Explicit solution for the expectation of inverse of the exponential of normal random variable

Question

Let $A$, $B$ and $C$ be some constants, and let $N$ be a random variable with standard normal distribution.

Question: is there an explicit solution to $$\mathbb{E}\left[\frac{1}{A+B e^{C N}}\right]?$$

If $A=0$, then the answer is $\frac{e^{\frac{C^2}{2}}}{B}$ which follows from the moment generating function. But if $A\neq 0$, using the density function of $N$, $$\mathbb{E}\left[\frac{1}{A+B e^{C N}}\right]=\int_{\mathbb{R}}\frac{1}{A+B e^{C x}}\cdot \frac{e^{-x^2/2}}{\sqrt{2 \pi}}dx$$ would lead nowhere.

Answer 1

You could expand the reciprocated expression as a geometric series $ \frac{ 1}{ 1+ m e^x}= \sum_k (-1)^k m^k e^{kx}$ and integrate termwise.

Answer 2

For $A=B \ne 0$: There is a solution, based on symmetry.

$$\int_{\mathbb{R}}\frac{1}{A+A e^{C x}}\cdot \frac{e^{-x^2/2}}{\sqrt{2 \pi}}dx = \frac1{2A} \int_{\mathbb{R}}\left(\frac{1}{1+e^{C x}}+ \frac{1}{1+e^{-C x}} \right)\cdot \frac{e^{-x^2/2}}{\sqrt{2 \pi}}dx = \frac1{2A} \int_{\mathbb{R}} \frac{e^{-x^2/2}}{\sqrt{2 \pi}}dx = \frac1{2A} $$

For $A \ne B$, I guess there is no closed-form solution, since my Mathematica just gave up on the integral. But of course, you can try some numerical integration, if you are facing a real-world problem.

Answer 3

For what it is worth, when $A=B$, you have a closed form solution. Consider $A=B=1$ to keep it simple:

\begin{align*} \frac{\partial}{\partial C} \int_{-\infty}^{\infty} \frac{\exp(-\frac{x^2}{2})}{1+\exp(Cx)}dx = \int_{-\infty}^{\infty} \frac{-x\exp(-\frac{x^2}{2})e^{Cx}}{(1+\exp(Cx))^2}dx = \int_{-\infty}^{\infty} \frac{-x\exp(-\frac{x^2}{2})}{\big(e^{-Cx/2}+e^{Cx/2}\big)^2}dx = 0 \end{align*} with the last equality due to the integrand being an odd (absolutely integrable) function.

When $C$ goes to $-\infty$, you find (with the dominated convergence theorem) that the common value is $\int_{0}^{\infty} \exp(-\frac{x^2}{2})dx = \sqrt{\frac{\pi}{2}}$. Hence: $$\int_{-\infty}^{\infty} \frac{\exp(-\frac{x^2}{2})}{1+\exp(Cx)}dx = \sqrt{\frac{\pi}{2}}$$