Integral Representation of a Double Sum 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?
 A: Hint
$$
\begin{array}{l}
 S(x,y) = \sum\limits_{n = 1}^\infty  {\frac{{x^n }}{{n!}}\sum\limits_{m = 1}^n {\frac{{\left( {m - 1} \right)!}}{{\left( {m + y} \right)!}}} }  =  \\ 
  = \sum\limits_{m = 1}^\infty  {\frac{{\left( {m - 1} \right)!}}{{\left( {m + y} \right)!}}\sum\limits_{n = m + 1}^\infty  {\frac{{x^n }}{{n!}}} }  =  \\ 
  = \sum\limits_{m = 1}^\infty  {\frac{{\left( {m - 1} \right)!}}{{\left( {m + y} \right)!}}\left( {e^x  - \sum\limits_{n = 0}^m {\frac{{x^n }}{{n!}}} } \right)}  =  \\ 
  = e^x \sum\limits_{m = 1}^\infty  {\frac{{\left( {m - 1} \right)!}}{{\left( {m + y} \right)!}}\left( {1 - \frac{1}{{e^x }}\sum\limits_{n = 0}^m {\frac{{x^n }}{{n!}}} } \right)}  =  \\ 
  = e^x \sum\limits_{m = 1}^\infty  {\frac{{\left( {m - 1} \right)!}}{{\left( {m + y} \right)!}}\left( {1 - \frac{{\Gamma \left( {m + 1,x} \right)}}{{\Gamma \left( {m + 1} \right)}}} \right)}  =  \\ 
  = e^x \sum\limits_{m = 1}^\infty  {\frac{{\Gamma \left( m \right)}}{{\Gamma \left( {m + 1 + y} \right)}}\frac{{\gamma \left( {m + 1,x} \right)}}{{\Gamma \left( {m + 1} \right)}}}  =  \\ 
  =  \cdots  \\ 
 \end{array}
$$
A: $$ \sum_{m = 1}^{n} \frac{\left( m - 1 \right)!}{\left( m + y \right)!}=\frac{1}{y^2\, \Gamma (y)}-\frac{\Gamma (n+1)}{y\, \Gamma (n+y+1)}$$
$$\sum_{n = 1}^{\infty} \frac{x^{n}}{n!} \sum_{m = 1}^{n} \frac{\left( m - 1 \right)!}{\left( m + y \right)!}=\frac{e^x-1}{y^2 \Gamma (y)}-\frac 1 y\sum_{n = 1}^{\infty}\frac{x^n}{(n+y)!}$$
$$\sum_{n = 1}^{\infty}\frac{x^n}{(n+y)!}=\frac 1 {x^y}\sum_{n = 1}^{\infty}\frac{x^{(n+y)}}{(n+y)!}=\frac {e^x} {x^y}\frac{ (y+1) (\Gamma (y+1)-\Gamma (y+1,x))}{\Gamma (y+2)}$$
$$\sum_{n = 1}^{\infty}\frac{x^n}{(n+y)!}=\frac {e^x} {x^y}\Bigg[1-\frac{\Gamma (y+1,x)}{\Gamma (y+1)} \Bigg]=\frac {e^x} {x^y}\Bigg[1-\frac {x^{1+y}}{y!}E_{-y}(x)\Bigg]$$ Recombining all the above
$$\sum_{n = 1}^{\infty} \frac{x^{n}}{n!} \sum_{m = 1}^{n} \frac{\left( m - 1 \right)!}{\left( m + y \right)!}=-\frac{e^x x^{-y}}{y}+\frac{e^x (x E_{-y}(x)+1)-1}{y^2 \Gamma (y)}$$
