Extended de Montmort-Prudnikov Sum

Abstract

Using a contour integral method, the de Montmort-Prudnikov sum is extended to derive a new series representation involving the incomplete gamma function. The series is uniformly convergent and completely analytical, which can be evaluated for general complex ranges of the parameters involved. Applications and evaluations of this formula are discussed.

1. Theoretical Background and Preliminaries

In this work we apply the contour integral method in [1] to the derangement polynomial generating function given by

$$\frac{e^{-t}}{1-t}{e}^{xt}=\sum _{n\ge 0}\left(\Gamma (n+1)\sum _{m\in [0,n]}\frac{{(x-1)}^m}{m!}\right)\frac{t^n}{\Gamma (n+1)}$$

(1)

which is listed as Equation (8) in [2]. Simplifying using Equations (6) and (12.1) in [3] and (16) in [4] yields the infinite sum of the incomplete gamma function Cauchy contour integral representation given by

$$\frac{1}{2\pi i}{\int }_C\sum _{n\ge 0}\frac{e^{x-1}{a}^w{w}^{-k+n-1}\Gamma (n+1,x-1)}{\Gamma (n+1)}dw=-\frac{1}{2\pi i}{\int }_C\frac{a^w{w}^{-k-1}{e}^{w(x-1)}}{w-1}dw$$

(2)

The work involved in counting derangements was first conducted by Pierre Rémond de Montmort in 1708 (see [5,6]). A derangement is a permutation of the elements of a set, where none of the elements appear in their original position, in other words, a permutation with no fixed points. The gamma function is used in mathematics, precision sciences, and engineering. In particular, solid-state physics and statistics use the incomplete gamma function, whereas discrete mathematics, number theory, and other scientific disciplines use the gamma function’s logarithm. P. Schlömilch [7] first introduced the name “incomplete gamma function” for integrals previously defined by A.M. Legendre [8] and Euler [9] who studied the incomplete Euler integrals. These functions were studied by J. Tannery [10], F. E. Prym [11], and M. Lerch [12] (who presented the incomplete gamma function as a series). N. Nielsen [13] and other mathematicians also had special interests in these functions. In this letter, we discuss the role the incomplete gamma function plays in string theory. Currently, the incomplete gamma function in the field of string theory is evaluated using multiple sums and asymptotic expansions [14]. In this work, we have taken a more direct and simplified approach by deriving the infinite sum of the incomplete gamma function in terms of the incomplete gamma function and other special functions. In our opinion, this could be viewed as an improvement on current methods of evaluation of this special function and, in turn, assist in producing more accurate evaluations where applicable in the field of String theory. The incomplete gamma function has applications in Cosmology. This function is used to evaluate quantities such as the Hawking radiation [15], mass, radius of a black hole and the black hole emission spectrum [16]. The incomplete gamma function is also used in the evaluation of the noncommutative version of the linearized Schwarzschild line element [17]. This special function is also used in the improved evaluation of the cross-section of small black holes where the impact parameter in the Parton-Parton collision is small (Equation (10), page 365 in [18]). The evaluation of the partition function in determining the 9-dimensional spatial volume in Gauge theory [19] is based on the incomplete gamma function. This function appears in the evaluation of the equation governing the energy density in the field of inflationary cosmological perturbations of quantum-mechanical origin [20]. The evaluation of weak turbulence in the spatial distribution of accelerated particles in Galactic cosmic rays and supernova remnants involves the incomplete gamma function where the first parameter is unity [21].

2. Introduction

In this paper, we derive the infinite sum involving the incomplete gamma function given by

$$\sum _{n\ge 1}\frac{{(-1)}^n{t}^n{(1-k)}_{n-2}\Gamma (n,x)}{\Gamma \left(n\right)}=\frac{e^{\frac{1}{t}}{t}^{k+1}\Gamma \left(k+1,x+\frac{1}{t}\right)}{k}$$

(3)

where the parameters $k,x,t$ are general complex numbers and $-1<Re\left(t\right)<1$. This infinite sum will be used to derive special cases in terms of special functions. The derivations follow the method used by us in [1]. This method involves using a form of the generalized Cauchy’s integral formula given by

$$\frac{y^k}{\Gamma (k+1)}=\frac{1}{2\pi i}{\int }_C\frac{e^{wy}}{w^{k+1}}dw,$$

(4)

where C is in general an open contour in the complex plane where the bilinear concomitant has the same value at the end points of the contour. We then multiply both sides by a function of then take an infinite sum of both sides. This yields an infinite sum in terms of a contour integral. Then we multiply both sides of Equation (4) by another function of y and take the infinite sum of both sides such that the contour integral of both equations are the same.

3. The Incomplete Gamma Function

The incomplete gamma functions Equation (8.4.13) in [22], $\gamma (a,z)$ and $\Gamma (a,z)$, are defined by

$$\gamma (a,z)={\int }_0^z{t}^{a-1}{e}^{-t}dt$$

(5)

and

$$\Gamma (a,z)={\int }_z^{\infty }{t}^{a-1}{e}^{-t}dt$$

(6)

where $Re\left(a\right)>0$. The incomplete gamma function has a recurrence relation given by

$$\gamma (a,z)+\Gamma (a,z)=\Gamma \left(a\right)$$

(7)

where $a\ne 0,-1,-2,\dots$ The incomplete gamma function is continued analytically by

$$\gamma (a,z{e}^{2m\pi i})=e^{2\pi mia}\gamma (a,z)$$

(8)

and

$$\Gamma (a,z{e}^{2m\pi i})=e^{2\pi mia}\Gamma (a,z)+(1-e^{2\pi mia})\Gamma \left(a\right)$$

(9)

where $m\in \mathbb{Z}$, ${\gamma }^{*}(a,z)=\frac{z^{-a}}{\Gamma \left(a\right)}\gamma (a,z)$ is entire in z and a. When $z\ne 0$, $\Gamma (a,z)$ is an entire function of a and $\gamma (a,z)$ is meromorphic with simple poles at $a=-n$ for $n=0,1,2,\dots$ with residue $\frac{{(-1)}^n}{\Gamma (n+1)}$. These definitions are listed in Section 8.2(i) and (ii) in [22]. The incomplete gamma functions are particular cases of the more general hypergeometric and Meijer G functions see Section (5.6) and Equation (6.9.2) in [23]. Some Meijer G representations we will use in this work are given by;

$$\Gamma (a,z)=\Gamma \left(a\right)-G_{1,2}^{1,1}\left(z\left|\begin{array}{c}1\\ a,0\end{array}\right.\right)$$

(10)

and

$$\Gamma (a,z)=G_{1,2}^{2,0}\left(z\left|\begin{array}{c}1\\ 0,a\end{array}\right.\right)$$

(11)

from Equations (2.4) and (2.6a) in [24]. We will also use the derivative notation given by;

$$\frac{\partial \Gamma (a,z)}{\partial a}=\Gamma (a,z)log\left(z\right)+G_{2,3}^{3,0}\left(z\left|\begin{array}{c}1,1\\ 0,0,a\end{array}\right.\right)$$

(12)

from Equations (2.19a) in [24], (9.31.3) in [25] and Equations (5.11.1), (6.2.11.1) and (6.2.11.2) in [26], and (6.36) in [27].

4. Left-Hand Side Contour Integral

We use the method in [1]. Using a generalization of Cauchy’s integral formula we form the infinite sum by replacing y by $log\left(a\right)$ and k by $k-n$ then multiply both sides by $\frac{e^{x-1}\Gamma (n+1,x-1)}{\Gamma (n+1)}$ then take the infinite sum of both sides over $n\in [0,\infty )$ to get

$$\begin{array}{c}\sum _{n\ge 0}\frac{e^{x-1}\Gamma (n+1,x-1){log}^{k-n}\left(a\right)}{\Gamma (n+1)(k-n)!}\hfill \\ =\frac{1}{2\pi i}\sum _{n\ge 0}{\int }_C\frac{e^{x-1}{a}^w{w}^{-k+n-1}\Gamma (n+1,x-1)}{\Gamma (n+1)}dw\\ =\frac{1}{2\pi i}{\int }_C\sum _{n\ge 0}\frac{e^{x-1}{a}^w{w}^{-k+n-1}\Gamma (n+1,x-1)}{\Gamma (n+1)}dw\\ \hfill =-\frac{1}{2\pi i}{\int }_C\frac{a^w{w}^{-k-1}{e}^{w(x-1)}}{w-1}dw\end{array}$$

(13)

from Equation (8) in [2] where $Re\left(w\right)>0$. We can switch the order of summation and integral using Tonelli’s theorem for sums and integrals see page 179 in [28], since the summation is of the bounded measure over the space $\mathbb{C}\times [0,\infty )$.

5. Right-Hand Side Contour Integral

Using a generalization of Cauchy’s integral formula we form the incomplete gamma function contour integral representation by replacing y by $y+log\left(a\right)$ then multiply both sides by $e^{-y}$ then take the infinite integral of both sides over $y\in [y,\infty )$ to get

$$\frac{a\Gamma (k+1,y+log(a\left)\right)}{\Gamma (k+1)}=-\frac{1}{2\pi i}{\int }_C\frac{a^w{w}^{-k-1}{e}^{(w-1)y}}{w-1}dw$$

(14)

Next we multiply both sides by $e^y$ and replace y by $x-1$ to get

$$\frac{a{e}^{x-1}\Gamma (k+1,x+log\left(a\right)-1)}{\Gamma (k+1)}=\frac{1}{2\pi i}{\int }_C\frac{a^w{w}^{-k-1}{e}^{w(x-1)}}{1-w}dw$$

(15)

from Equation (3.382.4) in [25] where $Im\left(w\right)>0$.

6. Main Results

In this section, we will derive the main theorem involving the incomplete gamma function. We will also evaluate this formula and show its application to interesting work in current literature. We also derive infinite sum formula to be added to Section (5.2.5) in [29] and provide a corrected version of one formula listed as Equation (5.2.5.1) in [29].

Theorem 1.

For all $k,a,x\in \mathbb{C}$ then,

$$\sum _{n\ge 0}\frac{{\left(-\frac{1}{a}\right)}^n\Gamma (1+n,x){(1-k)}_{n-1}}{\Gamma (n+1)}=-\frac{e^a\Gamma (1+k,a+x)}{k{a}^k}$$

(16)

Proof. 

Since the right-hand side of Equation (13) is equivalent to the right-hand side of Equation (15) we may equate the left-hand sides and simplify the Pochhammer symbol using Equation (5.2.5) in [22] and $\frac{1}{{\left(k\right)}_{1-n}}={(-1)}^{n+1}{(1-k)}_{n-1}$ to yield the stated result. □

Theorem 2.

For all $k,x,t$ with $-1<Re\left(t\right)<1$ and $-1<Im\left(t\right)<1$ then,

$$\sum _{n\ge 1}\frac{{(-1)}^n{t}^n{(1-k)}_{n-2}\Gamma (n,x)}{\Gamma \left(n\right)}=\frac{e^{\frac{1}{t}}{t}^{k+1}\Gamma \left(k+1,x+\frac{1}{t}\right)}{k}$$

(17)

Proof. 

Use Equation (16) and replace $n\to n-1,a\to 1/t$ and simplify. □

Example 1.

Derivation of entry (5.2.5.1) in [29].

$$\sum _{k\ge 1}{(-t)}^k\Gamma (k,x)=e^{\frac{1}{t}}\left(\mathit{Ei}\left(-\frac{tx+1}{t}\right)-i\pi \right)$$

(18)

Proof. 

Use Equation (16) and set $k=-1,a=1/t,n=k-1$ and simplify the right-hand side using Equations (8.4.4) and (8.4.13) in [22]. Note the current formula is in error. □

Example 2.

Derivation of new entry (5.2.5.10) in [29].

$$\sum _{n\ge 0}\frac{{\left(-\frac{1}{t}\right)}^n\Gamma (1+n-x)\Gamma (1+n,-t)}{\Gamma (1+n)\Gamma (2-x)}=-\frac{e^t{t}^{1-x}\Gamma \left(x\right)}{-1+x}$$

(19)

Proof. 

Use Equation (16) and set $x=-a,k=x-1,a=t$ and simplify using Equation (8.2.2) in [22]. □

Example 3.

Derivation of new entry (5.2.5.11) in [29].

$$\sum _{k\ge 1}k{(-t)}^k\Gamma (k,x)=-\frac{e^{-x}}{tx+1}+\frac{e^{\frac{1}{t}}\left(-\mathit{Ei}\left(-\frac{tx+1}{t}\right)-i\pi \right)}{t}$$

(20)

Proof. 

Use Equation (18) and take the first partial derivative with respect to t and multiply both sides by t and simplify. □

Example 4.

Derivation of new entry (5.2.5.12) in [29]. The double sum product of two incomplete gamma functions over independent variables.

$$\sum _{n\ge 1}\sum _{p\ge 1}{\left(-\frac{1}{t}\right)}^{n+p}\Gamma (n,x)\Gamma (p,y)=e^{2t}\Gamma (0,t+x)\Gamma (0,t+y)$$

(21)

Proof. 

Use Equation (16) and replace $n\to p,x\to y,k\to 1/k$ to form a second equation. Next, multiply both equations and set $k=-1,a=t,n=n-1,p=p-1$ and simplify. □

Example 5.

Derivation of a Mellin transform as an entry (5.2.5.13) in [29].

$$\sum _{n\ge 0}\frac{{(-1)}^n{t}^n\Gamma (1+n+s){(1-k)}_{-1+n}}{\Gamma (1+n)}=-\frac{s{e}^{1/t}{t}^k}{k}{\int }_0^{\infty }{x}^{-1+s}\Gamma \left(1+k,\frac{1}{t}+x\right)dx$$

(22)

Proof. 

Use Equation (16) and multiply both sides by $x^{s-1}$ and take the definite integral of both sides over $x\in [0,\infty )$ and simplify the left-hand side using Equation (8.14.4) in [22]. □

Example 6.

The infinite sum of the hypergeometric function.

$$\begin{array}{c}\sum _{n\ge 0}\frac{{\left(-\frac{1}{a}\right)}^n\Gamma (1+n+s){}_2{F}_1\left(1,1+n+s;1+s;\frac{s}{1+s}\right){(1-k)}_{-1+n}}{s\Gamma (1+n){(1+s)}^{n+s+1}}\hfill \\ \hfill =-\frac{e^a}{a^{k}k}{\int }_0^{\infty }{e}^{-sx}{x}^{-1+s}\Gamma (1+k,a+x)dx\end{array}$$

(23)

Proof. 

Use Equation (16) and multiply both sides by $x^{s-1}{e}^{-sx}$ and take the definite integral of both sides over $x\in [0,\infty )$ and simplify the left-hand side using Equation (8.14.6) in [22]. □

Example 7.

The Laplace transform of the incomplete gamma function.

$$\sum _{n\ge 0}{\left(-\frac{1}{t}\right)}^n\left(1-{(1+\beta )}^{-1-n}\right){(1-k)}_{-1+n}=-\frac{e^t\beta }{t^{k}k}{\int }_0^{\infty }{e}^{-x\beta }\Gamma (1+k,t+x)dx$$

(24)

Proof. 

Use Equation (16) and multiply both sides by $e^{-bx}$ and take the definite integral of both sides over $x\in [0,\infty )$ and simplify the left-hand side using Equation (8.14.2) in [22] and replace $a\to b,b\to \beta$. □

Example 8.

A special case of the derivative.

$$\sum _{k\ge 2}\frac{{(-1)}^k{t}^k\Gamma (k,x)}{k-1}=t{e}^{-x}\left(log\left(t\left(\frac{1}{t}+x\right)\right)+e^{\frac{1}{t}+x}\Gamma \left(0,x+\frac{1}{t}\right)\right)$$

(25)

Proof. 

Use Equation (16) to multiply both sides by k and take the first partial derivative with respect to k and set $k=0$ and replace $a\to 1/t,n\to k-1$ and simplify. □

Example 9.

A double sum representation for the incomplete gamma function.

$$\sum _{n\ge 0}\sum _{j\in [0,n]}\frac{{(-1)}^n{x}^j{t}^n{(1-k)}_{n-1}}{\Gamma (j+1)}=-\frac{t^k{e}^{\frac{1}{t}+x}\Gamma \left(k+1,x+\frac{1}{t}\right)}{k}$$

(26)

Proof. 

Use Equation (17) and simplify the left-hand side using Equation (8.4.11) in [22]. □

Example 10.

A double sum representation for the gamma function $\Gamma \left(x\right)$.

$$\begin{array}{c}\sum _{n\ge 0}\sum _{p\ge 0}\frac{{\left(-\frac{1}{t}\right)}^n{\left(-\frac{1}{s}\right)}^p\Gamma (n+1,-t)\Gamma (n-x+1)\Gamma (p+1,-s)\Gamma \left(p-x+\frac{1}{2}\right)}{\Gamma (n+1)\Gamma (p+1)\Gamma \left(\frac{3}{2}-x\right)\Gamma (2-x)}\hfill \\ \hfill =\sqrt{\pi }{2}^{3-2x}{e}^{s+t}{s}^{\frac{1}{2}-x}{t}^{1-x}\Gamma (2x-2)\end{array}$$

(27)

Proof. 

Use Equation (19) and form a second equation by replacing $x\to x+1/2,n\to p,t\to s$ and multiply and simplify. □

Example 11.

An infinite sum involving Euler’s constant γ.

$$\begin{array}{c}\sum _{k\ge 1}{(-1)}^k{t}^k({\psi }^{\left(0\right)}\left(k\right)+\gamma -1)\Gamma (k,x)\hfill \\ \hfill =e^{\frac{1}{t}}\left(G_{2,3}^{3,0}\left(x+\frac{1}{t}\left|\begin{array}{c}1,1\\ 0,0,0\end{array}\right.\right)+log\left((tx+1)e\right)\Gamma \left(0,x+\frac{1}{t}\right)\right)\end{array}$$

(28)

Proof. 

Use Equation (16) and take the first partial derivative with respect to k and set $k=-1$ and simplify using Equation (12). □

Example 12.

The ratio of infinite sums of the incomplete gamma function. examples and applications of such formulae are listed in [30,31].

$$\frac{{\displaystyle \sum _{n\ge 0}}\frac{{\left(-\frac{1}{a}\right)}^n\Gamma (n+1,-a){(2-k)}_{n-1}}{\Gamma (n+1)}}{{\displaystyle \sum _{p\ge 0}}\frac{{\left(-\frac{1}{b}\right)}^p\Gamma (p+1,-b){(1-k)}_{p-1}}{\Gamma (p+1)}}=\frac{e^{a-b}{a}^{1-k}{b}^k}{k-1}$$

(29)

Proof. 

Use Equation (16) and replace $x\to -a,k\to k-1$ to form a second equation. In the second equation replace $k\to k+1,n\to p,a\to b$ and divide both and simplify. □

7. Discussion

In this work, we proposed a method involving the derivation of the infinite sum of the incomplete gamma function in terms of the incomplete gamma function. We also provided a few examples we consider interesting in terms of infinite sums similar to those published by Prudnikov et al. [29]. This work was summarized in

Table 1. Summary Table of Infinite Sums.

${\displaystyle \sum _{n\ge 1}}\frac{{(-1)}^n{t}^n{(1-k)}_{n-2}\Gamma (n,x)}{\Gamma \left(n\right)}=\frac{e^{\frac{1}{t}}{t}^{k+1}\Gamma \left(k+1,x+\frac{1}{t}\right)}{k}$

${\displaystyle \sum _{k\ge 1}}{(-t)}^k\Gamma (k,x)=e^{\frac{1}{t}}\left(\mathrm{Ei}\left(-\frac{tx+1}{t}\right)-i\pi \right)$

${\displaystyle \sum _{n\ge 0}}\frac{{\left(-\frac{1}{t}\right)}^n\Gamma (1+n-x)\Gamma (1+n,-t)}{\Gamma (1+n)\Gamma (2-x)}=-\frac{e^t{t}^{1-x}\Gamma \left(x\right)}{-1+x}$

${\displaystyle \sum _{k\ge 1}}k{(-t)}^k\Gamma (k,x)=-\frac{e^{-x}}{tx+1}+\frac{e^{\frac{1}{t}}\left(-\mathrm{Ei}\left(-\frac{tx+1}{t}\right)-i\pi \right)}{t}$

${\displaystyle \sum _{n\ge 1}}{\displaystyle \sum _{p\ge 1}}{\left(-\frac{1}{t}\right)}^{n+p}\Gamma (n,x)\Gamma (p,y)=e^{2t}\Gamma (0,t+x)\Gamma (0,t+y)$

${\displaystyle \sum _{n\ge 0}}\frac{{(-1)}^n{t}^n\Gamma (1+n+s){(1-k)}_{-1+n}}{\Gamma (1+n)}=-{\int }_0^{\infty }\frac{s{e}^{1/t}{t}^k{x}^{-1+s}\Gamma \left(1+k,\frac{1}{t}+x\right)}{k}dx$

${\displaystyle \sum _{n\ge 0}}\frac{{\left(-\frac{1}{a}\right)}^n\Gamma (1+n+s){}_2{F}_1\left(1,1+n+s;1+s;\frac{s}{1+s}\right){(1-k)}_{-1+n}}{s\Gamma (1+n){(1+s)}^{n+s+1}}=-{\int }_0^{\infty }\frac{e^a{e}^{-sx}{x}^{-1+s}\Gamma (1+k,a+x)}{a^{k}k}dx$

${\displaystyle \sum _{n\ge 0}}{\left(-\frac{1}{t}\right)}^n\left(1-{(1+\beta )}^{-1-n}\right){(1-k)}_{-1+n}=-{\int }_0^{\infty }\frac{e^t{e}^{-x\beta }\beta \Gamma (1+k,t+x)}{t^{k}k}dx$

${\displaystyle \sum _{k\ge 2}}\frac{{(-1)}^k{t}^k\Gamma (k,x)}{k-1}=t{e}^{-x}\left(log\left(t\left(\frac{1}{t}+x\right)\right)+e^{\frac{1}{t}+x}\Gamma \left(0,x+\frac{1}{t}\right)\right)$

${\displaystyle \sum _{n\ge 0}}{\displaystyle \sum _{j\in [0,n]}}\frac{{(-1)}^n{x}^j{t}^n{(1-k)}_{n-1}}{\Gamma (j+1)}=-\frac{t^k{e}^{\frac{1}{t}+x}\Gamma \left(k+1,x+\frac{1}{t}\right)}{k}$

${\displaystyle \sum _{n\ge 0}}{\displaystyle \sum _{p\ge 0}}\frac{{\left(-\frac{1}{t}\right)}^n{\left(-\frac{1}{s}\right)}^p\Gamma (n+1,-t)\Gamma (n-x+1)\Gamma (p+1,-s)\Gamma \left(p-x+\frac{1}{2}\right)}{\Gamma (n+1)\Gamma (p+1)\Gamma \left(\frac{3}{2}-x\right)\Gamma (2-x)}=\sqrt{\pi }{2}^{3-2x}{e}^{s+t}{s}^{\frac{1}{2}-x}{t}^{1-x}\Gamma (2x-2)$

for easy reading and we made use of Mathematica by Wolfram to numerically verify our evaluations for both real and imaginary and complex values of the parameters involved. We will be applying this method to other generating function polynomials to derive more formulae in future work.