Let $ N \sim \mathrm{Poiss}(\lambda)$, then for $a > 1 $
Can expectation $\mathbb{E}\frac{1}{a + N } $ be explicitly calculated?
I've noticed that if one denotes the desired expectation as $\mathbb{E}\frac{1}{a + N } = e^{-\lambda} g(\lambda)$ then $g$ is a solution to the following differential equation: $$\lambda g'(\lambda) = e^{\lambda} - a g(\lambda)$$
But I don't know how to solve it (expect for $a = 1$). Any help greatly appreciated
Multiplying both sides in $\lambda^{a-1}$ we obtain$$\lambda^ag'(\lambda)+a\lambda^{a-1}g(\lambda)=\lambda^{a-1}e^{\lambda}$$integrating the above equation yields to $$\lambda^ag(\lambda)+C=\int_0^{\lambda}u^{a-1}e^{u}du$$by tending $\lambda\to 0$ we have $$C=0$$therefore $$g(\lambda)={1\over \lambda^a}\int_0^{\lambda}u^{a-1}e^{u}du$$