Some integrals involving k gamma and k digamma function

In this paper, some new integrals involving k gamma function and k 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.

The following properties of the k gamma function have been discussed in [5,6] Γ k z þ k ð Þ¼zΓ k z ð Þ; ð1:4Þ 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 Similarly, for the conjugate of p, we can write where θ and |p| are the principal argument and modulus of p, respectively, so that (2.4) reduces to By Euler's identity, we can write 1b), and using the relation (1.5) together with k = 1, we see that This reduces to a well-known integral As k → 1, the integral reduces to ; a > 0: ð2:5Þ In general, for m > 0 and z ¼ 1 The integral (2.5) can also be written as 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 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: Now if we let f ðtÞ ¼ − t k k þ z ln t and notice that its critical value is f 0 ðtÞ ¼ 0⇒t expand the function f(t) by Taylor series around its critical point, we get Since a is the critical point of the function, the second term of the series vanishes, and the rest simplifies to Substituting it into (3.3) and ignoring the higher order terms, we get Since the integrand of the integral in (3.4) is a Gaussian curve whose peak, the maximum value, lies at t ¼ z 1 k so at t < 0 the integral is negligible. Therefore, we can extend the lower limit to −∞ Using (2.6), we can write as 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] Taking the logarithmic derivative of the relation (1.4), we see that Using (4.1), we can write The relation (4.2) is sometimes called the functional equation of k digamma function. Remark 4.1 Notice that for k → 1, ψ k (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 And is given by where γ is the Euler-Mascheroni constant given by the following series form Next, we derive some integrals involving k digamma function. Theorem 4.2 Let s > 0 be a real number then for k > 0 ð4:6aÞ Proof Using the Taylor series of 1 1þx k , the LHS of (4.5) becomes

Interchanging integral and summation
Rearranging the sum into even and odd terms, we can write Adding and subtracting 1 2nk under the summation and factoring out 1 2 , we get Changing the index of the sum, we get Ahmed Journal of the Egyptian Mathematical Society (2020) 28:39 Page 6 of 10 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] Taking the logarithm of (4.8) Taking derivatives of (4.9) with respect to z and using the definition (4.1), we get Replacing z by s/2 in (4.10) Rearranging we get the second equality of (4.5).
To derive (4.6a), we replace s by s + k, s + 2k, s + 3k, ⋯s + nk in (4.5); we get ð4:11Þ Now observing that the RHS of (4.11) is a telescoping sum, so adding all the (n + 1) terms in (4.11), we get The series under the integral on LHS of (4.12) is a finite geometric series with common ratio x k , so summing it, we get (4.6a). In a similar fashion, we can derive (4.6b) by successively replacing s by s = 2 ; s = 2 2 ; s = 2 3 ; ⋯ s = 2 n in the second equality of (4.5). For the relation (4.7), first, we write the integrand as an exponential function By making the substitution kx ¼ − log ffiffiffiffi t k p ⇒dx ¼ − 1 2t dt in (4.13), we can write By using the first equality of (4.5), we can write After a bit simplification 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 Special values: Replacing k = 1 = s in the first equality of (4.5), we can write log 2 ð Þ ¼ 1 2 ψ 1 ð Þ − ψ 1 2