# 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. 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.. 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. 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.