# E(11+x2)E(\frac{1}{1+x^2}) under a Gaussian

This question is leading on from the following question.
https://math.stackexchange.com/questions/360275/e1-1×2-under-a-normal-distribution

Basically what is the $E\left(\frac{1}{1+x^2}\right)$ under a general Gaussian $\mathcal{N}(\mu,\sigma^2)$. I tried rewriting $\frac{1}{1+x^2}$ as a scalar mixture of Gaussians ($\propto \int\mathcal{N}(x|0,\tau^{-1})Ga(\tau|1/2,1/2)d\tau$). This also came to a halt, unless you folks have got a trick under your belt.

If this integral isn’t analytical any sensible bounds?

Let $f_\sigma(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{x^2}{2\sigma^2}\right)$ be the Normal$(0,\sigma)$ PDF and $g(x) = \frac{1}{\pi}\left(1+x^2\right)^{-1}$ be the PDF of a Student t distribution with one d.f. Because the PDF of a Normal$(\mu,\sigma)$ variable $X$ is $f_\sigma(x-\mu) = f_\sigma(\mu-x)$ (by symmetry), the expectation equals

This is the defining formula for the convolution $(f\star \pi g)(\mu)$. The most basic result of Fourier analysis is that the Fourier transform of a convolution is the product of Fourier transforms. Moreover, characteristic functions (c.f.) are (up to suitable multiples) Fourier transforms of PDFs. The c.f. of a Normal$(0,\sigma)$ distribution is

and the c.f. of this Student t distribution is

(Both can be obtained by elementary methods.) The value of the inverse Fourier transform of their product at $\mu$ is, by definition,

Its calculation is elementary: carry it out separately over the intervals $(-\infty,0]$ and $[0,\infty)$ to simplify $|t|$ to $-t$ and $t$, respectively, and complete the square each time. Integrals akin to the Normal CDF are obtained–but with complex arguments. One way to write the solution is

Here, $\text{erfc}(z) = 1 - \text{erf}(z)$ is the complementary error function where

A special case is $\mu=0, \sigma=1$ for which this expression reduces to

Here is contour plot of $\mathbb{E}_{\sigma,\mu}$ (on a logarithmic axis for $\sigma$).