# A tricky integral

Let $\alpha>0$ and $\beta\in\mathbb{R}$. I am looking for an explicit formula for the integral

$$\int_{-\infty}^{\infty} (1+x^2)^{-1/2}e^{-\alpha x^2}e^{-i \beta x}dx.$$

I tried several changes of variables, and contour integration doesn't seem to work.

Motivation comes from the following closely related kernel $$K(s,t)=e^{-\frac{(s-t)^2}{4}}\int_{-\infty}^{\infty}(1+x^2)^{-1/2}e^{-\frac{(s-t)^2}{4} x^2}e^{-i (s^2-t^2) x}dx,$$ which provides an example of a compact integral operator on $L^2(\mathbb{R})$ that is not Hilbert-Schmidt. I would like to check the details.

Thank you!

-
This looks like a Fourier-integral of a product of two functions. The convolution theorem could be helpful, but I don't have the time to work out the details right now... –  Dirk Nov 24 '10 at 5:19
Are you sure there is an explicit formula? By Dirk's comment, using the convolution theorem, your integral is fairly close to the convolution of a Whittaker function with a gaussian (there is some work needed, as the Fourier transform of the first function is not absolutely convergent, but it can be rewritten to be so). –  B R Nov 24 '10 at 6:45
Maybe this is not helpful, but $K$ solves the heat equation: $\frac{\partial^2 K}{\partial t^2} = \frac{\partial K}{\partial s}$. But unfortunately $s,t$ are the wrong way round: $s$ is the time variable, $t$ the space variable!! I would suggest swapping $s,t$ around to make the notation easier, unless you have some strong reason to keep $s,t$ as you've written. –  Zen Harper Nov 24 '10 at 10:45
You are told that $K$ provides an example of a compact integral operator? Is $K$ supposed to be the integral kernel (that would be suspicious since you restrict $s > 0$)? What is the operator you actually have and what is the Hilbert space you are working over? –  Willie Wong Nov 24 '10 at 11:34
"I am told", eh? Well, if they told it to you in this form, then presumably they do not know an explicit formula for it either. –  Gerald Edgar Nov 24 '10 at 15:22

Neither Maple nor Wolfram Alpha (http://www.wolframalpha.com) gives a closed form for this integral (even after inserting numeric values for $\alpha$ and $\beta$). That probably means that there isn't one. If $\beta=0$ then Wolfram Alpha gives an answer of $e^{\alpha/2}K_0(\alpha/2)$, where $K_0$ is a Bessel function.
You can express it as a (formal) sum over terms which involve the Hermite polynomials, in the form $HermiteH(n,\frac{\beta}{2\sqrt{\alpha}})$. You can compute the exact term from $$\int_{-\infty}^{\infty} x^n e^{-\alpha x^2}e^{-i\beta x}dx.$$
Note that for $\alpha=0$ this is $-\frac{\pi}{2}\left( Y_0(\beta i) + Y_0(-\beta i)\right)$ (again, Bessel function). So with Neil's answer for $\beta=0$ and the 'hint' that this integral satisfies the Heat equation (see Zen's comment), you get another equivalent formulation of the problem.