Gaussian type integral (but not a standard form)

[OmnipotentEntity]

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.):

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

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.

 

[fresh_42]

I thought that you could write the integral as twice of it from $0$ to infinity and then use the gamma function ($t=a^2x^2$), but the sum in the denominator makes trouble. So probably a first substitution $u=1+x^2$ and next $t=a^2x^2$ would be better. You said it's for a proof, so does the Taylor series is sufficient?

 

[OmnipotentEntity]

I managed to solve it with some help elsewhere. Here's the steps:

$$f(a) = \int_{-\infty}^{\infty} \frac{\mathrm{exp}(-a^2 y^2)}{1 + y^2} \mathrm{d}y$$

Then:

$$\begin{split}
f'(a) &= \int_{-\infty}^{\infty} \frac{\frac{\partial}{\partial a} \mathrm{exp}(-a^2 y^2)}{1 + y^2} \mathrm{d}y \\
&= \int_{-\infty}^{\infty} \frac{-2a y^2 \mathrm{exp}(-a^2 y^2)}{1 + y^2} \mathrm{d}y \\
&= \int_{-\infty}^{\infty} \frac{-2a ((1+y^2) - 1) \mathrm{exp}(-a^2 y^2)}{1 + y^2} \mathrm{d}y \\
&= \int_{-\infty}^{\infty} -2a \mathrm{exp}(-a^2 y^2) \mathrm{d}y + \int_{-\infty}^{\infty} \frac{2a \mathrm{exp}(-a^2 y^2)}{1 + y^2} \mathrm{d}y \\
&= -2 \sqrt{\pi} + 2a f(a) \\
\end{split}$$

Which is just an ODE, and we get:

$$f(a) = \pi e^{a^2} (C - \mathrm{erf}(a))$$

And because $f(0) = \pi$ we see that $C = 1$ and we have:

$$f(a) = \pi e^{a^2} \mathrm{erfc}(a)$$

 

[fresh_42]

I had the first part, too, as it is $\Gamma(\frac{1}{2})$ and the factor two for symmetry reasons, but didn't see the second summand. Thanks for posting.