Regularized Integral Representations of the Reciprocal Function

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Article Regularized Integral Representations of the Reciprocal Γ 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 Γ 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 Γ 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 Γ 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 Γ(z) n→∞ nz n!

Proceeding from the Euler’s reﬂection formula for negative arguments the reciprocal Gamma function is simply 1 sin πz = − Γ(z + 1) (1) Γ(−z) π

20 It is plot is presented in Fig.1.

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Figure 1. 1/Γ(−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 ∞ Γ(z) = e−ττz−1 dτ 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−τ − e (−τ) (− ) = n Γ z z+ dτ 0 τ 1

The Hankel’s representation of the Gamma function is given as

1 Z Γ(z) = eττz−1dτ, z 6= 0, −1, −2, . . . 2πi sin (πz) Ha−

The Heine’s complex representation for the reciprocal Gamma function is well known and is given below: 1 (−1)−z Z e−τ 1 Z eτ = = z dτ z dτ Γ(z) 2πi Ha+ τ 2πi Ha− τ Here Ha− denotes the Hankel contour in the complex ζ-plane with a cut along the negative real semi-axis arg ζ = π and Ha+ is its reﬂection. The contour is depicted in Fig.2. The integrand has a simple pole at τ = 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

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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 {z} :=

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 ∈/ Z and n = [z]. Then

1 sin πz Z ∞ e−x − e (−x) 1 Z 0 ex − e (x) = n−1 = n−1 z dx Im z dx Γ(z) π 0 x π −∞ 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

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Therefore,  0, z < n + 1  1 = + Lz = Γ(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 Γ k 1

Therefore,  ∞, z < n  1 = Mz = Γ(n+1) , z n   0, z > n Therefore, in order for both limits to vanish simultaneously n < z < n + 1. Therefore, n = [z]. Let {z} = 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 ( + ) = ( − ) = ({ }) = = In+1 z 1 I0 z n I0 z { } dx (z)n (z)n (z)n −∞ x z 1 Z ∞ e−x Γ(1 − {z}) = −iπ{z} { } dx e (z)n 0 (−x) z (z)n

Therefore,

e−iπ{z} Γ({z})In+1(z + 1) = Γ(1 − {z})Γ({z}) = (z)n π π e−iπ{z} = (cot π{z} − i) (z)n sin π{z} (z)n

by the Euler’s reﬂection formula. We take the imaginary part of the integral since Γ({z}) is real and the middle expression is imaginary. Therefore,

1 1 1 = = Im In+1(z + 1) Γ(z + 1) (z)nΓ({z}) π

Finally, 1 1 Z 0 ex − e (x) = n−1 Im z dx Γ(z) π −∞ x

Corollary 1. By change of variables it holds that

k ! Z ∞ 1 n−1 k 1 sin πz 1 − −u z (−1) u z = u z 2 e − du ( ) ∑ Γ z πz 0 k=0 k!

30 The latter result can be useful for computations with large arguments of Γ. Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 25 December 2018 doi:10.20944/preprints201812.0310.v1

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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 πz 1 − en−1 (log u) /u = z du = z du Γ(z) π 0 u (log u) π 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 ∈/ Z and n = [z]

1 1 Z eτ − e (τ) sin πz Z ∞ e−τ − e (−τ) = n−1 = n−1 z dτ z dτ Γ(z) 2πi Ha− τ π 0 τ

33 where τ ∈ R .

Proof. The proof technique follows [3]. We evaluate the line integral along the Hankel contour:

1 Z eτ − e (τ) ( ) = n−1 In z z dτ 2πi Ha− τ

with kernel τ e − e − (τ) Ker(τ) = n 1 , z > 0 τz The contour is depicted in Fig.2. The integral can be split in three parts Z Z Z Z Ker(τ)dτ = Ker(τ)dτ + Ker(τ)dτ + Ker(τ)dτ Ha AB BCD DE

along the rays AB, DE and the arch BCD, respectively. Along the ray AB where τ = reiδ the kernel becomes n−1 iδk−iδz k ! 1 iδ e r = e −iδzr − KerA z e ∑ r k=0 k! Along the ray DE where ξ = re−iδ the kernel becomes

n−1 −iδk+iδz k ! 1 −iδ e r = e +iδzr − KerB z e ∑ r k=0 k!

Therefore,

KerA − KerB = ! ! ! 2 i n−1 cos (δk) rk n−1 sin (δk) rk − cos (δ)r ( − ( ) ) − ( ) + ( ) z e sin δz sin δ r ∑ sin δz ∑ cos δz r k=0 k! k=0 k!

Therefore, 1 Z 0 sin(πz) Z ∞ e−z − e (−r) ( − ) = n−1 lim KerA KerB dr z dr δ→π 2πi ∞ π 0 r The integral on the arc BCD is given by the Cauchy Residue Theorem. By the above observation the residue at τ = 0 is given by the limit

1−z τ Res[Ker](τ) = lim τ (e − en−1(τ)) = lim τLz = 0, z ≤ n + 1 τ→0 τ→0 Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 25 December 2018 doi:10.20944/preprints201812.0310.v1

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since L = 0. Therefore, z Z Ker(τ)dτ = 0 BCD Furthermore, after integration by parts

τ Z τ e − en−1(τ) 1 e − en−2(τ) 1 In(z) = − + dτ = I − (z) z−1 − z−1 n 1 τ Ha− 2πi(z − 1) Ha τ z − 1

34 since Mz = 0. Therefore, the claim follows by reduction to n = 0.

35 4. Applications

36 4.1. Laplace transform pairs

Consider the Laplace transform Ls : f (t) ÷ F(s) As a concrete application of Th.2 consider the pair

k Γ(k + 1) Ls : t ÷ , k > 0 sk+1 The inverse Laplace transform can be calculated simply as

Z ts − ( ) −1 Γ(k + 1) 1 e e[k] ts Lt : + ÷ Γ(k + 1) + ds = sk 1 2πi Ha− sk 1 Z 0 ts − ( ) sin π(k + 1) e e[k] ts k Γ(k + 1) + ds = t πi −∞ sk 1

37 by change of variables. The latter result can be used for numerical inversion of Laplace transforms.

38 4.2. Ratios of Gamma functions

39 On a second place, the ratio of two Gamma functions can be represented as

Proposition 1. Let A, B > 0. Then

Γ(A) 1 Z 1 Z 1 1 − e (log u) /u = n−1 (− )A−1 B log v du dv Γ(B) π 0 0 (log u)

40 where n = [B].

Proof.

Γ(A + 1) 1 Z 1 1 − e (log u) /u Z 1 = n−1 (− )A = B du log v dv Γ(B) π 0 (log u) 0 1 Z 1 Z 1 1 − e (log u) /u n−1 (− )A B log v du dv π 0 0 (log u)

42 4.3. The Cauchy–Saalschütz integral

43 Finally, for negative arguments: Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 25 December 2018 doi:10.20944/preprints201812.0310.v1

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Proposition 2. For z > 0 it holds that

1 Z ∞ e−x − e (−x) (− ) = − n−1 Γ z z dx z 0 x

Proof. By the reﬂection formula π Γ(1 − z)Γ(z) = = −zΓ(−z)Γ(z) sin πz Therefore, π 1 1 Z ∞ e−x − e (−x) (− ) = − = − n−1 Γ z z dx z sin πz Γ(z) z 0 x

Remark 1. The latter result is equivalent to the classical Cauchy–Saalschütz integral representation. Indeed, by integration by parts

1 Z ∞ e−x − e (−x) 1 Z ∞ d(e−x − e (−x)) (− ) = − n−1 = n = Γ z z dx z z 0 x z 0 x e−x − e (−x) ∞ −z Z ∞ e−x − e (−x) Z ∞ e−x − e (−x) n − n = n z z+1 dx z+1 dx x 0 z 0 x 0 x

45 which is the Cauchy–Saalschütz integral.

46 5. Numerical Implementation

47 A reference implementation in the Computer Algebra System Maxima is given in Listings1 and2. 48 The numerical integration code uses the library Quadpack, distributed with Maxima. The reference 49 implementation given in this section uses a routine for semi-inﬁnite interval integration with tunable 50 relative error of approximation (i.e. the epsrel parameter). A plot of the reciprocal Γ function computed 51 from Listing1 is presented in Fig.3.

Figure 3. 1/Γ(z) computed from Th.1

52 Fig.4 represents a plot of Γ(−z) computed from Listing1. Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 25 December 2018 doi:10.20944/preprints201812.0310.v1

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Figure 4. Γ(−z) computed from Prop.2

Listing 1: The Maixma code corresponsing to Th.1

53 1 54 Kg( z ) : =block([ u, ret, k, n, fr:1, ss:0], 55 i f not numberp(z) then return(’Kg(z)) , 56 i f integerp(z) then if z=0 then return(0) else return(1/(z − 1)!) 57 e l s e( 58 6 f r : sin (%pi ∗z)/%pi , 59 n : f i x ( z ) , 60 ss : sum( (−u)^k/k!,k,0, n−1) , 61 r e t : f r ∗ first(quad_qagi ( (exp(−u)− ss)/ u^(z) , u, 0, inf, ’epsrel=1d−8)) 62 ), 6311 f l o a t ( r e t ) 64 );

Listing 2: The Maixma code corresponsing to Corr.2

65 66 Kgn( z ) : =block([ u, ret, k, n, fr:1, ss:0], 67 3 i f not numberp(z) then return(’Kgn(z)) , 68 i f integerp(z) then return(0) 69 e l s e( 70 i f z<0 then z:−z , 71 f r :−1/z , 72 8 n : f i x ( z ) , 73 ss : sum( (−u)^k/k!, k, 0, n−1) , 74 r e t : f r ∗ first(quad_qagi ( (exp(−u)− ss)/ u^(z) , u, 0, inf, ’epsrel=1d−8)) 75 ), 76 f l o a t ( r e t ) 7713 );

78 Acknowledgments: The work has been supported in part by a grants from Research Fund - Flanders (FWO), 79 contract number VS.097.16N and the COST Association Action CA16212 INDEPTH. Plots are prepared with the 80 computer algebra system Maxima.

81 References

82 1. J. M. Borwein and R. M. Corless. Gamma and factorial in the monthly. Am. Math. Monthly 2018, 125, 400–424. 83 2. P. J. Davis. Leonhard Euler’s integral: A historical pro

84 le of the Gamma function: In memo-riam: Milton Abramowitz. Am. Math. Monthly 1959, 66, 849–869. Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 25 December 2018 doi:10.20944/preprints201812.0310.v1

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85 3. Y. Luchko. Algorithms for evaluation of the Wright function for the real arguments’ values. Fract. Calc. Appl. 86 Anal. 2008, 11, 57–75. 87 4. F. Mainardi. Fractional Calculus and Waves in Linear Viscoelasticity; Imperial College Press, 2010. 88 5. D. Prodanov. Regularization of derivatives on non-differentiable points. Journal of Physics: Conference Series 89 2016, 701, 012031. 90 6. Temme. Special Functions; John Wiley & Sons, 1996.