What is the expected value of $\frac{1}{N+X}$ when $X$ is a poisson distributed random variabel?

Question

Assume that $X$ is a poisson distributed random variabel. That is, the probability of the event $X=i$ is $P(X=i)=\frac{\lambda^ie^{-\lambda}}{i!}$. Can anyone simplify the expression of $E[\frac{1}{N+X}]=\sum \limits_{i=0}^{\infty}\frac{1}{N+i} \frac{\lambda^ie^{-\lambda}}{i!} $ ?

Answer 1

Using an integration trick : \begin{align*} \mathbb E\left[ \frac{1}{N+X} \right]&=\sum_{i=0}^{\infty} \frac{1}{N+i} \frac{\lambda^i e^{-\lambda}}{i!}\\ &= \frac{e^{-\lambda}}{\lambda^N} \sum_{i=0}^{\infty} \frac{1}{N+i} \frac{\lambda^{N+i} }{i!}\\ &= \frac{e^{-\lambda}}{\lambda^N} \sum_{i=0}^{\infty} \int_0^\lambda \frac{t^{N+i-1} }{i!} dt\\ &= \frac{e^{-\lambda}}{\lambda^N} \int_0^\lambda t^{N-1} \sum_{i=0}^{\infty} \frac{t^{i} }{i!} dt\\ &= \frac{e^{-\lambda}}{\lambda^N} \int_0^\lambda t^{N-1} e^{-t} dt\\ &= \frac{e^{-\lambda}}{\lambda^N} (\Gamma(N,0)-\Gamma(N,\lambda))\\ &= \frac{e^{-\lambda}}{\lambda^N} (1-\Gamma(N,\lambda)) \end{align*} Where $\Gamma(\cdot\,,\cdot)$ is the incomplete Gamma function for which we have numerical values (especially if $N$ is an integer).