calculus - Gaussian type integral $int_{-infty}^{infty} frac{mathrm{e}^{-a^2 x^2}}{1 + x^2} mathrm{d}x$

When working a proof, I reached an expression similar to this:

$$\int_{-\infty}^{\infty} \frac{\mathrm{e}^{-a^2 x^2}}{1 + x^2} \mathrm{d}x$$

I've tried the following:

1. I tried squaring and combining and converting to polar coordinates, like one would solve a standard Gaussian. However, this yielded something which seems no more amenable to a solution:

$$\int_{\theta=0}^{\theta=2\pi} \int_{0}^{\infty} \frac{r \mathrm{e}^{-a^2 r^2}}{(1 + r^2 \sin^2(\theta))(1 + r^2 \cos^2(\theta))} \mathrm{d}r \mathrm{d}\theta$$

2. I tried doing a trig substitution, t = tan u, and I have no idea what to do from there.

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \mathrm{e}^{-a^2 \tan^2(u)} \mathrm{d}u$$

3. I looked into doing $u^2 = 1 + x^2$ but this gives us a ugly dx that I don't know how to handle, and moreover, I think I'm breaking my limits of integration (because Mathematica no longer solves it.):

$$u^2 = 1 + x^2$$

$$2 u \mathrm{d}u = 2 x \mathrm{d}x$$

$$\mathrm{d}x = \frac{u}{\sqrt{u^2 - 1}}$$
$$\mathrm{e}^{a^2} \int_{-\infty}^{\infty} \frac{\mathrm{e}^{-a^2 u^2}}{u \sqrt{u^2 - 1}} \mathrm{d}u$$

4. I looked into some form of differentiation under the integral, but that didn't seem to yield anything that looked promising. (I checked parameterizing x^2 to x^b in both places, and in either place, and nothing canceled cleanly.)

I have a solution from Mathematica, it's:

$$\pi e^{a^2} \text{erfc}(a)$$

But I'd like to know how to arrive at this. I'm sure it's something simple I'm missing.

Answer

Let $F$ be the function

$$F(a)=\int_{-\infty}^{\infty}\frac{e^{-a^{2}x^{2}}}{1+x^2}dx$$
We take the derivative w.r.t $a$

$$F^{\prime}(a)=\frac{d}{da}\left(\int_{-\infty}^{\infty}\frac{e^{-a^{2}x^{2}}}{1+x^2}dx\right)=\int_{-\infty}^{\infty}\frac{d}{da}\left(\frac{e^{-a^{2}x^{2}}}{1+x^2}\right)dx =\int_{-\infty}^{\infty}\frac{-2ax^{2}e^{-a^{2}x^{2}}}{1+x^2}dx$$
$$=\int_{-\infty}^{\infty}\frac{-2a\big((x^{2}+1)-1\big)e^{-a^{2}x^{2}}}{1+x^2}dx =-2a\int_{-\infty}^{\infty}e^{-a^{2}x^{2}}dx+2aF(a) =-2a\sqrt{\frac{\pi}{a^2}}+2aF(a)$$
Then
$$F^{\prime}(a)=2a\left(F(a)-\sqrt{\pi}\,\frac{1}{\vert{a}\vert}\right) =2aF(a)-2\sqrt{\pi}\mathrm{sign}(a).$$
Then you have a differential equation:
$$F^{\prime}(a)-2a\,F(a)=-2\sqrt{\pi}\mathrm{sign}(a)$$

with initial condition $F(0)=\pi$.
This fisrt order ode has integrant factor:
$$\mu(a)=\displaystyle{e^{\displaystyle{\int{-2ada}}}}=e^{-a^2}$$
Then
$$\left(e^{-a^2}F(a)\right)^{\prime}=-2\sqrt{\pi}\mathrm{sign}(a) e^{-a^2}$$
this implies
$$e^{-a^2}F(a)=-2\sqrt{\pi}\int{\mathrm{sign}(a) e^{-a^2}}da+C$$
Finaly
$$F(a)=e^{a^2}\left(C-2\sqrt{\pi}\mathrm{sign}(a)\int{e^{-a^2}da}\right)$$