Some Integrals Involving K Gamma and K Digamma Function Saeed Ahmed

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Ahmed Journal of the Egyptian Mathematical Society (2020) 28:39 Journal of the Egyptian https://doi.org/10.1186/s42787-020-00098-0 Mathematical Society

ORIGINAL RESEARCH Open Access Some integrals involving k gamma and k digamma function Saeed Ahmed

Correspondence: [email protected] Abstract Govt Shah Hussain Degree College, k k Lahore, Pakistan In this paper, some new integrals involving gamma function and digamma function have been established. An integral is established involving k gamma function, and its special values are discussed. Similarly, some new integrals have been established for k digamma function, and different elementary function is associated with it for different values of k. A nice representation of the Euler- Mascheroni constant and π in the form of k digamma function for different values of k is also obtained. Keywords: k gamma function, k digamma function Mathematics subject classification: 33B15, 41A58, 33C20

k gamma function The k gamma function is a generalization of the classical gamma function introduced by Diaz and Pariguan [1], denoted and defined as

z n k − 1 n!k ðÞnk − Γk ðÞ¼z lim ; k > 0; z∈ℂnkℤ : ð1:1Þ n→∞ ðÞz n;k

The symbol for (z)n, k is called Pochhammer’s k symbol [2] and is defined as

ðÞz n;k ¼ zzðÞþ k ðÞz þ 2k ⋯ðÞz þ ðÞn − 1 k : ð1:2Þ

Due to (1.2), we see that (1.1) has simple poles at 0, − k, − 2k, − 3k, ⋯. The residue of 1 k n n k gamma function at these simple poles is ð − 1Þ k n! , see [3]. The integral form of gamma function is denoted and defined as [4]

Z∞ k − t z − 1 Γk ðÞ¼z e k t dt: ð1:3Þ

The improper integral is convergent for Re(z)>0.. The k gamma function reduces to k the classical gamma function, i.e., Γk → Γ as k → 1. A simple change of variable t = ky reveals the relationship between k gamma function and classical gamma function

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z − z Γ ðÞ¼z kk 1Γ : k k

The following properties of the k gamma function have been discussed in [5, 6]

Γk ðÞ¼z þ k zΓk ðÞz ; ð1:4Þ π Γk ðÞz Γk ðÞ¼k − z πz : ð1:5Þ sin k

In the “An integral involving the k gamma function” section, we will establish an integral involving k gamma function and its special cases will also be discussed. In the “Stirling formula for the k gamma function” section, the Stirling formula will be derived for the k gamma function. In the “Some integrals representing digamma function” section, we will provide few integrals involving the k digamma function. Few special cases of the k digamma function will also be presented. In the “Euler-Mascheroni constant and k digamma function” section, we will find the relationship between the Euler- Mascheroni constant in the form k digamma function for different values of k.

An integral involving the k gamma function In this section, we will derive an interesting integral involving k gamma function. Theorem 2.1 Consider a complex number p of the form p = a + ib. Then

Z∞ Γk ðÞz zθ − aun n nz − ¼ e ðÞbu u k 1du; ð : Þ z z − cos cos 2 1a npjjk kk 1 k 0

Z∞ Γk ðÞz zθ − aun n nz − ¼ e ðÞbu u k 1du: ð : Þ z z − sin sin 2 1b npjjk kk 1 k 0

n 1 n 1 − Proof Making the substitution t ¼ðkpu Þk ⇒dt ¼ pnun − 1ðkpu Þk 1du into (1.3), we get

Z∞ Z∞ z − − pun n 1 n − 1 n 1 − 1 z z − 1 − pun nz − 1 Γk ðÞ¼z e ðÞkpu k pnu ðÞkpu k du ¼ npk kk e u k du;

0 0

so we get

Z∞ Γk ðÞz − pun nz − ¼ e u k 1du: ð : Þ z z − 2 2 npk kk 1 0

Similarly, for the conjugate of p, we can write

Z∞ Γk ðÞz − pun nz − ¼ e u k 1du: ð : Þ z z − 2 3 npk kk 1 0

Adding and simplifying (2.2) and (2.3), we get Ahmed Journal of the Egyptian Mathematical Society (2020) 28:39 Page 3 of 10

! Z∞ Γk ðÞz 1 1 − ðÞaþib un − ðÞa − ib un nz − þ ¼ e þ e u k 1du: ð : Þ z − 1 z izθ z − izθ 2 4 nkk jjp k e k jjp k e k 0

where θ and |p| are the principal argument and modulus of p, respectively, so that (2.4) reduces to

Z∞ ÀÁ Γk ðÞz 1 1 − aun − ibun ibun nz − þ ¼ e e þ e u k 1du: z z − izθ − izθ npjjk kk 1 e k e k 0

By Euler’s identity, we can write

Z∞ Γk ðÞz zθ − aun n nz − ¼ e ðÞðÞbu u k 1du: z z − 2 cos 2 cos npjjk kk 1 k 0

This yields the final integral (2.1a). Similarly, subtracting (2.2) and (2.3) and continu- ing in the same fashion, we get (2.1b). Corollary 2.2 a ¼ ; b ¼ ; n ¼ ⇒jpj¼ ; θ ¼ π Take 0 1 1 1 2 in (2.1b), and using the relation (1.5) together with k = 1, we see that

Z∞ sinðÞzπ=2 πz π zπ − lim lim ¼ sinðÞu u 1du: z→0 zπ=2 z→0 sinðÞπz ΓðÞ1 2πz 0

This reduces to a well-known integral

Z∞ u π sin du ¼ : u 2 0

Corollary 2.3 z ¼ 1 ; a ¼ ; b ¼ ; n ¼ ⇒p ¼ ; θ ¼ π Take 2 0 1 2 1 2 into (2.1b) Z∞ Γ 1 ÀÁ k 1 − 2 π sin u2 uk 1du ¼ sin : 1 − 1 k 2k2k 4 0

As k → 1, the integral reduces to

Z∞ pffiffiffiffiffiffi ÀÁ 2π sin u2 du ¼ : 4 0

Corollary 2.4 z ¼ 1 ; k ¼ ; b ¼ ; n ¼ Take 2 1 0 2 into (2.1a); it turns out the Gaussian integral

Z∞ rffiffiffi − au2 1 π e du ¼ ; a > 0: ð2:5Þ 2 a 0

1 In general, for m > 0 and z ¼ m ; k ¼ 1; a ¼ 1; b ¼ 0 into (2.1a), we get Ahmed Journal of the Egyptian Mathematical Society (2020) 28:39 Page 4 of 10

Z∞ e − um du ¼ 1 Γ 1 : m m 0

The integral (2.5) can also be written as Z∞ rffiffiffi π e − au2 du ¼ ; a > : ð : Þ a 0 2 6 −∞

Stirling formula for the k gamma function The Stirling formula is an approximation of the factorial for large n. It associates an appropriate function to the growth of n! which is given as pffiffiffiffiffiffiffiffi n − n ΓðÞ¼n þ 1 n! ≈ n e 2πn; n∈ℕ: ð3:1Þ

In fact, it is quite accurate even for small n; for example, the Stirling formula gives 99% accuracy when compared with the value of 10! A formula similar to the Stirling formula can be obtained for the k gamma function as follows: Theorem 3.1 For k > 0, Re(z)>0, z sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi z =k 2π Γk ðÞ¼z þ 1 : ð3:2Þ e − 2 kz1 k

Proof Consider Z∞ Z∞ k k − t z − t þz ln t Γk ðÞ¼z þ 1 e k t dt ¼ e k dt: ð3:3Þ

0 0

k 0 Now if we let f ðtÞ¼ − t þ z lnt and notice that its critical value is f ðtÞ¼0⇒t k 0 1 0 1 − 2 ¼ z k ¼ awhich gives maximum value f ðaÞ¼ − kz k < 0 fork > 0: Now if we expand the function f(t) by Taylor series around its critical point, we get

2 0 0 ðÞt − a 0 ÀÁ ftðÞ¼faðÞþðÞt − a f ðÞþa f ðÞþa OtðÞ− a 3 : 2! Since a is the critical point of the function, the second term of the series vanishes, and the rest simplifies to ! 2 3 z z k 1 − 2 1 1 ftðÞ¼− þ lnðÞz − z k t − z k þ Ot− z k : k k 2

Substituting it into (3.3) and ignoring the higher order terms, we get 2 Z∞ 1 − 2 1 z − kz k t − z k = − z 2 Γk ðÞ¼z þ 1 z k e k e dt:

Substituting 1 t − z k ¼ y Ahmed Journal of the Egyptian Mathematical Society (2020) 28:39 Page 5 of 10

Z∞ z=k − z − kz1 − 2=k ðÞt − z1=k 2 Γk ðÞ¼z þ 1 z e k e 2 dt: ð3:4Þ

− z1=k

Since the integrand of the integral in (3.4) is a Gaussian curve whose peak, the max- 1 imum value, lies at t ¼ z k so at t < 0 the integral is negligible. Therefore, we can ex- tend the lower limit to −∞ 2 Z∞ 1 − 2 1 z − kz k t − z k = − z 2 Γk ðÞz þ 1 ≈ z k e k e dt: −∞

Using (2.6), we can write as vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi z u π = − z u Γk ðÞ¼z þ 1 z k e k : ð3:5Þ tk 2 z1 − k 2

This simplifies to (3.2) as claimed. Notice that for k → 1, (3.5) reduces to (3.1).

Some integrals representing digamma function The logarithmic derivative of the k gamma function for Re(z), k > 0 is known as k digamma function, denoted and defined as [7]

0 ∂ Γk ðÞz ψk ðÞ¼z logΓkðÞ¼z : ð4:1Þ ∂z Γk ðÞz

Taking the logarithmic derivative of the relation (1.4), we see that ∂ ∂ ∂ Γ ðÞ¼z þ k z þ Γ ðÞz : ∂z log k ∂z log ∂z log k

Using (4.1), we can write

ψ ðÞ¼z þ k ψ ðÞþz 1 : ð : Þ k k z 4 2

The relation (4.2) is sometimes called the functional equation of k digamma function. Remark 4.1 Notice that for k → 1, ψk(z) → ψ(z). A series representation of k digamma function is derived in [3] by taking the logarithmic derivative of the inverse of k—analogue Weierstrass form of the k gamma function Y∞ − z − zγ nk z Γ ðÞ¼z z 1kk e k enk : k z þ nk n¼1

And is given by γ X∞ ψ ðÞ¼z − 1 þ 1 k − þ 1 − 1 : ð : Þ k z k log k nk z þ nk 4 3 n¼1

where γ is the Euler-Mascheroni constant given by the following series form Ahmed Journal of the Egyptian Mathematical Society (2020) 28:39 Page 6 of 10

X∞ γ ¼ 1 − ðÞ¼n : : n log 0 5772156649 n¼1

We can rearrange the series (4.3) to write γ X∞ ψ ðÞ¼z 1 k − þ 1 − 1 : ð : Þ k k log k ðÞn þ k nk þ z 4 4 n¼0 1

Next, we derive some integrals involving k digamma function. Theorem 4.2 Let s > 0 be a real number then for k >0

Z1 xs − 1 1 s þ k s s 1 dx ¼ ψk − ψk ¼ ψk ðÞs − ψk − logðÞ 2 ; ð4:5Þ 1 þ xk 2 2 2 2 k 0 ÀÁ Z1 s − ðÞnþ k x 1 1 − x 1 1 s þ ðÞn þ 1 k s dx ¼ ψk − ψk ; ð4:6aÞ 1 − x2k 2 2 2 0 Z1 X∞ s s n þ 1 n 1 x2 dx ¼ ψ ðÞs − ψ − logðÞ 2 ; ð4:6bÞ xðÞþ xk k k nþ1 k 1 n¼0 2 0

Z∞ − sx 1 s þ 2k s 2 tanhðÞkx e dx ¼ ψk − ψk − : ð4:7Þ 2 4 4 s 0

Proof 1 Using the Taylor series of 1þxk , the LHS of (4.5) becomes

Z1 Z1 xs − 1 X∞ dx ¼ ðÞ− n − 1xkn − kþs − 1dx: þ xk 1 1 n¼1 0 0

Interchanging integral and summation

X∞ Z1 X∞ ¼ ðÞ− n − 1 xkn − kþs − 1dx ¼ ðÞ− n − 1 1 : 1 1 kn − k þ s n¼1 n¼1 0

Rearranging the sum into even and odd terms, we can write X∞ ¼ 1 − 1 : nk − k þ s nk − k þ s n¼1 2 2 2

1 1 Adding and subtracting 2nk under the summation and factoring out 2, we get 0 1 0 1 X∞ B C X∞ B C ¼ 1 @ 1 − 1 A − 1 @ 1 − 1 A: nk s − k nk s − 2k 2 n¼1 nk þ 2 n¼1 nk þ 2 2

Changing the index of the sum, we get Ahmed Journal of the Egyptian Mathematical Society (2020) 28:39 Page 7 of 10

0 1 0 1 X∞ X∞ 1 B 1 1 C 1 B 1 1 C ¼ @ − A − @ − A: ðÞn þ k s þ k ðÞn þ k s 2 n¼0 1 nk þ 2 n¼0 1 nk þ 2 2 Adding and subtracting (logk)/k − γ/k and using (4.4), we get the first equality of (4.5). To prove the second equality of (4.5), we take Legendre duplication k analogous formula r = 2 in corollary 3.14 in [8] 2z − 1 1 − 1 k Γk ðÞ¼2z 2 k 2k2ðÞ2π 2Γk ðÞz Γk z þ : ð4:8Þ 2

Taking the logarithm of (4.8) 2z 1 1 1 logΓkðÞ¼2z − log2 þ logk − logðÞþ 2π logΓkðÞz k 22 2 k þ logΓk z þ : ð4:9Þ 2

Taking derivatives of (4.9) with respect to z and using the definition (4.1), we get 1 k 1 ψk ðÞ¼2z ψk ðÞþz ψk z þ þ logðÞ 2 : ð4:10Þ 2 2 k

Replacing z by s/2 in (4.10) 1 s s þ k 1 ψk ðÞ¼s ψk þ ψk þ logðÞ 2 : 2 2 2 k

Rearranging we get the second equality of (4.5). 1 s þ k s s 1 ψk − ψk ¼ ψk ðÞs − ψk − logðÞ 2 : 2 2 2 2 k

To derive (4.6a), we replace s by s + k, s +2k, s +3k, ⋯s + nk in (4.5); we get

Z1 xsþk − 1 1 s þ 2k s þ k dx ¼ ψk − ψk 1 þ xk 2 2 2 0 Z1 xsþ2k − 1 1 s þ 3k s þ 2k dx ¼ ψk − ψk ð : Þ 1 þ xk 2 2 2 4 11 0 MM Z1 xsþnk − 1 1 s þ ðÞn þ 1 k s þ nk dx ¼ ψk − ψk : 1 þ xk 2 2 2 0

Now observing that the RHS of (4.11) is a telescoping sum, so adding all the (n +1) terms in (4.11), we get ÀÁ Z1 s − k k nk x 1 1 þ x þ x2 þ ⋯x 1 s þ ðÞn þ 1 k s dx ¼ ψk − ψk : ð4:12Þ 1 þ xk 2 2 2 0

The series under the integral on LHS of (4.12) is a finite geometric series with com- k mon ratio x , so summing it, we get (4.6a). In a similar fashion, we can derive (4.6b)by s s s s successively replacing s by = ; = ; = ; ⋯ = n in the second equality of (4.5). 2 22 23 2 Ahmed Journal of the Egyptian Mathematical Society (2020) 28:39 Page 8 of 10

P∞ s Remark 4.3 x n The ratio test shows that the infinite series n¼0 2 is convergent for all real value of x as long as s >0. For the relation (4.7), first, we write the integrand as an exponential function Z∞ Z∞ − 2kx − sx 1 − e − sx tanhðÞkx e dx ¼ e dx: ð4:13Þ 1 þ e − 2kx 0 0 pffiffiffiffi kx ¼ − tk⇒dx ¼ − 1 dt By making the substitution log 2t in (4.13), we can write

Z∞ Z1 Z1 Z1 − tk t s − 1 t sþk − 1 − sx 1 1 s − 1 1 2 1 2 tanhðÞkx e dx ¼ t2 dt ¼ dt − dt: 2 1 þ tk 2 1 þ tk 2 1 þ tk 0 0 0 0

By using the first equality of (4.5), we can write Z∞ − sx 1 1 s þ 2k s 1 s þ 4k s þ 2k tanhðÞkx e dx ¼ ψk − ψk − ψk − ψk : 2 2 4 4 2 4 4 0

After a bit simplification Z∞ − sx 1 s þ 2k 1 s 1 s tanhðÞkx e dx ¼ ψk − ψk − ψk k þ : ð4:14Þ 2 4 2 4 2 4 0

Using the relation (4.2), Eq. (4.14) reduces to the required result (4.7) Remark 4.4 Equation (4.7) can be also be written in the form of Laplace transform of tanh(kx), that is 1 s þ 2k s 2 LðÞ¼tanhðÞkx ψk − ψk − : 2 4 4 s

Special values: Replacing k =1=s in the first equality of (4.5), we can write 1 1 logðÞ¼ 2 ψðÞ1 − ψ : 2 2

Replacing k =2,s = 1 and for k =4,s = 2, we get from (4.5) π 3 1 ¼ ψ − ψ ¼ 2ðÞψ ðÞ3 − ψ ðÞ1 : ð4:15Þ 2 2 2 2 2 4 4

Replacing k = 1/2, s = 1, we get from (4.5)

Z1 1 1 3 1 pffiffiffi dx ¼ 2 − 2 logðÞ¼ 2 ψ1 − ψ1 : 1 þ x 2 2 4 2 2 0

Euler-Mascheroni constant and k digamma function In this section, we represent the Euler-Mascheroni constant in the form k digamma function for different values of k. Proposition 5.1 Substituting k = 4 and z = 1 in (4.3), we see that Ahmed Journal of the Egyptian Mathematical Society (2020) 28:39 Page 9 of 10

γ X∞ ψ ðÞ¼− þ 1 − þ 1 − 1 : ð : Þ 4 1 1 log4 n n þ 5 1 4 4 n¼1 4 4 1

The series in (5.1) is not hard to sum. First, observe that this series can be written as

Z1 Z1 X∞ X∞ ÀÁX∞ ÀÁ 1 − 1 ¼ x4n − 1 − x4n dx ¼ x4n − 1 − x4n dx n n þ n¼1 4 4 1 n¼1 n¼1 0 0 Z1 ! X∞ ÀÁ X∞ ÀÁ ¼ x3 x4 n − 1 − x4 x4 n − 1 dx n¼1 n¼1 0

Now employing the Taylor series

Z1 Z1 x3 x4 x3 ¼ − dx ¼ dx 1 − x4 1 − x4 ðÞ1 þ x ðÞ1 þ x2 0 0 Z1 1 x þ 1 ¼ 1 − − dx: 21ðÞþ x 21ðÞþ x2 0

Evaluating the integral, we get the sum of the series X∞ π 1 − 1 ¼ − − 3 ðÞ: ð : Þ n n þ 1 log 2 5 2 n¼1 4 4 1 4 4

Substituting (5.2) into (5.1), we get a nice representation of the Euler-Mascheroni constant in the form of 4—digamma function π γ ¼ − 4ψ ðÞ1 − − logðÞ 2 : ð5:3Þ 4 2 Or more elegantly π þ γ ¼ − ψ ðÞ− ðÞ: 2 8 4 1 2 log 2

From (4.15) and (5.3), we get some more such representations π γ ¼ − 4ψ ðÞþ3 − logðÞ 2 ; 4 2 γ ¼ − ψ ðÞþψ 3 − ψ 1 − ðÞ; 4 4 3 2 2 log 2 ð5:4Þ 2 2 γ ¼ − ψ ðÞþψ 3 − ψ 1 þ 1 ψ 3 − 1 ψ 1 − : 4 4 3 2 2 1 1 1 2 2 4 2 4 4 2 2

Adding (5.3) and (5.4), we get such representation for π π ¼ ½ψ ðÞ− ψ ðÞ: 4 4 3 4 1

Conclusion In this paper, we established and investigated few new definite integrals involving k gamma function and k digamma function. Known results of the classical gamma function and classical digamma function were obtained as special cases of k gamma function and k digamma function. We also established nice representations of π and of the Euler-Mascheroni constant which were generated, and many more can still be obtained. Ahmed Journal of the Egyptian Mathematical Society (2020) 28:39 Page 10 of 10

Acknowledgements Not applicable.

Author’s contributions Ahmed S. contributed the whole research article and approved the final manuscript.

Funding There are no funding sources for this manuscript.

Availability of data and materials Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

Competing interests The author declares that there are no competing interests.

Received: 31 March 2020 Accepted: 8 July 2020

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