Integral Representation of a Double Sum

Question

Let us assume we know the value of $x$ and $y$. I'm trying to write the following double sum as an integral. I went through many pages and saw various methods but I'm completely lost with my problem.

$$\sum_{n = 1}^{\infty} \frac{x^{n}}{n!} \sum_{m = 1}^{n} \frac{\left( m - 1 \right)!}{\left( m + y \right)!}$$

Could anyone please give me a hint so I can go about solving it myself?

Answer 1

Hints: Recall \begin{align*} \binom{n}{k}^{-1}=(n+1)\int_{0}^1z^k(1-z)^{n-k}\,dz\tag{1.1} \end{align*} and consider \begin{align*} \frac{\left( m - 1 \right)!}{\left( m + y \right)!}=\frac{1}{(y+1)!}\binom{m+y}{m-1}^{-1}\tag{1.2} \end{align*} We also need \begin{align*} \sum_{m=1}^nz^{m-1}=\frac{1-z^n}{1-z}\qquad\text{and}\qquad\sum_{n=1}^{\infty}\frac{x^n}{n!}=e^x-1\tag{1.3} \end{align*}

Added [2021-10-26]: Here is according to a comment from OP a rather detailed derivation which uses (1.1) to (1.3). It's admittedly not the shortest derivation, but here it is:

We obtain \begin{align*} \color{blue}{\sum_{n=1}^\infty}&\color{blue}{\frac{x^n}{n!}\sum_{m=1}^n\frac{(m-1)!}{(m+y)!}}\\ &=\frac{1}{(y+1)!}\sum_{n=1}^\infty\frac{x^n}{n!}\sum_{m=1}^n\binom{m+y}{m-1}^{-1}\tag{2.1}\\ &=\frac{1}{(y+1)!}\sum_{n=1}^\infty\frac{x^n}{n!}\sum_{m=1}^n(m+y+1)\int_{0}^1z^{m-1}(1-z)^{y+1}\,dz\tag{2.2}\\ &=\frac{1}{(y+1)!}\int_{0}^1(1-z)^{y+1}\sum_{n=1}^\infty\frac{x^n}{n!}\sum_{m=1}^n(m+y+1)z^{m-1}\,dz\tag{2.3}\\ &=\frac{1}{(y+1)!}\int_{0}^1(1-z)^{y+1}\\ &\qquad\cdot\sum_{n=1}^\infty\frac{x^n}{n!} \left(\frac{1-z^n}{(1-z)^2}-\frac{nz^n}{1-z}+(y+1)\frac{1-z^n}{1-z}\right)\,dz\tag{2.4}\\ &=\frac{1}{(y+1)!}\int_{0}^1(1-z)^{y+1}\\ &\qquad\cdot\left(\frac{e^x-e^{xz}}{(1-z)^2}-\frac{xze^{xz}}{1-z}+(y+1)\frac{e^x-e^{xz}}{1-z}\right)\,dz\tag{2.5}\\ &=\frac{e^x}{(y+1)!}\int_{0}^1(1-z)^{y-1}\left(1-e^{x(z-1)}\right)\,dz\\ &\qquad-\frac{xe^x}{(y+1)!}\int_{0}^1(1-z)^{y}ze^{x(z-1)}\,dz\\ &\qquad+\frac{e^x}{y!}\int_{0}^1(1-z)^{y}\left(1-e^{x(z-1)}\right)\,dz\tag{2.6}\\ &=\frac{e^x}{\Gamma(y+2)}\left(\frac{1}{y}-\frac{\Gamma(y)-\Gamma(y,x)}{x^y}\right)\\ &\qquad+\frac{xe^x}{\Gamma(y+2)}\left(\frac{\Gamma(y+2)-\Gamma(y+2,x)}{x^{y+2}}\right)\\ &\qquad-\frac{xe^x}{\Gamma(y+2)}\left(\frac{\Gamma(y+1)-\Gamma(y+1,x)}{x^{y+1}}\right)\\ &\qquad+\frac{e^x}{\Gamma(y+1)}\left(\frac{1}{y+1}-\frac{\Gamma(y+1)-\Gamma(y+1,x)}{x^{y+1}}\right)\tag{2.7}\\ &=\frac{e^x}{y\Gamma(y+1)}-\frac{e^x}{yx^y}-\frac{1}{y\Gamma(y+1)}+\frac{e^x\Gamma(y+1,x)}{yx^y\Gamma(y+1)}\tag{2.8}\\ &\,\,\color{blue}{=\frac{e^x-1}{y\Gamma(y+1)}+\frac{e^x}{yx^y}\left(\frac{\Gamma(y+1,x)}{\Gamma(y+1)}-1\right)}\tag{2.9} \end{align*}

Cross-check:

We perform a cross-check with the Mathematica generated result stated in the comment section above by @user64494.

We obtain \begin{align*} &\color{blue}{\frac{x^{-y}\left(\left(e^x(y+1)-x-y-1\right)x^y+e^x\left(\Gamma(y+2,x)-\Gamma(y+2)\right)\right)}{y\Gamma(y+2)}}\\ &\qquad=\frac{e^x(y+1)-x-y-1}{y\Gamma(y+2)}+\frac{e^x}{yx^y}\left(\frac{\Gamma(y+2,x)}{\Gamma(y+2)}-1\right)\\ &\qquad=\frac{e^x(y+1)}{y\Gamma(y+2)}-\frac{x}{y\Gamma(y+2)}-\frac{y+1}{y\Gamma(y+2)}\\ &\qquad\qquad+\frac{e^x}{yx^y}\left(\frac{(y+1)\Gamma(y+1,x)+x^{y+1}e^{-x}}{\Gamma(y+2)}-1\right)\tag{3.1}\\ &\qquad=\frac{e^x}{y\Gamma(y+1)}-\frac{x}{y\Gamma(y+2)}-\frac{1}{y\Gamma(y+1)}\\ &\qquad\qquad+\frac{e^x}{yx^y}\frac{\Gamma(y+1,x)}{\Gamma(y+1)}+\frac{x}{y\Gamma(y+2)}-\frac{e^x}{yx^y}\\ &\qquad\color{blue}{=\frac{e^x-1}{y\Gamma(y+1)}+\frac{e^x}{yx^y}\left(\frac{\Gamma(y+1,x)}{\Gamma(y+1)}-1\right)} \end{align*} in accordance with the derivation (2.9).

Comment:

• In (2.1) we use the binomial identity (1.2).

• In (2.2) we use the integral representation (1.1).

• In (2.3) we do some rearrangements.

• In (2.4) we apply the geometric series expansion addressed in (1.3) in the form \begin{align*} \sum_{m=1}^nz^{m-1}=\frac{1-z^n}{(1-z)}, \qquad \sum_{m=1}^nmz^{m-1}=\frac{1-z^n}{(1-z)^2}-\frac{nz^n}{1-z}\\ \end{align*}

• In (2.5) we use the exponential series expansion addressed in (1.3).

• In (2.6) we separate the terms and factor out $e^x$.

• In (2.7) we evaluate the integrals (with some help of Wolfram Alpha) using the Gamma function $\Gamma(y)$ and the Incomplete Gamma function $\Gamma(y,x)$.

• In (2.8) we collect terms and simplify using the recurrence relations \begin{align*} \Gamma(y+1)=y\Gamma(y), \qquad \Gamma(y+1,x)=y\Gamma(y,x)+\frac{x^y}{e^x} \end{align*}

• In (2.9) we make a final simplification.

• In (3.1) we use again the recurrence relation given in (2.8).

Answer 2

Hint

$$ \begin{array}{l} S(x,y) = \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n!}}\sum\limits_{m = 1}^n {\frac{{\left( {m - 1} \right)!}}{{\left( {m + y} \right)!}}} } = \\ = \sum\limits_{m = 1}^\infty {\frac{{\left( {m - 1} \right)!}}{{\left( {m + y} \right)!}}\sum\limits_{n = m + 1}^\infty {\frac{{x^n }}{{n!}}} } = \\ = \sum\limits_{m = 1}^\infty {\frac{{\left( {m - 1} \right)!}}{{\left( {m + y} \right)!}}\left( {e^x - \sum\limits_{n = 0}^m {\frac{{x^n }}{{n!}}} } \right)} = \\ = e^x \sum\limits_{m = 1}^\infty {\frac{{\left( {m - 1} \right)!}}{{\left( {m + y} \right)!}}\left( {1 - \frac{1}{{e^x }}\sum\limits_{n = 0}^m {\frac{{x^n }}{{n!}}} } \right)} = \\ = e^x \sum\limits_{m = 1}^\infty {\frac{{\left( {m - 1} \right)!}}{{\left( {m + y} \right)!}}\left( {1 - \frac{{\Gamma \left( {m + 1,x} \right)}}{{\Gamma \left( {m + 1} \right)}}} \right)} = \\ = e^x \sum\limits_{m = 1}^\infty {\frac{{\Gamma \left( m \right)}}{{\Gamma \left( {m + 1 + y} \right)}}\frac{{\gamma \left( {m + 1,x} \right)}}{{\Gamma \left( {m + 1} \right)}}} = \\ = \cdots \\ \end{array} $$

Answer 3

$$ \sum_{m = 1}^{n} \frac{\left( m - 1 \right)!}{\left( m + y \right)!}=\frac{1}{y^2\, \Gamma (y)}-\frac{\Gamma (n+1)}{y\, \Gamma (n+y+1)}$$ $$\sum_{n = 1}^{\infty} \frac{x^{n}}{n!} \sum_{m = 1}^{n} \frac{\left( m - 1 \right)!}{\left( m + y \right)!}=\frac{e^x-1}{y^2 \Gamma (y)}-\frac 1 y\sum_{n = 1}^{\infty}\frac{x^n}{(n+y)!}$$ $$\sum_{n = 1}^{\infty}\frac{x^n}{(n+y)!}=\frac 1 {x^y}\sum_{n = 1}^{\infty}\frac{x^{(n+y)}}{(n+y)!}=\frac {e^x} {x^y}\frac{ (y+1) (\Gamma (y+1)-\Gamma (y+1,x))}{\Gamma (y+2)}$$ $$\sum_{n = 1}^{\infty}\frac{x^n}{(n+y)!}=\frac {e^x} {x^y}\Bigg[1-\frac{\Gamma (y+1,x)}{\Gamma (y+1)} \Bigg]=\frac {e^x} {x^y}\Bigg[1-\frac {x^{1+y}}{y!}E_{-y}(x)\Bigg]$$ Recombining all the above $$\sum_{n = 1}^{\infty} \frac{x^{n}}{n!} \sum_{m = 1}^{n} \frac{\left( m - 1 \right)!}{\left( m + y \right)!}=-\frac{e^x x^{-y}}{y}+\frac{e^x (x E_{-y}(x)+1)-1}{y^2 \Gamma (y)}$$