx in a very elementary 0

manner. Referring to (3.8) and (3.10) we see that 1

2 ∫ log Γ( x) dx =

0

=

1 ∞ ⎛ 1 1 [γ + log(2kπ )]2 ⎞ Γ + + log ( x ) dx 2 ⎜ ⎟ ∑ 2 2 ∫0 4π 2 k 2 k =1 ⎝ 16k ⎠

∞ ⎛ 1 1 γ 2 + log 2 (2π ) + 2γ log(2π ) + 2[γ + log(2π )]log k + log 2 k ⎞ + log(2π ) + 2∑ ⎜ ⎟ 2 4 4π 2 k 2 k =1 ⎝ 16k ⎠

107

=

1 1 1 ς ′(2) ς ′′(2) log(2π ) + ς (2) + [γ + log(2π )]2 − [γ + log(2π )] 2 + π 4 8 12 2π 2

in agreement with (3.32) above. Amdeberhan et al. [4] have recently given further consideration to integrals of the 1

form ∫ log n Γ( x) dx . 0

□ Berndt [10] used Kummer’s formula (3.12) to derive Euler’s reflection formula in the following elementary manner. We have log Γ( x) =

∞ 1 1 (γ + log 2π n) sin 2π nx (0 < x < 1) log π − log sin π x + ∑ 2 2 πn n =1

Letting x → 1 − x we get log Γ(1 − x) =

=

∞ 1 1 (γ + log 2π n) sin 2π n(1 − x) log π − log sin π (1 − x) + ∑ 2 2 πn n =1 ∞ 1 1 (γ + log 2π n) sin 2π nx log π − log sin π x − ∑ 2 2 πn n =1

and hence log Γ( x) + log Γ(1 − x) = log

π sin π x

We then obtain Euler’s reflection formula Γ( x)Γ(1 − x) =

π . sin π x

Alternatively, we may subtract the above two equations to obtain the following formula which appears in Ramanujan’s Notebooks [11, Vol.1, p.199] log

Γ( x) 2 ∞ log n + [γ + log(2π )](2 x − 1) = ∑ sin 2π nx π n =1 n Γ(1 − x) sin 2π nx 1 = − x which is valid for 0 < x < 1. 2 nπ n =1 ∞

where we have used the Fourier series

∑

Integration then results in

108

u

u

0

0

2 ∫ log Γ( x)dx − ∫ log Γ(1 − x)dx + [γ + log(2π )](u − u )

=

1

∞

log n 1 − 2 ∑ π n =1 n 2 π 2

∞

log n cos 2π nu 2 n =1 n

∑

We have u

−u

0

0

∫ log Γ(1 − x)dx = − ∫ log Γ(1 + x)dx and we note Alexeiewsky’s theorem [52, p.32] (a further derivation of which is contained in (1.115) and also in equation (4.3.85) of [20]) u

∫ log Γ (1 + x ) dx = 0

1 u2 [log(2π ) − 1] u − + u log Γ (1 + u ) − log G (1 + u ) 2 2

in terms of the Barnes double gamma function G (u ) (see Appendix C). Hence we have u

u

u

−u

0

0

0

0

∫ log Γ( x)dx − ∫ log Γ(1 − x)dx = ∫ log Γ( x)dx + ∫ log Γ(1 + x)dx u

−u

u

0

0

0

= ∫ log Γ(1 + x)dx + ∫ log Γ(1 + x)dx − ∫ log x dx

We see that u

−u

0

0

∫ log Γ(1 + x)dx + ∫ log Γ(1 + x)dx = −u 2 + u log Γ (1 + u ) − u log Γ (1 − u ) − log G (1 + u ) − log G (1 − u )

and with u = 1/ 2 we get 1

2

−1

2

∫ log Γ(1 + x)dx + ∫ log Γ(1 + x)dx 0

0

109

1 1 1 = − + log Γ ( 3 / 2 ) − log Γ (1/ 2 ) − log G (3 / 2) − log G (1/ 2) 4 2 2 1 1 = − − log 2 − log Γ (1/ 2 ) − 2 log G (1/ 2) 4 2 Therefore we obtain 1 1 1 1 − − log 2 − log Γ (1/ 2 ) − 2 log G (1/ 2) − [γ + log(2π )] = 2 π 4 2 4

=−

(

∞

log n 1 − 2 ∑ 2 π n =1 n

1

π2

∞

∑ (−1) n =1

n

log n n2

[ς ′(2) + ς a′ (2)]

)

Since ς a′ ( s ) = 1 − 21− s ς ′( s ) + 21− s log 2.ς ( s ) we have 1 2

1 2

ς a′ (2) = ς ′(2) + log 2.ς (2) and this gives us 1 1 1 1 − − log 2 − log Γ (1/ 2 ) − 2 log G (1/ 2) − [γ + log(2π )] = − 2 π 4 2 4 =− The derivative of the Riemann functional equation gives us log A −

1 1 [γ + log(2π )] = − 2 ς ′(2) 12 2π

and we then obtain (3.34)

3 1 1 1 log G (1/ 2) = − log A − log π + + log 2 2 4 8 24

as originally determined by Barnes [9] in 1899.

110

1 ⎡3 ′ ⎤ ⎢⎣ 2 ς (2) + 2 log 2.ς (2) ⎥⎦

3 2π

2

ς ′(2) −

1 log 2 12

In the following two sections we generalise the analysis to address integrals of the form 1

∫ log Γ( x) e

iπ px

dx .

0

1

4. The ∫ log Γ( x) cos pπ x dx integral 0

1

In this part we first of all consider the integral ∫ log Γ( x) cos pπ x dx ; the corresponding 0

integral with sin pπ x in the integrand is considered later in Section 5 of this paper. Provided a ≠ b we readily determine that 1 ⎡ cos(a + b) x cos(a − b) x ⎤ + +c a+b a − b ⎥⎦

∫ sin ax cos bx dx = − 2 ⎢⎣ and hence we have 1

1⎡

1

1

⎤ 1 ⎡ cos(2n + p )π cos(2n − p )π ⎤ + (2n + p )π (2n − p )π ⎥⎦

∫ sin 2π nx cos pπ x dx = 2 ⎣⎢ (2n + p)π + (2n − p)π ⎦⎥ − 2 ⎣⎢ 0

=

(4.1)

2n 1 − cos pπ π 4n 2 − p 2

As before, provided a ≠ b we readily determine that 1 ⎡ sin(a + b) x sin(a − b) x ⎤ + +c a+b a − b ⎥⎦

∫ cos ax cos bx dx = 2 ⎢⎣ and hence we have 1

1 ⎡ sin(2n + p )π sin(2n − p )π ⎤ + (2n + p)π (2n − p)π ⎥⎦

∫ cos 2π nx cos pπ x dx = 2 ⎢⎣ 0

(4.2)

=−

p sin pπ π 4n 2 − p 2

We now multiply Hurwitz’s identity (3.1) by cos pπ x and integrate to obtain

111

1

∫ ς (s, x) cos pπ x dx

(4.3)

0

⎡ 1 p sin pπ 1 2n 1 − cos pπ ⎤ ⎛πs ⎞ ∞ ⎛πs ⎞ ∞ cos = 2Γ(1 − s ) ⎢ − sin ⎜ ⎟ ∑ + ⎜ ⎟∑ 1− s 2 2 1− s 2 2 ⎥ ⎝ 2 ⎠ n =1 (2π n) π 4n − p ⎝ 2 ⎠ n =1 (2π n) π 4n − p ⎦ ⎣ Differentiation with respect to s gives us 1

∫ ς ′(s, x) cos pπ x dx

(4.4)

0

⎡ 1 p sin pπ ⎤ ⎛ π s ⎞ ∞ log(2π n) p sin pπ π ⎛πs ⎞ ∞ cos = 2Γ(1 − s ) ⎢ − sin ⎜ ⎟ ∑ − ⎜ ⎟∑ 1− s 2 2 1− s 2 2 ⎥ 2 ⎝ 2 ⎠ n =1 (2π n) π 4n − p ⎝ 2 ⎠ n =1 (2π n) π 4n − p ⎦ ⎣ ⎡ ⎛ π s ⎞ ∞ log(2π n) 2n 1 − cos pπ π 1 2n 1 − cos pπ ⎤ ⎛πs ⎞ ∞ +2Γ(1 − s ) ⎢ cos ⎜ ⎟ ∑ − sin ⎜ ⎟ ∑ 1− s 2 2 1− s 2 2 ⎥ π 4n − p 2 ⎝ 2 ⎠ n =1 (2π n) π 4n − p ⎦ ⎣ ⎝ 2 ⎠ n =1 (2π n) ⎡ 1 p sin pπ 1 2n 1 − cos pπ ⎤ ⎛πs ⎞ ∞ ⎛πs ⎞ ∞ cos −2Γ′(1 − s ) ⎢ − sin ⎜ ⎟ ∑ + ⎜ ⎟∑ 1− s 2 2 1− s 2 2 ⎥ ⎝ 2 ⎠ n =1 (2π n) π 4n − p ⎝ 2 ⎠ n =1 (2π n) π 4n − p ⎦ ⎣ and with s = 0 we have using Lerch’s identity (1.46.1) 1

(4.5)

∫ log Γ( x) cos pπ x dx 0

=

1 sin pπ p sin pπ log(2π ) − 2 pπ 2π

1 1 2(1 − cos pπ ) ∞ γ + log(2π n) + ∑ ∑ 2 2 π2 4n 2 − p 2 n =1 n 4n − p n =1 ∞

and, as expected, with p = 0 we recover Raabe’s integral (2.11). Using (2.15) and (4.7) this may be expressed as 1

(4.5.1)

∫ log Γ( x) cos pπ x dx 0

=

1 sin pπ sin pπ + log(2π ) 2 pπ 4 pπ

⎡ ⎛ ⎤ p⎞ p⎞ ⎛ ⎢ψ ⎜1 + 2 ⎟ +ψ ⎜1 − 2 ⎟ + 2γ ⎥ ⎠ ⎝ ⎠ ⎣ ⎝ ⎦

112

+

2(1 − cos pπ )[γ + log(2π )] ⎡ 1 π ⎛ pπ ⎞ ⎤ 2(1 − cos pπ ) ∞ log n cot − ∑ ⎜ ⎟⎥ + ⎢ 2 p2 4 p 2 2 π2 π2 ⎝ 2 ⎠⎦ n =1 4n − p ⎣

=

sin pπ [γ + log(2π )] sin pπ + 2 pπ 4 pπ

+

2(1 − cos pπ )[γ + log(2π )] ⎡ 1 π ⎛ pπ ⎞ ⎤ 2(1 − cos pπ ) ∞ log n cot − ∑ ⎜ ⎟⎥ + ⎢ 2 p2 4 p 2 2 π2 π2 ⎝ 2 ⎠⎦ n =1 4n − p ⎣

⎡ ⎛ p⎞ ⎛ p ⎞⎤ ⎢ψ ⎜ 2 ⎟ +ψ ⎜ − 2 ⎟ ⎥ ⎝ ⎠⎦ ⎣ ⎝ ⎠

and with p = 1 we have 1

∫ logΓ( x) cos π x dx =

(4.5.2)

0

∞ 2 ⎡ log n ⎤ log(2 ) 2 + + π γ ∑ 2 ⎢ 2 ⎥ π ⎣ n =1 4n − 1 ⎦

With p = 2k + 1 in (4.5) we obtain 1

∫ logΓ( x) cos(2k + 1)π x dx =

(4.6)

0

4

π

2

∞

γ + log(2π n)

∑ 4n n =1

2

− (2k + 1) 2

We have the well known identity [8, p.345] which is easily derived via Fourier series (in fact, a different derivation of this is obtained as a by-product in Section 6 below)

π cot απ =

(4.7)

1

α

∞

1 2 n =1 n − α

− 2α ∑

2

and with 2α = 2k + 1 we obtain [47, No. 5.1.25.4] ∞

∑ 4n

(4.8)

n =1

2

1 1 = 2 2(2k + 1) 2 − (2k + 1)

which then gives us 1

(4.9)

∫ logΓ( x) cos(2k + 1)π x dx = 0

∞ ⎤ 2 ⎡ log(2π ) + γ log n + 2 ∑ ⎢ 2 ⎥ 2 2 2 π ⎣ (2k + 1) n =1 4 n − (2 k + 1) ⎦

as originally determined by Kölbig [40] using Kummer’s Fourier series for log Γ( x) . As noted by Kölbig [40], this integral is incorrectly reported in Nielsen’s book [45, p.203]. The result was corrected in the 5th edition of Gradshteyn and Ryzhik [33, p.650, 6.443 4]. 113

p sin pπ With p = 2k in (4.5), and applying L’Hôpital’s rule to 2π obtain (2.28) again. 1

∞

1

∑ n 4n n =1

2

1 , we − p2

1

∫ log Γ( x) cos 2k π x dx = 4k 0

Letting k = 0 in (4.9) results in (4.5.2) again. □ Letting p = 1/ 2 in (4.5) results in 1

∫ log Γ( x) cos(π x / 2) dx = 0

1

π +

log(2π ) −

1

∞

1

n =1

8

π

2

∞

1

∑ n 16n π

2

−1

+

8 [γ + log(2π )]

π

2

∞

1

∑ 16n n =1

2

−1

log n 2 −1

∑ 16n n =1

With α = 1/ 4 in (4.7) we obtain ∞

1

∑ 16n n =1

−1

2

=

4 −π 8

and using (2.14) ∞

1

1

∑ n 16n n =1

2

−1

= 3log 2 − 2

we obtain 1

(4.10)

1

∫ log Γ( x) cos(π x / 2) dx = π log(2π ) −

3log 2 − 2 [γ + log(2π )](4 − π ) + 2

0

+

8

π

2

∞

π

π

log n 2 −1

∑ 16n n =1

□ Using integration by parts we have

114

1

log Γ( x) sin(2k + 1)π x 1 − ∫0 logΓ( x) cos(2k + 1)π x dx = ∫0 ψ ( x) sin(2k + 1)π x dx k (2k + 1)π (2 + 1) π 0 1

1

We see that

log Γ( x) sin(2k + 1)π x = [log Γ(1 + x) − log x]sin(2k + 1)π x = [ x log Γ(1 + x) − x log x]

sin(2k + 1)π x x

and it is therefore seen that the integrated part vanishes. Hence we have 1

(4.11)

1

1 ∫0 logΓ( x) cos(2k + 1)π x dx = − (2k + 1)π ∫0 ψ ( x) sin(2k + 1)π x dx

and, as pointed out by Kölbig [40], letting k = 0 we obtain 1

(4.12)

∫ψ ( x) sin π x dx = − 0

∞ 2⎡ log n ⎤ log(2 ) 2 + + π γ ∑ 2 ⎢ ⎥ π⎣ n =1 4n − 1 ⎦

Kölbig [40] states that the infinite series in (4.12) “does not seem to be expressible in terms of well-known functions”. Letting k = 0 in (4.11) gives us 1

∫ logΓ( x) cos π x dx = − 0

1

π

1

∫ψ ( x) sin π x dx 0

An alternative derivation of (4.12) is shown below. We recall Lerch’s trigonometric series expansion for the digamma function for 0 < x < 1 (see for example Gronwall’s paper [34, p.105] and Nielsen’s book [45, p.204]) (4.13)

ψ ( x) sin π x +

π 2

∞

cos π x + (γ + log 2π ) sin π x = −∑ sin(2n + 1)π x ⋅ log n =1

Integration of (4.13) gives us 1

(4.14)

∫ψ ( x) sin π x dx = − 0

∞ 2⎡ 1 n + 1⎤ log(2 ) log + + γ π ∑ ⎢ n ⎥⎦ π⎣ n =1 2n + 1

115

n +1 n

We have ∞

∞ log n 1 ⎤ ⎡ 1 = − log n ∑ 2 ⎢ 2n + 1 ⎥⎦ n =1 4n − 1 n =1 ⎣ 2n − 1

2∑

and consider the finite sum N

log(m + 1) 2m + 1 m =0 N −1

log n

∑ 2n − 1 = ∑ n =1

log(n + 1) log( N + 1) − 2n + 1 2N +1 n =1 N

=∑ Hence we have

N 1 ⎤ ⎡ 1 ⎡ log(n + 1) log n ⎤ log( N + 1) − log n = − − ∑ ∑ ⎢⎣ 2n − 1 2n + 1 ⎦⎥ ⎢ 2n + 1 ⎦⎥ 2N +1 n =1 n =1 ⎣ 2n + 1 N

Therefore, as N → ∞ we see that ∞ log n 1 n +1 = log ∑ 2 n n =1 4n − 1 n =1 2n + 1 ∞

2∑

(4.15)

and we have another proof of (4.12). □ In passing, letting x = 1/ 2 in (4.13) we obtain ∞

ψ (1/ 2) + γ + log(2π ) = ∑ (−1) n +1 log n =1

n +1 n

and, since [52, p.20] ψ (1/ 2) = −γ − 2 log 2 , we have ∞

∑ (−1) n =1

n +1

log

n +1 π = log n 2

as previously reported by Sondow [51]. □ Integration of (4.13) gives us 1

1 1 1 ∞ 1 n +1 log ∫0 ψ ( x) sin π x dx = 2 − π [γ + log(2π )] − π ∑ n n =1 2 n + 1 2

116

□ We may also write (4.13) as

ψ ( x) +

(4.16)

π 2

sin(2n + 1)π x n +1 ⋅ log n sin π x n =1 ∞

cot π x + (γ + log 2π ) = −∑

The above formula suggests that integration may be fruitful employing the following representation n sin(2n + 1)π x = 1 + 2∑ cos 2kπ x sin π x k =1

The integral

∫ x ψ ( x) may also produce interesting results. p

□ We multiply (4.13) by log Γ( x) and integrate to obtain

π

1

1

1

0

0

∫ log Γ( x)ψ ( x) sin π x dx + 2 ∫ log Γ( x) cos π x dx + (γ + log 2π ) ∫ log Γ( x) sin π x dx 0

∞ 1

= −∑ ∫ log Γ( x) sin(2n + 1)π x ⋅ log n =1 0

n +1 dx n

Integration by parts gives us 1

1 π 2 2 ∫0 log Γ( x)ψ ( x) sin π x dx = 2 log Γ( x) sin π x 0 − 2 ∫0 log Γ( x) cos π x dx 1

and with regard to the integrated part we see that lim log 2 Γ( x) sin π x = lim x log 2 Γ( x) x →0

x →0

sin π x x

We have log 2 Γ( x) x →0 1/ x

lim x log 2 Γ( x) = lim x →0

2 log Γ( x)ψ ( x) x →0 −1/ x 2

= lim

117

1

= −2 lim x log Γ( x) lim xψ ( x) = 0 x →0

x →0

and we thereby obtain

π

1

1

∫ log Γ( x)ψ ( x) sin π x dx = − 2 ∫ log 0

−

π 2

2

Γ( x) cos π x dx

0

1

2 ∫ log Γ( x) cos π x dx + 0

∞ 1⎡ log n ⎤ γ + log 2π log(2 π ) + γ + 2 ∑ 2 ⎢ ⎥+ π⎣ π n =1 4 n − 1 ⎦

⎡ π ⎤ ⎢ log 2 + 1⎥ ⎣ ⎦

n −1 ⎡ ⎛π ⎞ 1 1 1 ⎤ n +1 + 2∑ ⎢ log ⎜ ⎟ + ⎥ log n ⎝ 2 ⎠ 2n + 1 n =1 (2n + 1)π ⎣ j =0 2 j + 1 ⎦ ∞

= −∑

Hence we have (4.17)

π2

n +1 ⎡ π ⎤ ∞ 1 2 log Γ ( x ) cos π x dx = log(2 π ) + γ + log + 1⎥ ∑ log ∫ ⎢ 2 0 2 ⎦ n =1 2n + 1 n ⎣ 1

n −1 1 ⎤ n +1 ⎡ π ⎤ ∞ 1 ⎡ 1 + (γ + log 2π ) ⎢log + 1⎥ + ∑ + 2∑ ⎢ ⎥ log 2 ⎦ n =1 2n + 1 ⎣ 2n + 1 n ⎣ j =0 2 j + 1 ⎦

where we have used (5.6) and (5.7). □ With reference to (4.6) we make the summation 1

cos(2k + 1)π x 2[log(2π ) + γ ] ∞ 1 dx = ∑ 2 2 4 (2k + 1) π k =0 k = 0 (2k + 1) ∞

∫ logΓ( x)∑ 0

+

∞ 1 log n 2 ∑ 2∑ 2 π k =0 (2k + 1) n =1 4n − (2k + 1) 2

4

∞

where we have assumed that the interchange of the order of integration and summation is valid. We have the Fourier series [55, p.149] which is valid for 0 ≤ x ≤ 1 cos(2k + 1)π x π 2 = (1 − 2 x) (2k + 1) 2 8 k =0 ∞

∑

118

so that cos(2k + 1)π x π2 log Γ ( ) = (1 − 2 x) log Γ( x) dx x dx ∑ ∫0 (2k + 1) 2 8 ∫0 k =0 1

1

∞

Using (C.2) this becomes =

π 2 ⎡3

⎤ log 2π − log A⎥ ⎢ 8 ⎣4 ⎦

and hence we determine that ∞ 1 log n π 2 ⎡3 ⎤ (4.18) log(2π ) − log A⎥ − 48[log(2π ) + γ ]π 2 = 2 ∑ 2∑ 2 2 ⎢ 8 ⎣4 π k =0 (2k + 1) n =1 4n − (2k + 1) ⎦ ∞

4

Similarly, one could also consider the representation [55, p.149] cos(2k + 1)π x 1 πx = − log tan 2k + 1 2 2 k =0 ∞

∑

but it seems that this may give rise to convergence issues. □ Differentiating (4.5) gives us 1

(4.19)

−π ∫ x log Γ( x) sin pπ x dx 0

=

1 pπ cos pπ − sin pπ p sin pπ log(2π ) − 2 2π p 2π +

4 p(1 − cos pπ )

π

2

∞

∑ n =1

γ + log(2π n) (4n − p ) 2

2 2

+

1 2p pπ cos pπ + sin pπ − ∑ 2 2 2 2π n =1 n (4n − p ) ∞

2sin pπ

π

∞

∑

γ + log(2π n) 4n 2 − p 2

n =1

and with p = 1 we have 1

−π ∫ x log Γ( x) sin π x dx 0

1 1 ∞ 1 1 8 = − log(2π ) + ∑ + 2 2 2 2 n =1 n 4n − 1 π

119

∞

∑ n =1

γ + log(2π n) (4n 2 − 1) 2

∞

1

∑ n 4n n =1

2

1 − p2

1 1 ∞ 1 1 8[γ + log(2π )] ∞ 1 8 = − log(2π ) + ∑ + + 2 ∑ 2 2 2 2 2 2 n =1 n 4n − 1 π π n =1 (4n − 1)

∞

log n 2 − 1) 2

∑ (4n n =1

Differentiating (4.7) results in

π 2 cosec 2απ =

∞

1

α2

∞ 1 1 + 2 ∑ 2 2 2 2 2 n =1 ( n − α ) n =1 n − α

+ 4α 2 ∑

and with α = 1/ 2 we have ∞

∞ 1 1 + 8 ∑ 2 2 2 n =1 (4n − 1) n =1 4n − 1

π 2 = 4 + 16∑ ∞

1 +4 2 n =1 (4n − 1)

= 4 + 16∑

2

so that 1 π 2 −8 = ∑ 2 2 16 n =1 (4n − 1) ∞

as reported in [33, p.9]. Therefore using (2.15) we obtain (4.20) 1

1 2

1 2

π ∫ x log Γ( x) sin π x dx = log(2π ) − (2 log 2 − 1) − 0

[γ + log(2π )] π 2 − 8 8 − 2 π2 2 π

∞

log n 2 − 1) 2

∑ (4n n =1

□ Applying Lerch’s identity [10] 1 2

ς ′(0, x) = log Γ( x) − log(2π ) we see that 1

1

1

∂ 1 ς ( s, x)ς ( p, x) cos qπ x dx = ∫ ς ( p, x) log Γ( x) cos qπ x dx − log(2π ) ∫ ς ( p, x) cos qπ x dx ∫ ∂s 0 2 0 0 s =0

120

With p = −n where n ≥ 0 is a positive integer we have [6, p.264] in terms of the Bernoulli polynomials

ς ( − n, x ) = −

Bn +1 ( x) n +1

so that 1

1

∂ 1 (n + 1) ∫ ς ( s, x)ς (−n, x) cos qπ x dx = log(2π ) ∫ Bn +1 ( x) cos qπ x dx ∂s 0 2 0 s =0 1

− ∫ Bn +1 ( x) log Γ( x) cos qπ x dx 0

The integral on the left-hand side may then be evaluated by using Hurwitz’s formula (3.1). □ We now let p → ip where i = −1 and hence we have 1

1

0

0

1

1

0

0

∫ log Γ( x) cos ipπ x dx = ∫ log Γ( x) cosh pπ x dx ∫ log Γ( x) sin ipπ x dx = i ∫ log Γ( x) sinh pπ x dx With α → iα in (4.7) we see that

π coth απ =

1

α

∞

1 2 n =1 n + α

+ 2α ∑

2

1

(4.21)

∫ log Γ( x) cosh pπ x dx 0

=

1 sinh pπ p sinh pπ log(2π ) + 2 pπ 2π

1 1 2(1 − cosh pπ ) ∞ γ + log(2π n) + ∑ ∑ 2 2 π2 4n 2 + p 2 n =1 n 4n + p n =1 ∞

Espinosa and Moll [30] showed that 1

∫ e ς (s, x) dx = 2(e tx

0

t

− 1)Γ(1 − s )(2π )

s −2

⎛ t ⎞ ⎛π ⎞ (−1) ⎜ ∑ ⎟ ς (n + 2 − s ) cos ⎜ [ s − n] ⎟ ⎝ 2π ⎠ ⎝2 ⎠ n =0 n

∞

121

n

and differentiating this with respect to s will give us integrals involving hyperbolic functions. 1

5. The ∫ log Γ( x) sin pπ x dx integral 0

Provided a ≠ b we readily determine that 1 ⎡ sin(a − b) x sin(a + b) x ⎤ +c − a −b a + b ⎥⎦

∫ sin ax sin bx dx = 2 ⎢⎣ and hence we have 1

∫ sin 2π nx sin pπ x dx = 0

1 ⎡ sin(2n − p )π sin(2n + p )π ⎤ − 2 ⎢⎣ (2n − p )π (2n + p )π ⎥⎦

=−

(5.1)

2n sin pπ

π

1 4n − p 2 2

Similarly we have 1 ⎡ cos(a − b) x cos(a + b) x ⎤ +c − a −b a + b ⎥⎦

∫ cos ax sin bx dx = 2 ⎢⎣ which gives us

⎤ 1 ⎡ cos(2n − p )π cos(2n + p )π ⎤ 1 ⎡ 1 1 − − ⎢ − ⎥ (2n − p)π (2n + p )π ⎦ 2 ⎣ (2n − p )π (2n + p )π ⎥⎦

1

∫ cos 2π nx sin pπ x dx = 2 ⎢⎣ 0

=−

(5.2)

p 1 − cos pπ π 4n 2 − p 2

We now multiply (3.1) by sin pπ x and integrate to get 1

(5.3)

∫ ς (s, x) sin pπ x dx 0

⎡ ⎛πs ⎞ ∞ ⎤ 1 p 1 − cos pπ 1 2n sin pπ 1 ⎛πs ⎞ ∞ cos = −2Γ(1 − s ) ⎢sin ⎜ ⎟ ∑ + ⎜ ⎟∑ 1− s 2 2 1− s 2 2 ⎥ π 4n − p ⎦ ⎝ 2 ⎠ n =1 (2π n) ⎣ ⎝ 2 ⎠ n =1 (2π n) π 4n − p

122

Differentiation with respect to s gives us 1

∫ ς ′(s, x) sin pπ x dx

(5.4)

0

⎡ ⎛ π s ⎞ ∞ log(2π n) p 1 − cos pπ π 1 p 1 − cos pπ ⎤ ⎛πs ⎞ ∞ = −2Γ(1 − s ) ⎢sin ⎜ ⎟ ∑ + cos ⎜ ⎟∑ 1− s 2 2 1− s 2 2 ⎥ 2 ⎝ 2 ⎠ n =1 (2π n) π 4n − p ⎦ ⎣ ⎝ 2 ⎠ n =1 (2π n) π 4n − p ⎡ ⎛ π s ⎞ ∞ log(2π n) 2n sin pπ ⎤ 1 π 1 2n sin pπ 1 ⎛πs ⎞ ∞ −2Γ(1 − s ) ⎢cos ⎜ ⎟ ∑ − sin ⎜ ⎟∑ 1− s 2 2 1− s 2 2⎥ π 4n − p 2 π 4n − p ⎦ ⎝ 2 ⎠ n =1 (2π n) ⎣ ⎝ 2 ⎠ n =1 (2π n) ⎡ ⎛πs ⎞ ∞ ⎤ 1 p 1 − cos pπ 1 2n sin pπ 1 ⎛πs ⎞ ∞ +2Γ′(1 − s ) ⎢sin ⎜ ⎟ ∑ + cos ⎜ ⎟∑ 1− s 2 2 1− s 2 2⎥ π 4n − p ⎦ ⎝ 2 ⎠ n =1 (2π n) ⎣ ⎝ 2 ⎠ n =1 (2π n) π 4n − p

and with s = 0 we have 1

∫ log Γ( x) sin pπ x dx

(5.5)

0

=

1 1 − cos pπ p(1 − cos pπ ) ∞ 1 1 2sin pπ log(2π ) − − ∑ 2 2 pπ 2 2π π2 n =1 n 4n − p

∞

∑ n =1

γ + log(2π n) 4n 2 − p 2

With a little algebra and using (2.15) and (4.7) this may be expressed as (5.5.1) 1

∫ log Γ( x) sin pπ x dx = 0

−

(1 − cos pπ )[γ + log(2π )] (1 − cos pπ ) ⎡ ⎛ p⎞ p ⎞⎤ ⎛ ψ ⎜1 + ⎟ +ψ ⎜1 − ⎟ ⎥ + ⎢ 2 pπ 4 pπ 2⎠ ⎝ 2 ⎠⎦ ⎣ ⎝

2sin pπ [γ + log(2π )] ⎡ 1 π ⎛ pπ ⎞ ⎤ 2sin pπ cot ⎜ − ⎟⎥ − ⎢ 2 2 π π2 ⎝ 2 ⎠⎦ ⎣2p 4p

With p = 1 we have 1

∫ log Γ( x) sin π x dx = 0

[γ + log(2π )]

π

+

1 2π

⎡ ⎛ 1⎞ ⎛ 1 ⎞⎤ ⎢ψ ⎜ 1 + 2 ⎟ +ψ ⎜ 2 ⎟ ⎥ ⎠ ⎝ ⎠⎦ ⎣ ⎝

or

123

∞

log n 2 − p2

∑ 4n n =1

1

π

1⎡

⎤

∫ log Γ( x) sin π x dx = π ⎢⎣log 2 + 1⎥⎦ 0

With p = 1/ 2 we have

γ + log(2π ) 1 + π 2π

1

∫ log Γ( x) sin(π x / 2) dx = 0

−

2[γ + log(2π )] ⎡ π ⎤ 8 ⎢⎣ 2 − 2 ⎥⎦ − π 2 π2

∞

n =1

γ + log(2π ) 1 + [ 2 − γ − 3log 2] π π

−

2[γ + log(2π )] ⎡ π ⎤ 8 ⎢⎣ 2 − 2 ⎥⎦ − π 2 π2 1

∫ log Γ( x) sin(π x / 2) dx = π [ 2 − γ − 3log 2] − 0

log n 2 −1

∑ 16n

=

1

(5.5.1)

⎡ ⎛ 1⎞ ⎛ 3 ⎞⎤ ⎢ψ ⎜1 + 4 ⎟ + ψ ⎜ 4 ⎟ ⎥ ⎠ ⎝ ⎠⎦ ⎣ ⎝

∞

log n 2 −1

∑ 16n n =1

4[γ + log(2π )]

π

2

−

8

π

2

∞

log n 2 −1

∑ 16n n =1

With p = 2k + 1 in (5.5) we obtain 1

∫ logΓ( x) sin(2k + 1)π x dx = 0

1 (2k + 1) ∞ 1 1 log(2π ) − ∑ 2 2 π (2k + 1)π n =1 n 4n − (2k + 1)

and using (2.21) we have the well-known integral [33, p.650, 6.443.2] 1

(5.6)

∫ logΓ( x) sin(2k + 1)π x dx = 0

k −1 ⎡ ⎛π ⎞ 1 1 1 ⎤ + 2∑ ⎢ log ⎜ ⎟ + ⎥ (2k + 1)π ⎣ ⎝ 2 ⎠ 2k + 1 j =0 2 j + 1 ⎦

As noted by Kölbig [40], this integral is also incorrectly reported in Nielsen’s book [45, p.203]. The result was corrected in the 5th edition of Gradshteyn and Ryzhik [33, p.650, 6.443 2]). A different proof of (5.6) is contained in [23] We obtain with k = 0 1

(5.7)

1⎡

⎛π ⎞

⎤

∫ logΓ( x) sin π x dx = π ⎢⎣log ⎜⎝ 2 ⎟⎠ + 1⎥⎦ 0

124

Letting p → 2k in (5.5) and using (2.21.1) and (2.22.1) we easily obtain another derivation of (1.5) 1

∫ logΓ( x) sin 2kπ x dx = 0

γ + log(2π k ) 2π k □

We have 1

1

0

0

I = ∫ log Γ( x) sin(2k + 1)π x dx = ∫ log Γ(1 − x) sin(2k + 1)π x dx

so that using Euler’s reflection formula we have 1

∫ [log π − log sin π x]sin(2k + 1)π x dx = 2 I 0

Using (5.6) we then obtain 1

(5.8)

∫ log sin π x sin(2k + 1)π x dx = 0

2 (2k + 1)π

k −1 ⎡ 1 1 ⎤ − − log 2 2 ⎢ ⎥ ∑ 2k + 1 j =0 2 j + 1 ⎦ ⎣

which is reported in [33, p.577]. A different derivation is given in Section 6. □ Using integration by parts we have 1

log Γ( x) cos 2π x 1 + ∫a log Γ( x) sin 2π x dx = − ∫a ψ ( x) cos 2π x dx 2π 2 π a 1

1

log Γ(a )[cos 2π a − 1] 1 ψ ( x)[cos 2π x − 1] dx + 2π 2π ∫a 1

=

We see that log Γ(a )[cos 2π a − 1] = [log Γ(a + 1) − log a ][cos 2π a − 1] ⎡ cos 2π a − 1 ⎤ = log Γ(a + 1)[cos 2π a − 1] − a log a ⎢ ⎥⎦ a ⎣

125

and taking the limit as a → 0 we find that 1

∫ log Γ( x) sin 2π x dx = 0

1

1 ψ ( x)[cos 2π x − 1] dx 2π ∫0

=−

1

π

1

∫ψ ( x) sin

2

2π x dx

0

Since from (1.5) 1

∫ log Γ( x) sin 2π x dx = 0

γ + log 2π 2π

we obtain 1

∫ψ ( x) sin

(5.9)

2

2π x dx = −

0

γ + log 2π 2

as noted in [33, p.652, 6.468] and [45, p.204]. □ We make the following summation by reference to (5.6) ∞ sin(2k + 1)π x 1 dx = ∑ 2 2k + 1 k =0 k = 0 (2k + 1) π

1

∞

∫ logΓ( x)∑ 0

k −1 ⎡ ⎛π ⎞ 1 1 ⎤ + + log 2 ⎢ ⎜ ⎟ ⎥ ∑ j =0 2 j + 1 ⎦ ⎣ ⎝ 2 ⎠ 2k + 1

and using the Fourier series [55, p.149] which is valid for 0 < x < 1 sin(2k + 1)π x π = 2k + 1 4 k =0 ∞

∑

together with the well known formula for the Riemann zeta function ∞

1

∑ (2k + 1) k =0

s

= (1 − 2− s )ς ( s )

we obtain

π2

∞ k −1 3 1 1 ⎛π ⎞ 7 ς ς x dx log Γ( ) = (2) log ⎜ ⎟ + (3) + 2∑ 2∑ ∫ 4 0 4 ⎝2⎠ 8 k = 0 (2k + 1) j = 0 2 j + 1

1

126

Using Raabe’s integral (2.11) we have the Euler sum k −1 1 1 π2 7 = log 2 − ς (3) ∑ 2∑ 8 16 k = 0 (2k + 1) j = 0 2 j + 1 ∞

which was previously reported in [23]. We note that [52, p.20] ⎛ ⎝

n −1

1⎞

⎛1⎞ ⎝ ⎠

1

n −1

1

ψ ⎜ n + ⎟ = −γ − 2 log 2 + 2∑ = ψ ⎜ ⎟ + 2∑ 2 j +1 2 2 j +1 2 ⎠

j =0

j =0

and we therefore have ∞

1 ψ ⎛⎜ n + ⎞⎟

1

∑ (2⎝n + 1)2 ⎠ = − 8 [γπ 2

n =0

2

+ 7ς (3)] □

Using Euler’s reflection formula Γ( x)Γ(1 − x) =

π we obtain from (3.12) sin π x

∞ 1 Γ( x) (γ + log 2π n) sin 2π nx log =∑ πn 2 Γ(1 − x) n =1

(5.10)

We now multiply this by cos pπ x and integrate to obtain (γ + log 2π n) ∫0 log Γ( x) cos pπ x dx − ∫0 log Γ(1 − x) cos pπ x dx = 2∑ ∫0 sin 2π nx cos pπ x dx πn n =1 1

1

1

∞

and using the integral (4.1) we have 1

1

∫ log Γ( x) cos pπ x dx − ∫ log Γ(1 − x) cos pπ x dx = 0

0

4(1 − cos pπ )

π

2

(γ + log 2π n) 4n 2 − p 2 n =1 ∞

∑

and using (2.9) 1

1

∫ log Γ( x) cos pπ x dx + ∫ log Γ(1 − x) cos pπ x dx = 0

0

we obtain (4.5) again

127

sin pπ p sin pπ log(2π ) − π pπ

∞

1

∑ n 4n n =1

2

1 − p2

1

2 ∫ log Γ( x) cos pπ x dx = 0

sin pπ p sin pπ log(2π ) − π pπ

1 1 4(1 − cos pπ ) ∞ (γ + log 2π n) + ∑ ∑ 2 2 π2 4n 2 − p 2 n =1 n 4n − p n =1 ∞

6. Some integrals involving log sin π x

We note from (4.2) that p sin pπ 4n 2 − p 2

1

∫ cos 2π nx cos pπ x dx = − π 0

and make the summation 1 p ∞ 1 sin pπ π π = − cos 2 nx cos p x dx ∑ ∑ ∫ π n =1 n 4n 2 − p 2 n =1 n 0 1

∞

Assuming that it is valid to interchange the order of the integration and summation ∞ 1 cos 2π nx π π = cos 2 nx cos p x dx cos pπ x dx ∑ ∑ ∫ ∫ n n =1 n 0 0 n =1

1

∞

1

and using the Fourier series cos 2π nx = − log(2sin π x) n n =1 ∞

∑

0< x 0

0

We now make a further summation 1

∞

1 1 log sin π x cos 2kπ x dx = − ς (2) ∑ ∫ 2 k =1 k 0

and assuming that ∞

1

∑k k =1

1

1

0

0

cos 2kπ x dx k k =1 ∞

∫ log sin π x cos 2kπ x dx = ∫ log sin π x∑

we have 1

= − ∫ log sin π x log(2sin π x) dx 0

1

1

= − ∫ log sin π x dx − log 2 ∫ log sin π x dx 2

0

0

1

= − ∫ log 2 sin π x dx + log 2 2 0

129

Accordingly we obtain the well known integral 1

∫ log

(6.3)

2

1

sin π x dx = ς (2) + log 2 2

0

2

which is a problem posed by Bremekamp [13] more than fifty years ago in 1957. There are many ways to evaluate this integral; for example, we could also have employed Parseval’s theorem [8, p.343] π

1

∫

π

f ( x) g ( x)dx =

−π

∞ 1 a0α 0 + ∑ (anα n + bn β n ) 2 n =1

to evaluate it by utilising the known Fourier coefficients (2.29), (2.31) and (2.32). With p = 2k +1 in (6.1) we immediately determine (2.30) 1

∫ log sin π x cos(2k + 1)π x dx = 0 0

This may also be derived as follows. We designate I as 1

I = ∫ log Γ( x) cos(2k + 1)π x dx 0

and with the substitution x → 1 − x we see that 1

I = − ∫ log Γ(1 − x) cos(2k + 1)π x dx 0

Accordingly we have 1

∫ [logΓ( x) + Γ(1 − x)]cos(2k + 1)π x dx = 0 0

or equivalently 1

∫ [logπ − log sin π x]cos(2k + 1)π x dx = 0 0

We then deduce that 130

1

∫ log sin π x cos(2k + 1)π x dx = 0 0

We have the particular integral with k = 0 which may be directly evaluated as 1

1

∫ log sin π x ⋅ cos π x dx = π sin π x[log sin π x − 1] 0

1

=0 0

because using L’Hôpital’s rule we see that log sin π x π cot π x = − lim =0 x →1 1/ sin π x x →1 π cos π x / sin 2 π x

lim sin π x log sin π x = lim x →1

□ We see from (6.1) that with p → 2 p ∞ 2 pπ 1 1 2 π π = log(2sin x ) cos 2 p x dx p ∑ 2 2 ∫ sin 2 pπ 0 n =1 n n − p

1

and it is easily seen that 2 pπ 2 pπ I= log(sin π x) cos 2 pπ x dx = log(2sin π x) cos 2 pπ x dx − log 2 ∫ sin 2 pπ 0 sin 2 pπ ∫0 1

1

∞

1 1 − log 2 2 2 n =1 n n − p

= p2 ∑ From (A.3) we have ∞

1 1 2 2 n =1 n n − p

ψ (1 + p ) +ψ (1 − p) + 2γ = −2 p 2 ∑ and therefore we have

1 I = − [ψ (1 + p ) +ψ (1 − p) + 2γ ] − log 2 2 1 = − [ψ (1 + p ) +ψ (1 − p) + 2γ + 2 log 2] 2

131

1 = − [ψ ( p ) + ψ (− p ) + 2γ + 2 log 2] 2 This corresponds with the formula given by Dwilewicz and Mináč [28] 2 pπ 1 log(sin π x) cos 2 pπ x dx = − [ψ ( p) + ψ (− p) + 2γ + 2 log 2] ∫ sin 2 pπ 0 2 1

(6.4)

□ We note from (5.2) that p 1 − cos pπ 4n 2 − p 2

1

∫ cos 2π nx sin pπ x dx = − π 0

and following the above procedure we obtain 1

(6.5)

∫ log(2sin π x) sin pπ x dx = 0

p

1 1 − cos pπ 4n 2 − p 2 n =1 ∞

∑n π

As above, we may also show that (6.5) is equivalent to 2 pπ 1 log(sin π x) sin 2 pπ x dx = − [ψ (1 + p ) + ψ (1 − p ) + 2γ + 2 log 2] ∫ 1 − cos 2 pπ 0 2 1

(6.6)

which was derived in a very different manner by Dwilewicz and Mináč [28]. With p = 2k we have 1 − cos pπ 4k 2 − p 2

1

∫ log(2sin π x) sin 2k π x dx = 2 lim

p →2 k

0

= −2 lim

p →2 k

sin pπ 2p

Hence we obtain (2.32) 1

∫0 log sin π x sin 2kπ x dx = 0

k ≥0

With p = 2k +1 in (6.5) we immediately obtain

132

1

2(2k + 1)

∫ log(2sin π x) sin(2k + 1)π x dx = −

π

0

∞

1

∑ n 4n n =1

2

1 − (2k + 1) 2

and using (2.18) this becomes (2.33) 1

∫ log sin π x sin(2k + 1)π x dx = 0

k −1 ⎡ 2 1 1 ⎤ − 2∑ ⎢log 2 − ⎥ (2k + 1)π ⎣ 2k + 1 j =0 2 j + 1 ⎦

□ We now consider the limit of (6.6) as p → 0 ; we have lim p →0

h( p ) 1 = − lim[ψ (1 + p) +ψ (1 − p) + 2γ + 2 log 2] = − log 2 1 − cos 2 pπ 2 p →0

where 1

h( p ) = 2π ∫ log(sin π x) p sin 2 pπ x dx 0

Applying L’Hôpital’s rule twice, we easily find that 1

h( p ) = 2 ∫ x log(sin π x) dx p →0 1 − cos 2 pπ 0

lim

so that we have 1

1

∫ x log(sin π x) dx = − 2 log 2

(6.7)

0

in agreement with [30]. □ We note from (5.1) that 1

∫ sin 2π nx sin pπ x dx = − 0

2n sin pπ

π

1 4n − p 2 2

and make the summation 1 2sin pπ sin 2π nx sin pπ x dx = − ∑ ∫ π n =1 n 0 ∞

1

∞

∑ 4n n =1

133

2

1 − p2

We again assume that it is valid to interchange the order of the integration and summation ∞ 1 sin 2π nx π π = sin 2 nx sin p x dx sin pπ x dx ∑ ∑ ∫ ∫ n n =1 n 0 n = 1 0 ∞

1

1

and use the Fourier series sin 2π nx π = (1 − 2 x) n 2 n =1 ∞

∑

0< x