If $X$ is Poisson, find the expectation of $\frac{1}{a+X}$

Question

If $X$ is a Poisson random variable with $\Pr(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}$ and $a>0$ then find the expectation of $\frac{1}{a+X}$

If I make use of $\displaystyle\frac1{a+x}=\int_0^{\infty}e^{-(a+x)t}dt$

then I obtain;

$\displaystyle\mathbf E\bigg(\frac1{a+X}\bigg)=\sum\limits_{x=0}^{\infty}\bigg(\int_0^{\infty}e^{-(a+x)t}dt\bigg)e^{-\lambda}\frac{\lambda^x}{x!}$

now I can interchange sum and integral, but I don't see how to proceed.

Answer 1

The result that I am obtaining using Maple is

$$\sum _{k=0}^{\infty }{\frac {{{\rm e}^{-\lambda}}{\lambda}^{k}}{ \left( a+k \right) k!}}={\frac {\Gamma \left( a+1 \right) -a\Gamma \left( a,-\lambda \right) }{{{\rm e}^{\lambda}}a \left( -\lambda \right) ^{a}}} $$

or

$$\sum _{k=0}^{\infty }{\frac {{{\rm e}^{-\lambda}}{\lambda}^{k}}{ \left( a+k \right) k!}}={\frac {{_{1}F_1(a;\,a+1;\,\lambda)}}{{{\rm e}^{ \lambda}}{a}}} $$

or

$$\sum _{k=0}^{\infty }{\frac {{{\rm e}^{-\lambda}}{\lambda}^{k}}{ \left( a+k \right) k!}}={\frac { {{\rm M}\left(a,\,a+1,\,\lambda\right)}}{{{\rm e}^{\lambda}}{a}}} $$

Other related results are:

$$\sum _{k=0}^{\infty }{\frac {{{\rm e}^{-\lambda}}{\lambda}^{k}}{ \left( a+k \right) ^{2}k!}}={\frac {{_2F_2(a,a;\,a+1,a+1;\,\lambda)}}{ {{\rm e}^{\lambda}}{a}^{2}}} $$

$$\sum _{k=0}^{\infty }{\frac {{{\rm e}^{-\lambda}}{\lambda}^{k}}{ \left( a+k \right) ^{3}k!}}={\frac { {_3F_3(a,a,a;\,a+1,a+1,a+1;\,\lambda)}}{{{\rm e}^{\lambda}}{a}^{3}}} $$

$$\sum _{k=0}^{\infty }{\frac {{{\rm e}^{-\lambda}}{\lambda}^{k}}{ \left( a+k \right) ^{n}k!}}={\frac { {_nF_n(a,..,a;\,a+1,..,a+1;\,\lambda)}}{{{\rm e}^{\lambda}}{a}^{n }}} $$