Defining Incomplete Gamma Type Function with Negative Arguments ...

arXiv:1407.0349v1 [math.CA] 27 Jun 2014

Defining Incomplete Gamma Type Function with Negative Arguments and Polygamma functions ψ (n)(−m) ¨ a˘ ˙ Emin Ozc . g and Inci Ege Abstract. In this paper the incomplete gamma function γ(α, x) and its derivative is considered for negative values of α and the incomplete gamma type function γ∗ (α, x− ) is introduced. Further the polygamma functions ψ (n) (x) are defined for negative integers via the neutrix setting.

AMS Mathematics Subject Classification (2010): 33B15, 33B20, 33D05 Key words and phrases: Gamma function, incomplete gamma function, Digamma Function, Polygamma Function, neutrix, neutrix limit.

1. Introduction The incomplete gamma function of real variable and its complement are defined via integrals Z x γ(α, x) = uα−1 e−u du (1) 0

and Γ(α, x) =

Z

∞

uα−1 e−u du

(2)

x

respectively, for α > 0 and x ≥ 0 see [1, 8]. Note that the definition of γ(α, x) does not valid for negative α, but it can be extended by the identity γ(α, x) = Γ(α) − Γ(α, x). The incomplete gamma function does not exists for negative integers α or zero [13]. However, Fisher et. al. [4] defined γ(0, x) by Z x γ(0, x) = u−1 (e−u − 1) du + ln x. 0

The problem of evaluating incomplete gamma function is subject of some earlier articles. The general problem in which α and x are complex was considered by Winitzki in [15], but no method is found to be satisfactory in cases where Re[x] < 0. The incomplete gamma function with negative arguments are difficult to compute, see [6]. In [14] Thompson gave the algorithm for accurately computing the incomplete gamma function γ(α, x) in the cases where α = n + 1/2, n ∈ Z and x < 0.

1

2. Incomplete Gamma Function A series expansion of incomplete gamma function can be obtained by replacing the exponential in (1) with its Maclaurin series, the result is that γ(α, x) =

∞ X k=0

(−1)k α+k x . k!(α + k)

(3)

A single integration by parts in (1) yields the recurrence relation γ(α + 1, x) = αγ(α, x) − xα e−x

(4)

and thus the incomplete gamma function can be defined for α < 0 and α 6= −1, −2, . . . . By repeatedly applying recurrence relation, we obtain α−1 −x

γ(α, x) = x

e

∞ X xk 1 + γ(α + m, x) (α) (α) k m k=0

where (.)k is Pochhamer’s symbol [1]. The last term on the right-hand side disappears in the limit m → ∞. By regularization we have Z x m−1 m−1 h X (−1)i xα+i X (−u)i i α−1 −u du + . γ(α, x) = u e − i! (α + i)i! 0 i=0 i=0

(5)

for −m < α < −m + 1 and x > 0. It follows from the definition of gamma function that lim γ(α, x) = Γ(α) x→∞

for α 6= 0, −1, −2, . . . , see [7, 11]. In the following we let N be the neutrix [2, 3, 4] having domain N ′ = {ε : 0 < ε < ∞} and range N ′′ the real numbers, with negligible functions finite linear sums of the functions ελ lnr−1 ε, lnr ε (λ < 0, r ∈ Z+ ) (6) and all functions of ε which converge to zero in the normal sense as ε tends to zero. If f (ε) is a real (or complex) valued function defined on N ′ and if it is possible to find a constant c such that f (ε) − c is in N, then c is called the neutrix limit of f (ε) as ε → 0 and we write N−limε→0 f (ε) = c. Note that if a function f (ε) tends to c in the normal sense as ε tends to zero, it converges to c in the neutrix sense. On using equation (5), the incomplete gamma function γ(α, x) was also defined by Z x uα−1 e−u du (7) γ(α, x) = N−lim ε→0

ε

2

for all α ∈ R and x > 0, and it was shown that limx→∞ γ(−m, x) = Γ(−m) for m ∈ N, see [4]. Further, the r-th derivative of γ(α, x) was similarly defined by Z x (r) uα−1 lnr ue−u du γ (α, x) = N−lim ε→0

(8)

ε

for all α and r = 0, 1, 2, . . . , provided that the neutrix limit exists, see [11]. By using the neutrix N given in equation(6), the interesting formula γ(−m, x) +

(−1)m 1 1 γ(−m + 1, x) = − e−x x−m m mm! m

was obtained in [11] for m ∈ N. It can be easily seen that equation (3) is also valid for α < 0 and α 6= −1, −2, . . . . In fact it follows from the definition that Z x Z ∞ hX (−1)k x α+k−1 i α−1 −u γ(α, x) = N−lim u e du = N−lim u du ε→0 ε→0 k! ε ε k=0 = N−lim ε→0

=

∞ X k=0

∞ X k=0

 (−1)k  α+k x − εα+k k!(α + k)

(−1)k α+k x . k!(α + k)

Now assume that α ∈ Z− , then writing Z

x

u ε

−m−1 −u

e

Z Z ∞ X (−1)k x −m+k−1 (−1)m x −1 du = u du + u du k! m! ε ε k=0, =

k6=m ∞ X

k=0, k6=m

(−1)k (−1)m k−m x + ln x + O(ε) k!(k − m) m!

where O(ε) is negligible function, and taking the neutrix limit we obtain γ(−m, x) =

∞ X

(−1)k (−1)m xk−m + ln x k!(k − m) m! k=0, k6=m

for x > 0 and m ∈ N. If m = 0 we particularly have γ(0, x) =

∞ X (−x)k k=1

3

kk!

+ ln x

and thus it follows by the calculation of Mathematica that γ(0, 1/2) ≈ −1.13699,

γ(0, 3/4) ≈ −0.917556,

γ(0, 1) ≈ −0.7966.

Similar equation for γ (r) (α, x) can be obtained using the following lemma. Lemma 2.1 N−lim ε→0

Z

N−lim ε→0

x

lnr+1 x r+1 r−1 X (−1)k r!xα+1 lnr−k x

u−1 lnr u du =

ε

Z

x

uα lnr u du =

ε

(α + 1)k+1 (r − k)!

k=0

+

(−1)r r!xα+1 (α + 1)r+1

for all α 6= −1, x > 0 and r ∈ N. Proof. Straight forward. Now let α < 0 and α 6= 0, −1, −2, . . . . Then it follows from lemma 2.1 that Z x (r) uα−1 lnr ue−u du γ (α, x) = N−lim ε→0

ε

= N−lim ε→0

Z ∞ X (−1)k

∞ X r−1 X

=

k!

k=0

k=0 i=0

x

uα+k−1 lnr u du

ε

(−1)i+k r! xα+k lnr−i x + k!(r − i)!(α + k)i+1 ∞ X (−1)k+r r! α+k x + k!(α + k)r+1 k=0

and similarly (r)

γ (−m, x) = N−lim ε→0

=

Z

r−1 ∞ X X

k=0, i=0 k6=m

+

x

u−m−1 lnr ue−u du

ε

(−1)i+k r! xk−m lnr−i x + i+1 k!(r − i)!(m − k) ∞ X

k=0, k6=m

(−1)k+r r! k−m (−1)m x + lnr+1 x k!(m − k)r+1 (r + 1)m!

for r = 1, 2, . . . and m ∈ N. It was indicated in [3] that equation (1) could be replaced by the equation Z x γ(α, x) = |u|α−1e−u du 0

4

(9)

and this equation was used to define γ(α, x) for all x and α > 0, the integral again diverging for α ≤ 0. The locally summable function γ∗ (α, x− ) for α > 0 was then defined by  R x α−1 −u |u| e du, x ≤ 0, 0 γ∗ (α, x− ) = 0, x>0 Z −x− = |u|α−1 e−u du.

(10)

0

If α > 0, then the recurrence relation γ∗ (α + 1, x− ) = −αγ∗ (α, x− ) − xα− e−x .

(11)

holds. So we can use equation (11) to extend the definition of γ∗ (α, x− ) to negative non-integer values of α. More generally it can be easily proved that γ∗ (α, x− ) =

Z

−x−

0

m−1 m−1 h X xα+i X (−u)i i − du − . |u|α−1 e−u − i! (α + i)i! i=0 i=0

(12)

if −m < α < −m + 1 for m = 1, 2, . . . . Now if −m < α < −m + 1, m ∈ N and x < 0, we have Z

−x−

|u|α−1e−u du =

−ε

Z

−x− −ε

h i α+i α+i m−1 m−1 h i −ε X (−u)i X |x| |u|α−1 e−u − du − i! (α + i)i! i=0 i=0

and thus γ∗ (α, x− ) = N−lim ε→0

Z

−x−

|u|α−1e−u du

−ε

on using equation (12). This suggests that we define γ∗ (−m, x− ) by Z −x− |u|−m−1e−u du γ∗ (−m, x− ) = N−lim ε→0

−ε

for x < 0 and m ∈ N. If x < 0, then we simply write Z −x− Z −1 −u |u| e du = −ε

−x−

|u|−1(e−u − 1) du − ln |x| + ln ε

−ε

thus we have N−lim ε→0

Z

−x−

−ε

−1 −u

|u| e

du =

Z

−x−

|u|−1 (e−u − 1) du − ln x−

0

= γ∗ (0, x− ). 5

(13)

Next, writing similarly Z −x− Z −m−1 −u |u| e du =

m h X (−u)i i −u du |u| e − i! −ε i=0 i h m−1 i X |x|i−m − εi−m 1 h − ln |x| − ln ε − (m − i)i! m! i=0

−ε

−x−

−m−1

for x < 0 and m ∈ N, then it follows that Z −x− |u|−m−1 e−u du γ∗ (−m, x− ) = N−lim ε→0

−ε

m−1 m h −x− X xi−m X 1 (−u)i i − −m−1 −u du − − ln x− . |u| e − = i! (m − i)i! m! 0 i=0 i=0

Z

(14)

Equation (14) can be regarded as the definition of incomplete gamma function γ∗ (−m, x− ) for negative integers and also written in the form Z −1 m h X (−u)i i −m−1 −u γ∗ (−m, x− ) = e − |u| du i! 0 i=0 Z −x− m−1 X 1 −m−1 −u + |u| e du + (15) (m − i)i! −1 i=0 Replacing e−x by its Maclaurin series yields in equation (13) γ∗ (−m, x− ) =

∞ X

k−m x− 1 − ln x− (m − k)k! m! k=0,

(m ∈ N).

(16)

k6=m

If m = 0 we particular have γ∗ (0, x− ) = −

∞ X xk− − ln x− . kk! k=1

and by Mathematica γ∗ (0, −1/3) ≈ 0.735308 γ∗ (0, −1/2) ≈ 0.122996 γ∗ (0, −3/4) ≈ −0.630117 γ∗ (0, −1) ≈ −1.3179. Differentiating equation (12) we get γ∗′ (α, x− )

=

Z

0

−x−

m−1 h X (−u)i i du |u|α−1 ln |u| e−u − i! i=0

−

m−1 X i=0

α+i (α + i)xα+i − ln x− − x− (α + i)2 i!

6

(17)

for −m < α < −m + 1 and x < 0. On the other hand, arranging the following integral and taking the neutrix limit Z

−x−

|u|

α−1

−u

ln |u|e

du =

−ε

m−1 h X (−u)i i du |u|α−1 ln |u| e−u − i! i=0 Z m−1 −x − X1 + |u|α+i−1 ln |u| du i! −ε i=0

Z

−x−

Z

−x−

−ε

=

|u|

α−1

−ε

−

m−1 X

h i 1 α+i α+i x ln x− − ε ln ε i!(α + i) −

i=0

+

m−1 X i=0

Thus γ∗′ (α, x− )

= N−lim ε→0

m−1 h X (−u)i i −u ln |u| e − du i! i=0

Z

−x−

Z

−x−

h i 1 α+i α+i x −ε . i!(α + i)2 − |u|α−1 ln |u|e−u du

−ε

for α 6= 0, −1, −2, . . . and x < 0. More generally it can be shown that γ∗(r) (α, x− )

= N−lim ε→0

|u|α−1 lnr |u|e−u du

(18)

−ε

for α 6= 0, −1, −2, . . . , r = 0, 1, 2, . . . and x < 0. This suggests the following definition. (r)

Definition 2.2 The r-th derivative of incomplete gamma function γ∗ (−m, x− ) is defined by Z −x− (r) |u|−m−1 lnr |u|e−u du (19) γ∗ (−m, x− ) = N−lim ε→0

−ε

for r, m = 0, 1, 2, . . . and x < 0 provided that the neutrix limit exists. (r)

Equation(18) will then define γ∗ (α, x− ) for all α and r = 0, 1, 2, . . . . To prove the neutrix limit above exists we need the following lemma. Lemma 2.3 Z

r

ln |u| du =

r−1 X (−1)i r! i=0

u lnr−i |u| + (−1)r r!u

(20)

r! s−i−1 u−s lnr−i |u| − r!s−r−1 u−s (r − i)!

(21)

(r − i)!

and Z

u−s−1 lnr |u| du = −

r−1 X i=0

7

for r, s = 1, 2, . . . . Proof. Equation (20) follows by induction and equation (21) follows on using equation (20) and making the substitution w = u−s . (r)

(r)

Theorem 2.4 The functions γ∗ (0, x− ) and γ∗ (−m, x− ) exist for x < 0 and Z −x− Z −1 r (r) −1 −u γ∗ (0, x− ) = |u| ln |u|e du + |u|−1 lnr |u|[e−u − 1] du (22) 0

−1

for r = 0, 1, 2, . . . and Z (r) γ∗ (−m, x− ) =

−x−

|u|−m−1 lnr |u|e−u du +

−1

+

Z

−1

|u|

−m−1

0

m−1 m h X r!(m − i)−r−1 X (−u)i i −u du + ln |u| e − i! i! i=0 i=0 r

(23)

for r, m = 1, 2, . . . . Proof. We have Z −x− |u|−1 lnr |u|e−u du = −ε Z −x− Z −1 Z −1 r r −1 −u −1 −u = |u| ln |u|e du + |u| ln |u|[e − 1] du + |u|−1 lnr |u| du −1 −ε −ε Z −x− Z −1 lnr+1 ε r r −1 −u −1 −u = |u| ln |u|e du + |u| ln |u|[e − 1] du + . r+1 −1 −ε for r = 0, 1, 2, . . . . It follows that Z −x− |u|−1 lnr |u|e−u du = N−lim ε→0 −ε Z −x− Z r −1 −u = |u| ln |u|e du +

−1

|u|−1 lnr |u|[e−u − 1] du.

0

−1

and so equation (22) follows. Now let us consider Z −x− Z r −m−1 −u |u| ln |u| e du = −ε

−x−

−1 −1

+

Z

−ε

|u|−m−1 lnr |u|e−u du +

m h X (−u)i i −u du |u| ln |u| e − i! i=0 Z m X (−1)i −1 −m+i−1 r + |u| ln |u| du i! −ε i=0 −m−1

8

r

=

Z

−x−

|u|

−m−1

r

−u

ln |u|e

du +

−1

Z

−1

|u|

−m−1

−ε

r−1 m−1 XX

+

(−1)i−m−1 r! (m − i)−j−1 εm−i lnr−j ε + i!(r − j)!

i=0 j=0

−

m h X (−u)i i −u du + ln |u| e − i! i=0 r

m−1 X i=0

i lnr+1 ε (−1)i−m−1 (m − i)−r−1 r! h (−1)m−i − εm−i ) + . i! m!(r + 1)

Thus γ∗(r) (−m, x− ) =

Z

= N−lim ε→0

−x−

|u|

−x−

Z

|u|−m−1 lnr |u|e−u du

−ε

−m−1

r

−u

ln |u|e

du +

−1

|u|

−m−1

0

−1

+

Z

m−1 X i=0

m h X (−u)i i −u ln |u| e − du i! i=0 r

r!(m − i)−r−1 . i!

for r, m = 1, 2, . . . . Equation (23) follows.

3. Polygamma Functions ψ (n) (−m) The polygamma function is defined by dn dn+1 ψ(x) = ln Γ(x) dxn dxn+1

ψ (n) (x) =

(x > 0).

(24)

It may be represented as ψ

(n)

(x) = (−1)

n+1

Z

∞

0

tn e−xt 1 − e−t

which holds for x > 0, and ψ

(n)

(x) = −

Z

0

lnn t dt. 1−t

1 x−1

t

(25)

It satisfies the recurrence relation ψ (n) (x + 1) = ψ (n) (x) +

(−1)n n! xn+1

(26)

see [8]. This is used to define the polygamma function for negative non-integer values of x. Thus if −m < x < −m + 1, m = 1, 2, . . . , then ψ

(n)

(x) = −

Z

1 x+m−1

t

0

1−t

lnn t dt −

m−1 X k=0

9

(−1)n n! . (x + k)n+1

(27)

R1 K¨olbig gave the formulae for the integral 0 tλ−1 (1−t)−ν lnm t dt for integer and halfinteger values of λ and ν in [10]. As the integral representation of the polygamma function is similar to the integral mentioned above, we prove the existence of the integral in equation (25) for all values of x by using the neutrix limit. Now we let N be a neutrix having domain the open interval {ǫ : 0 < ε < 12 } with the same negligible functions as in equation (6). We first of all need the following lemma. Lemma 3.1. The neutrix limits as ε tends to zero of the functions Z 1/2 Z 1−ε n r x t ln t ln (1 − t) dt, (1 − t)x lnn t lnr (1 − t) dt ε

1/2

exists for n, r = 0, 1, 2, . . . and all x. Proof. Suppose first of all that n = r = 0. Then  2−x−1 −εx+1 Z 1/2 , x 6= −1, x x+1 t dt = − ln 2 − ln ε, x = −1 ε R 1/2 and so N−lim ε tx dt exists for all x. ε→0

Now suppose that r = 0 and that N−lim integer n and all x. Then   Z 1/2  n+1 x t ln t dt =  ε 

ε→0

R 1/2 ε

tx lnn t dt exists for some nonnegative

−2−x−1 lnn+1 2−εx+1 lnn+1 ε x+1

−

n+1 x+1

R 1/2 ε

tx lnn t dt, x 6= −1,

(−1)n lnn+2 2−lnn+2 ε , m+2

and it follows by induction that N−lim ε→0

all x.

R 1/2 ǫ

x = −1

tx lnn t dt exists for n = 0, 1, 2, . . . and

Finally we note that we can write lnr (1 − t) =

∞ X

αir ti

i=1

for r = 1, 2, . . . , the expansion being valid for |t| < 1. Choosing a positive integer k such that x + k > −1, we have Z 1/2 tx lnn t lnr (1 − t) dt = ε

=

k−1 X i=1

αin

Z

1/2 x+i

t

n

ln t dt +

ε

∞ X i=k

αin

Z

1/2

tx+i lnn t dt.

ε

It follows from what we have just proved that Z 1/2 k−1 X αin tx+i lnn t dt N−lim ε→0

i=1

ε

10

exists and further N−lim ε→0

∞ X i=k

αin

1/2

Z

x+i

t

n

ln t dt = lim

ε→0

ε

=

∞ X

∞ X

αin

αin

Z

i=k

i=k

Z

1/2

tx+i lnn t dt

ε

1/2

tx+i lnn t dt,

0

R 1/2 proving that the neutrix limit of ε tx lnn t lnr (1 − t) dt exists for n, r = 0, 1, 2, . . . and all x. Making the substitution 1 − t = u in Z 1−ε (1 − t)x lnn t lnr t dt, 1/2

R 1−ε

(1 − t)x lnn t lnr t dt also exits for n, r = 0, 1, 2, . . . and all x. R1 Lemma 3.2. The neutrix limit as ε → 0 of the integral ε t−m−1 lnn t dt exists for m, n = 1, 2, . . . and Z 1 n! (28) N−lim t−m−1 lnn t dt = − n+1 . m ε→0 ε

it follows that

1/2

Proof. Integrating by parts, we have Z 1 Z −m−1 −1 −m −1 t ln t dt = m ε ln ε + m ε

1

t−m−1 dt

ε

and so N−lim ε→0

Z

1

t−m−1 ln t dt = −

ε

1 n2

proving equation (28) for n = 1 and m = 1, 2, . . . . Now assume that equation (28) holds for some m and n = 1, 2, . . . . Then Z

ε

1 −m−2

t

Z 1 n ln t dt = (m + 1) ε ln ε + t−m−2 lnn−1 t dt m+1 ε n −(n − 1)! = (m + 1)−1 ε−m−1 lnn ε + m + 1 (m + 1)n n

−1 −m−1

n

and it follows that N−lim ε→0

Z

1

t−m−2 lnn t dt = −

ε

n! (m + 1)n+1

proving equation (28) for m + 1 and n = 1, 2, . . . . Using the regularization and the neutrix limit, we prove the following theorem. Theorem 3.3. The function ψ (n) (x) exists for n = 0, 1, 2, . . . , and all x.

11

Proof. Choose positive integer r such that x > −r. Then we can write Z

ε

1−ε

r−1 i h 1 X − (−1)i ti dt t ln t 1 − t i=0 ε Z 1/2 Z 1−ε x−1 r−1 X t n x+i−1 i t ln t dt + + (−1) lnn t dt. 1 − t ε 1/2 i=0

tx−1 n ln t dt = 1−t

We have Z lim ε→0

1/2 ε

Z

1/2

n

x−1

r−1 i X tx−1 n h 1 i i ln t − (−1) t dt = 1−t 1 − t i=0 Z 1/2 x−1 r−1 i h 1 X t n i i = ln t − (−1) t dt 1−t 1 − t i=0 0

and lim

ε→0

Z

1−ε

1/2

tx−1 n ln t dt = 1−t

Z

1 1/2

tx−1 n ln t dt 1−t

the integrals being convergent. Further, from the Lemma 3.1 we see that the neutrix limit of the function Z 1/2 r−1 X i tx+i−1 lnn t dt (−1) ε

i=0

exists and implying that N−lim ε→0

Z

ε

1−ε

tx−1 n ln t dt 1−t

exists. This proves the existence of the function ψ (n) (x) for n = 0, 1, 2, . . . , and all x. Before giving our main theorem, we note that Z 1 x−1 t (n) lnn t dt ψ (x) = − N−lim 1 − t ε→0 ε since the integral is convergent in the neighborhood of the point t = 1. Theorem 3.4. The function ψ (n) (−m) exists and ψ

(n)

m X n! (−m) = + (−1)n+1 n!ζ(n + 1) n+1 i i=1

for n = 1, 2, . . . and m = 0, 1, 2, . . . , where ζ(n) denotes zeta function.

12

(29)

Proof. From Theorem 3.3, we have ψ

(n)

lnn t dt ε→0 1−t ε Z 1 hm+1 i X = − N−lim t−i + (1 − t)−1 lnn t dt.

(−m) = − N−lim

Z

ε→0

1 −m−1

t

ε

(30)

i=1

R1 We first of all evaluate the neutrix limit of integral ε t−i lnn t dt for i = 1, 2, . . . and n = 1, 2, . . . . It follows from Lemma 3.2 that Z 1 n! t−i lnn t dt = − N−lim (i > 1). (31) ε→0 (i − 1)n+1 ε For i = 1, we have Z

1

t−1 lnn t dt = O(ε).

ε

Next Z

1

ε

lnn t dt = 1−t =

Z

1

0

∞ X k=0

∞

X lnn t dt = 1−t k=0

Z

1

tk lnn t dt

0

(−1)n n! (k + 1)n+1

= (−1)n n!ζ(n + 1), where

Z

1

tk lnn t dt =

0

(32)

(−1)n n! . (k + 1)n+1

It now follows from equations (30), (31) and (32) that Z 1 −m−1 n t ln t (n) dt ψ (−m) = − N−lim ε→0 1−t ε Z Z 1 m+1 X n −i t ln t dt − N−lim = − N−lim i=1

=

m X i=1

n! in+1

ε→0

ε→0

ε

ε

1

lnn t dt. (1 − t)

+ (−1)n+1 n!ζ(n + 1)

implying equation (29). Note that the digamma function ψ(x) can be defined by Z 1 1 − tx−1 dt ψ(x) = −γ + N−lim 1−t ε→0 ε 13

(33)

for all x. Also note that using Lemma 3.1 we have Z

1

ε

m+1 Z 1

X 1 − t−m−1 dt = − 1−t i=1

t−i dt

ε

= −[ln 1 − ln ε] −

m+1 X i=2

[1 − ε−i+1] −i + 1

and it follows from equation (33) that ψ(−m) = −γ + N−lim ε→0

Z

1

ε

m

X 1 − t−m−1 i−1 dt = −γ + 1−t i=1

= −γ + φ(m)

(34)

which was also obtained in [5] and [9]. Equation (34) can be regarded as the definition of the digamma function for negative integers.

References [1] M. Abramowitz and I. Stegun, Hand of Mathematical Functions, Dover Publications, Newyork, 1965. [2] J.G. van der Corput, Introduction to the neutrix calculus, J. Analyse Math., 7(1959), 291-398. [3] B. Fisher, On defining the incomplete gamma function γ(−m, x− ), Integral Trans. Spec. Funct., 15(6)(2004), 467-476. [4] B. Fisher, B. Jolevska-Tuneska and A. Kılıc.man, On defining the incomplete gamma function, Integral Trans. Spec. Funct., 14(4)(2003), 293-299. [5] Fisher, B. and Kuribayashi, Y., Some results on the Gamma function, J. Fac. Ed. Tottori Univ. Mat. Sci., 3(2)(1988) 111-117. [6] W. Gautschi, the incoplete gamma function since Tricomi. Tricomi’s Ideas Contemporary Applied Mathematics. Number 147 in Atti dei Convegni Lincei. Roma : Accedemia nazionale dei Lincel, Roma, Italy, 1998, 203-237. [7] I. M. Gel’fand and G. E. Shilov, Generalized Functions, Vol. I., Academic Press, Newyork/London, (1964). [8] I. S. Gradshhteyn, and I. M. Ryzhik, Tables of integrals, Series, and Products, Academic Press, San Diego, 2000.

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[9] Jolevska-Tuneska, B. and Jolevski, I., Some results on the digamma function, Appl. Math. Inf. Sci., 7(2013) 167-170. R1 [10] K¨olbig, K. S., On the integral 0 xν−1 (1 − x)−λ lnm x dx, J. Comput. Appl. Math., 18(1987) 369-394. ¨ . a˘g, I. ˙ Ege, H. G¨ [11] E. Ozc urc.ay and B. Jolevska-Tuneska, Some remarks on the incomplete gamma function, in: Kenan Ta¸s et al. (Eds.), Mathematical Methods in Engineering, Springer, Dordrecht, 2007, pp. 97-108. ¨ . a˘g, E., Ege, I., ˙ G¨ [12] Ozc urc.ay, H. and Jolevska-Tuneska, B., On partial derivatives of the incomplete beta function, Appl. Math. Lett., 21(2008) 675-681. [13] I. Thompson, A note on real zeros of the incomplete gamma function, Integral Trans. Spec. Funct., 23(6)(2012), 445-453. [14] I. Thompson, Algorithm 926: Incomplete Gamma Function with Negative Arguments, ACM Trans. Math. Softw. 39(2)(2013), 9 pp. [15] S. Winitzki, computing the incomplete gamma function to arbitrary precision, In Proceedings of International Conference on Computational Science and Its Applications (ICCSA’03). Lecturer Notes in Computer Science, Vol. 2667 (2003), 790-798. ¨ . a˘ Emin Ozc g Department of Mathematics Hacettepe University, Beytepe, Ankara, Turkey e-mail: [email protected] ˙ Inci Ege Department of Mathemacis Adnan Menderes University, Aydın, Turkey e-mail : [email protected]

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