# Contour integration problem from probability

Can integrals of the form $$\int_{-\infty}^{\infty}{\exp\left(-\left[x - c\right]^{2}\right) \over 1 + x^{2}}\, {\rm d}x$$ be computed in closed form using contour integration (or any other technique)? If $c = 0$, the integral is $\pi{\rm e\ erfc}\left(1\right)$, but I'm interested in $c$ real and non-zero.

( In probability terms, the integrand is a product of normal and Cauchy densities. )

-
Just for the sake of completeness, I'll mention that Wolfram Alpha chokes on this. – Steve Huntsman Aug 30 '10 at 2:07
You can do a good deal of reduction by using Parseval's theorem. You know that (I'm dropping constants) the fourier transform of $(1+x^2)^{-1}$ is $e^{-|\xi|}$ and its easy to find the fourier transform of a shifted gaussian. I think you get something like your integral is some constant times the inverse fourier transform of $e^{-(|\xi| +4)^2}$ where the inverse fourier transform goes from $\xi$ to $c$ variables.. Not quite sure how to evaluate that, so this is just a comment.. – Otis Chodosh Aug 30 '10 at 3:02
The Fourier transform of a convolution is the product of the Fourier transforms, and the normal density is its own transform, so the transform of the integral is exp( - |x| - x^2 ) give or take some constants. So you could phrase the question as finding the inverse transform of that expression. But I'm not sure how to make use of this. – John D. Cook Aug 30 '10 at 3:55

$$J(c)=\int_{-\infty}^{\infty}\frac{\exp[-(x-c)^2]}{1+x^2}dx=e^{-c^2}\int_{-\infty}^{\infty}\frac{\exp[-x^2]}{1+x^2} e^{2cx}dx$$ The integral on the right can be treated as the Fourier transform $\mathcal{F}(\exp[-x^2]/(1+x^2))$, with the transform parameter equal to $\mbox{i}2c$. The function is actually symmetric wrt $x$, thus, it is the cosine Fourier transform we are talking about. The necessary transform is available in Vol. 1 of Bateman & Erdelyi's "Tables of Integral Transforms" (1954). I used a shortcut and computed the transform using Maple. The resulting expression is: $$J(c)=\frac{\pi\mbox{e}}{2}\left( \mbox{erfc}(1+\mbox{i}c)e^{\mbox{i}2c}+\mbox{erfc}(1-\mbox{i}c)e^{-\mbox{i}2c} \right)$$ It is easy to check that this answer satisfies the ODE obtained by fedja. Written as the sum of conjugate terms, the function $J(c)$ is clearly real-valued for real $c$. It remains an open question whether this is a "nicer" form compared to what you had originally!
Now, since you call erfc(1) "a closed form expression", I should confess I do not understand the rules of this game. What's the big difference between $\int_1^\infty e^{-x^2/2} dx$ and the original integral? Or, do you ask if it is an elementary function of the parameter $c$?
If the latter, note that the function $J(c)=e^{c^2}\int_{-\infty}^\infty\frac{e^{-(x-c)^2}}{1+x^2}dx$ satisfies the equation $J''+4J=4\sqrt\pi e^{c^2}$, which, if you try to solve it by the method of variation of parameters, leads to the indefinite integrals like $\int e^{c^2}\cos 2c\ dc$. Those are not elementary, but not much worse than your erfc.