Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 25 December 2018 doi:10.20944/preprints201812.0310.v1 Peer-reviewed version available at Fractal Fract 2019, 3, 1; doi:10.3390/fractalfract3010001 Article Regularized Integral Representations of the Reciprocal G Function Dimiter Prodanov Correspondence: Environment, Health and Safety, IMEC vzw, Kapeldreef 75, 3001 Leuven, Belgium; [email protected]; [email protected] Version December 25, 2018 submitted to Preprints 1 Abstract: This paper establishes a real integral representation of the reciprocal G function in terms 2 of a regularized hypersingular integral. The equivalence with the usual complex representation is 3 demonstrated. A regularized complex representation along the usual Hankel path is derived. 4 Keywords: gamma function; reciprocal gamma function; integral equation 5 MSC: 33B15; 65D20, 30D10 6 1. Introduction 7 Applications of the Gamma function are ubiquitous in fractional calculus and the special function 8 theory. It has numerous interesting properties summarized in [1]. It is indispensable in the theory of 9 Laplace transforms. The history of the Gamma function is surveyed in [2]. In a previous note I have 10 investigated an approach to regularize derivatives at points, where the ordinary limit diverges [5]. 11 This paper exploits the same approach for the purposes of numerical computation of singular integrals, 12 such as the Euler G integrals for negative arguments. The paper also builds on an observations in [4]. 13 The present paper proves a real singular integral representation of the reciprocal G function. The 14 algorithm is implemented in the Computer Algebra System Maxima for reference and demonstration 15 purposes. 16 As a second contribution, the paper provides an integral representation of the Gamma function for 17 negative numbers. Finally, the paper demonstrates an equivalent regularized complex representation 18 based on the regularization of the Heine integral. 19 2. Preliminaries and notation The reciprocal Gamma function is an entire function. Starting from the Euler’s infinite product definition, the reciprocal Gamma function can be defined by the infinite product: 1 z (z + 1) ... (z + n) := lim G(z) n!¥ nz n! Proceeding from the Euler’s reflection formula for negative arguments the reciprocal Gamma function is simply 1 sin pz = − G(z + 1) (1) G(−z) p 20 It is plot is presented in Fig.1. © 2018 by the author(s). Distributed under a Creative Commons CC BY license. Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 25 December 2018 doi:10.20944/preprints201812.0310.v1 Peer-reviewed version available at Fractal Fract 2019, 3, 1; doi:10.3390/fractalfract3010001 2 of 9 Figure 1. 1/G(−z) computed from Eq.1 The Euler’s Gamma function integral representation is valid for real z > 0 or complex numbers, such that Re z > 0 Z ¥ G(z) = e−ttz−1 dt 0 however for negative z the integral diverges. A less well-known integral representations for Re z < 0 is the Cauchy–Saalschütz integral [6][Ch. 3]: Z ¥ e−t − e (−t) (− ) = n G z z+ dt 0 t 1 The Hankel’s representation of the Gamma function is given as 1 Z G(z) = ettz−1dt, z 6= 0, −1, −2, . 2pi sin (pz) Ha− The Heine’s complex representation for the reciprocal Gamma function is well known and is given below: 1 (−1)−z Z e−t 1 Z et = = z dt z dt G(z) 2pi Ha+ t 2pi Ha− t Here Ha− denotes the Hankel contour in the complex z-plane with a cut along the negative real semi-axis arg z = p and Ha+ is its reflection. The contour is depicted in Fig.2. The integrand has a simple pole at t = 0. The Hölder exponent at 0 can be computed in the closed interval [0, e] as log eee−z e lim = −z + lim = −z e!0 log e e!0 log e k e −z 21 Therefore, for k > [z] it holds that lim e e e = 0 and the order of the residue is [z]. This observation e!0 22 is indicative for the statement of the main result of the paper. Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 25 December 2018 doi:10.20944/preprints201812.0310.v1 Peer-reviewed version available at Fractal Fract 2019, 3, 1; doi:10.3390/fractalfract3010001 3 of 9 y A B x E D C Figure 2. The Hankel contour Ha−. 23 2.1. Auxiliary notation 24 Definition 1. For a real number z the notation [z] will mean the integral part of the number, while fzg := 25 z − [z] will denote the non-integral remaining part. Definition 2. The falling factorial is defined as n−1 (z)n := ∏ z − k k=0 Definition 3. Let n xk en(x) := ∑ k=0 k! 26 be the truncated exponent polynomial sum under the convention e−1(x) = 0. 27 3. Theoretical Results Theorem 1 (Real Reciprocal Gamma representation). Let z > 0, z 2/ Z and n = [z]. Then 1 sin pz Z ¥ e−x − e (−x) 1 Z 0 ex − e (x) = n−1 = n−1 z dx Im z dx G(z) p 0 x p −¥ x 28 where the integrals are over the real axis. Proof. First, we establish two preliminary results. Consider the following limit of the form 0/0 and apply n times l’Hôpital s’rule: ex − e (x) 1 ex − 1 = n = Lz lim z lim z−n x!0 x (z)n x!0 x Another application of l’Hôpital s’rule leads to 1 x n+1−z Lz = lim e x (z)n (z − n) x!0 Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 25 December 2018 doi:10.20944/preprints201812.0310.v1 Peer-reviewed version available at Fractal Fract 2019, 3, 1; doi:10.3390/fractalfract3010001 4 of 9 Therefore, 8 0, z < n + 1 <> 1 = + Lz = G(n+1) , z n 1 > : ¥, z > n + 1 Secondly, consider the limit x x n k−z e − en(x) e x Mz = lim = Mz = lim − lim x!−¥ z x!−¥ z ∑ x!−¥ ( + ) x x k=0 G k 1 Therefore, 8 ¥, z < n <> 1 = Mz = G(n+1) , z n > : 0, z > n Therefore, in order for both limits to vanish simultaneously n < z < n + 1. Therefore, n = [z]. Let fzg = z − [z]. Then 0 Z 0 ex − e (x) ex − e (x) ¥ 1 Z 0 (ex − e (x)) ( + ) = n = n + n = In+1 z 1 z+1 dx z z dx −¥ x zx 0 z −¥ x 1 Z 0 ex − e (x) 1 n−1 = ( ) z dx In z z −¥ x z by the above results. Therefore, by reduction 1 1 1 Z 0 ex ( + ) = ( − ) = (f g) = = In+1 z 1 I0 z n I0 z f g dx (z)n (z)n (z)n −¥ x z 1 Z ¥ e−x G(1 − fzg) = −ipfzg f g dx e (z)n 0 (−x) z (z)n Therefore, e−ipfzg G(fzg)In+1(z + 1) = G(1 − fzg)G(fzg) = (z)n p p e−ipfzg = (cot pfzg − i) (z)n sin pfzg (z)n by the Euler’s reflection formula. We take the imaginary part of the integral since G(fzg) is real and the middle expression is imaginary. Therefore, 1 1 1 = = Im In+1(z + 1) G(z + 1) (z)nG(fzg) p Finally, 1 1 Z 0 ex − e (x) = n−1 Im z dx G(z) p −¥ x 29 Corollary 1. By change of variables it holds that k ! Z ¥ 1 n−1 k 1 sin pz 1 − −u z (−1) u z = u z 2 e − du ( ) ∑ G z pz 0 k=0 k! 30 The latter result can be useful for computations with large arguments of G. Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 25 December 2018 doi:10.20944/preprints201812.0310.v1 Peer-reviewed version available at Fractal Fract 2019, 3, 1; doi:10.3390/fractalfract3010001 5 of 9 Corollary 2 (Modified Euler Integral of the second kind). By change of variables it holds that for z > 0 n−1 ( )k u − log u Z 1 ∑ k! Z 1 1 1 k=0 sin pz 1 − en−1 (log u) /u = z du = z du G(z) p 0 u (log u) p 0 (log 1/u) 31 Finally, it is instructive to demonstrate the correspondence between the complex-analytical 32 representation and the hyper-singular representation. Theorem 2 (Regularized complex reciprocal Gamma representation). For z > 0, z 2/ Z and n = [z] 1 1 Z et − e (t) sin pz Z ¥ e−t − e (−t) = n−1 = n−1 z dt z dt G(z) 2pi Ha− t p 0 t 33 where t 2 R . Proof. The proof technique follows [3]. We evaluate the line integral along the Hankel contour: 1 Z et − e (t) ( ) = n−1 In z z dt 2pi Ha− t with kernel t e − e − (t) Ker(t) = n 1 , z > 0 tz The contour is depicted in Fig.2. The integral can be split in three parts Z Z Z Z Ker(t)dt = Ker(t)dt + Ker(t)dt + Ker(t)dt Ha AB BCD DE along the rays AB, DE and the arch BCD, respectively. Along the ray AB where t = reid the kernel becomes n−1 idk−idz k ! 1 id e r = e −idzr − KerA z e ∑ r k=0 k! Along the ray DE where x = re−id the kernel becomes n−1 −idk+idz k ! 1 −id e r = e +idzr − KerB z e ∑ r k=0 k! Therefore, KerA − KerB = ! ! ! 2 i n−1 cos (dk) rk n−1 sin (dk) rk − cos (d)r ( − ( ) ) − ( ) + ( ) z e sin dz sin d r ∑ sin dz ∑ cos dz r k=0 k! k=0 k! Therefore, 1 Z 0 sin(pz) Z ¥ e−z − e (−r) ( − ) = n−1 lim KerA KerB dr z dr d!p 2pi ¥ p 0 r The integral on the arc BCD is given by the Cauchy Residue Theorem.