arXiv:1005.2940v1 [math.CA] 17 May 2010
SCIENTIA Series A: Mathematical Sciences, Vol. ?? (2010), ?? Universidad T´ ecnica Federico Santa Mar´ıa Valpara´ıso, Chile ISSN 0716-8446 c Universidad T´
ecnica Federico Santa Mar´ıa 2010
The integrals in Gradshteyn and Ryzhik. Part 15: Frullani integrals Matthew Albano, Tewodros Amdeberhan, Erin Beyerstedt, and Victor H. Moll Abstract. The table of Gradshteyn and Ryzhik contains some integrals that can be reduced to the Frullani type. We present a selection of them.
1. Introduction The table of integrals [3] contains many evaluations of the form Z ∞ f (ax) − f (bx) b (1.1) . dx = [f (0) − f (∞)] ln x a 0 Expressions of this type are called Frullani integrals. Conditions that guarantee the validity of this formula are given in [1] and [4]. In particular, the continuity of f ′ and the convergence of the integral are sufficient for (1.1) to hold. 2. A list of examples Many of the entries in [3] are simply particular cases of (1.1). Example 2.1. The evaluation of 3.434.2 in [3]: Z ∞ −ax e − e−bx b (2.1) dx = ln x a 0 corresponds to the function f (x) = e−x . Example 2.2. The change of variables t = e−x in Example 2.1 yields Z 1 b−1 a t − ta−1 (2.2) . dt = ln ln t b 0 This is 4.267.8 in [3]. 2000 Mathematics Subject Classification. Primary 33. Key words and phrases. Integrals, Frullani integrals. 1
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M. ALBANO, T. AMDEBERHAN, E. BEYERSTEDT, AND V. MOLL
Example 2.3. A generalization of the previous example appears as entry 3.476.1 in [3]: Z ∞ dx p p 1 u . = ln e−vx − e−ux (2.3) x p v 0 This comes from Frullani’s result with a simple additional scaling. Example 2.4. The choice (2.4)
f (x) =
e−qx − e−px , x
with p, q > 0 satisfies f (∞) = 0 and (2.5)
e−qx − e−px = p − q. x→0 x
f (0) = lim
Then Frullani’s theorem yields Z ∞ −aqx e − e−apx e−bqx − e−bpx dx b , − = (p − q) ln ax bx x a 0 that can be written as Z ∞ −aqx e − e−apx e−bqx − e−bpx dx b . = (p − q) ln − 2 a b x a 0 This is entry 3.436 in [3]. Example 2.5. Now choose x exp(−cex ). 1 − e−x Then Frullani’s theorem yields entry 3.329 of [3], in view of f (0) = e−c and f (∞) = 0: Z ∞ a exp(−ceax ) b exp(−cebx ) b −c . dx = e ln − −ax −bx 1 − e 1 − e a 0
(2.6)
f (x) =
Example 2.6. The next example uses (2.7)
f (x) = (x + c)−µ ,
with c, µ > 0, to produce Z ∞ b (ax + c)−µ − (bx + c)−µ . dx = c−µ ln (2.8) x a 0 This is 3.232 in [3]. Example 2.7. Entry 4.536.2 in [3] is Z ∞ tan−1 (px) − tan−1 (qx) π p (2.9) . dx = ln x 2 q 0 This follows directly from (1.1) by choosing f (x) = tan−1 x.
FRULLANI INTEGRALS
3
Example 2.8. The function f (x) = ln(a + be−x ) gives the evaluation of entry 4.319.3 of [3]: Z ∞ ln(a + be−px ) − ln(a + be−qx ) p a (2.10) ln . dx = ln x a+b q 0 Example 2.9. The function f (x) = ab ln(1 + x)/x produces entry 4.297.7 of [3]: Z ∞ b b ln(1 + ax) − a ln(1 + bx) . dx = ab ln (2.11) 2 x a 0 Example 2.10. Entry 3.484: qx px Z ∞ a dx a q 1+ (2.12) , − 1+ = (ea − 1) ln qx px x p 0 is obtained by choosing f (x) = (1 + a/x)x in (1.1). Example 2.11. The final example in this section corresponds to the function (2.13)
f (x) =
a + be−x cex + g + he−x
that produces entry 3.412.1 of [3]: Z ∞ a + be−px a + be−qx dx a+b q . (2.14) − qx = ln px −px −qx ce + g + he ce + g + he x c+g+h p 0 3. A separate source of examples The list presented in this section contains integrals of Frullani type that were found in volume 1 of Ramanujan’s Notebooks [2]. Example 3.1. Z
∞
0
Example 3.2. Z
0
∞
π a tan−1 ax − tan−1 bx dx = ln x 2 b
p + qe−ax dx q b ln = ln 1 + ln p + qe−bx x p a
Example 3.3. Z
0
∞
ax + p ax + q
n
−
bx + p bx + q
n
dx = x
a pn 1 − n ln q b
where a, b, p, q are all positive. Example 3.4. Z
0
∞
cos ax − cos bx b dx = ln x a
Example 3.5. Z ∞ Z ∞ (b − a)x (b + a)x dx 1 b cos ax − cos bx sin sin = dx = ln 2 2 x 2x 2 a 0 0
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M. ALBANO, T. AMDEBERHAN, E. BEYERSTEDT, AND V. MOLL
Example 3.6. Z ∞ Z ∞ 1 p+q cos[(p − q)x] − cos[(p + q)x] dx = dx = ln sin px sin qx x 2x 2 p−q 0 0 Example 3.7. The evaluation of Z ∞ ln ab ln(1 + n)2 1 + 2n cos ax + n2 dx 2 ln = 1 + 2n cos bx + n2 x ln ab ln 1 + n1 0
n2 < 1 n2 > 1
is more delicate and is given in detail in the next section. Example 3.8. The value Z ∞ 0
e−ax sin ax − e−bx sin bx dx = 0 x
follows directly from (1.1) since, in this case f (x) = e−x sin x satisfies f (∞) = f (0) = 0. Example 3.9. Z
0
∞
e−ax cos ax − e−bx cos bx b dx = ln . x a 4. A more delicate example
Entry 4.324.2 of [3] states that Z ∞ dx ln(1 + 2a cos px + a2 ) − ln(1 + 2a cos qx + a2 ) (4.1) = x 0 2 ln q ln(1 + a) −1 < a 6 1 p q 2 ln a < −1 or a > 1. p ln(1 + 1/a)
This requires a different approach since the obvious candidate for a direct application of Frullani’s theorem, namely f (x) = ln(1 + 2a cos x + a2 ), does not have a limit at infinity. In order to evaluate this entry, start with Z 1 1 (4.2) xy dx = , y+1 0 so Z 1 Z 1 Z 1 Z 1 Z 1 Z 1 x−1 dy (4.3) dy xy dx = dx xy dy = dx = = ln 2. ln x y +1 0 0 0 0 0 0 This is now generalized for arbitrary symbols α and β as Z ∞ αt β e − eβt . dt = ln (4.4) t α 0
FRULLANI INTEGRALS
5
To prove (4.4), make the substitution u = e−t that turns the integral into
Z
0
1
u−1−β − u−1−α du ln u
=
Z
1
Z
−1−β
du
0
= =
−1−α Z −1−β −1−α
=
−1−β
Z
−1−α
dw
Z
1
uw dw uw du
0
dw w+1
β ln . α
cos(rx) Now observe that | 2a1+a | 6 1, therefore it is legitimate to expand the logarithmic 2 P k terms as infinite series using ln(1 + z) = k (−1)k−1 zk . The outcome reads
Z
∞
0
dx X (−1)k−1 Ak (cosk px − cosk qx) = x k k>1 X (−1)k−1 Ak Z ∞ (eipx + e−ipx )k − (eiqx + e−iqx )k dx; 2k k x 0 k>1
where A = 2a/(1+a2). The inner integral is evaluated using some binomial expansions. That is, (4.5) Z ∞ ipx k Z ∞ (2r−k)ipx X k e − e(2r−k)iqx (e + e−ipx )k − (eiqx + e−iqx )k dx = dx. r x x 0 0 r=0
It is time to employ equation (4.4). A closer look at (4.5) shows that care must be exercised. The integrals are sensitive to the parity of k. More precisely, the quantity 2r − k vanishes if and only if k is even and r = k/2, in which case there is a zero contribution to summation. Otherwise, the second integral in (4.5) is always equal to ln(q/p). Therefore,
k Z X k r=0
r
∞ 0
i dx 2k ln q h p = e(2r−k)ipx − e(2r−k)iqx q 2k − k x ln p k/2
if k is odd, if k is even.
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M. ALBANO, T. AMDEBERHAN, E. BEYERSTEDT, AND V. MOLL
Combining the results obtained thus far yields (4.6) I
=
Z
∞
=
Z
∞
0
0
= = =
=
ln(1 + 2a cos(px) + a2 ) − ln(1 + 2a cos(qx) + a2 ) dx x dx X (−1)k−1 Ak (cosk px − cosk qx) x k k>1
k Z ∞ (2r−k)ipx X (−1)k−1 Ak X k e − e(2r−k)iqx dx k2k x r 0 r=0 k>1 X X 1 k (−1)k−1 Ak (−1)k−1 Ak q q 1− k + ln ln p k p k 2 k/2 k odd k even X 2k 2k (−1)k−1 Ak q q X 1 A ln + ln p k p 2k 2 k k>1 k>1 X 2 k 1 q q 2k 1 A ln(1 + A) + ln ln . p 2 p k k 22 k>1
The last step utilizes the Taylor series i h X 2k Qk p = −2 ln 21 1 + 1 − 4Q (4.7) k k k>1
k √ P This follows from the binomial series k>0 2k k R = 1/ 1 − 4R after rearranging in the manner X 2k 4 1 1 √ − = √ , Rk−1 = √ R k R 1 − 4R 1 − 4R(1 + 1 − 4R) k>1 and then integrating by parts (from 0 to Q) Z Q Z √1−4Q i h X 2k Qk p −2 · du 4 · dR √ √ = = −2 ln 12 1 + 1 − 4Q . = k k 1+u 1 − 4R(1 + 1 − 4R) 1 0 k>1 Formula (4.7) applied to equation (4.6) leading to h i p q q I = ln ln(1 + A) − ln ln 12 1 + 1 − 4Q . p p It remains to replace Q = A2 /22 = a2 /(1 + a2 )2 and use the identity 1 − 4Q = Observe that the expression for proof is complete.
√
(a2 − 1)2 . (a2 + 1)2
1 − 4Q depends on whether |a| > 1 or not. The
FRULLANI INTEGRALS
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Acknowledgments. Matthew Albano and Erin Beyerstedt were partially supported as students by NSF-DMS 0713836. The work of the last author was also partially supported by the same grant. References [1] J. Arias-de Reyna. On the theorem of Frullani. Proc. Amer. Math. Soc., 109:165–175, 1990. [2] B. Berndt. Ramanujan’s Notebooks, Part I. Springer-Verlag, New York, 1985. [3] I. S. Gradshteyn and I. M. Ryzhik. Table of Integrals, Series, and Products. Edited by A. Jeffrey and D. Zwillinger. Academic Press, New York, 7th edition, 2007. [4] A. M. Ostrowski. On Cauchy-Frullani integrals. Comment. Math. Helvetici, 51:57–91, 1976. Department of Mathematics, New Jersey Institute of Technology, Newark, NJ 07102 E-mail address:
[email protected] Department of Mathematics, Tulane University, New Orleans, LA 70118 E-mail address:
[email protected] Department of Mathematics, Tulane University, New Orleans, LA 70118 E-mail address:
[email protected] Department of Mathematics, Tulane University, New Orleans, LA 70118 E-mail address:
[email protected] Received 4 20 2010 revised ?? ´ tica Departamento de Matema Universidad T´ ecnica Federico Santa Mar´ıa Casilla 110-V, Valpara´ıso, Chile
