Evaluating $\int_{-\infty}^{\infty}\frac{\exp\left(-a x^2\right)}{x^2+b^2}dx$ 
I would like to evaluate the following integral $(a>0)$
$$\int_{-\infty}^{\infty}\frac{\exp\left(-a x^2\right)}{x^2+b^2}dx.$$

I've have tried integration by parts, putting $e^{-ax^2}=u$, but I come across with this integral
$$\int_{-\infty}^{\infty}\exp\left(-a x^2\right)\arctan\left(\frac{x}{b}\right),$$
and I don't know how to do it. Could you help me?
 A: Hint
By using the Schwinger parametrization
$$\int_{-\infty}^{\infty}\frac{\exp\left(-a x^2\right)}{x^2+b^2}dx=\int_{-\infty}^{\infty}dx\, e^{-ax^2}\int_{0}^{\infty}dt\,e^{-t(x^2+b^2)}.$$
The integral over $x-$ variable is a Gaussian integral. To evaluate the integral over $t-$ variable look at the Error function.
A: Call your integral $f(a,\,b)$ so $f(0,\,b)=\frac{\pi}{b}$ and $\partial_a f+Pf=Q$ with $P=-b^2,\,Q=-\sqrt{\pi}a^{-1/2}$. Defining $R:=\exp\int P da=\exp (-ab^2)$,$$f=R^{-1}\int RQ da=-\sqrt{\pi}\exp (ab^2)\int a^{-1/2}\exp (-ab^2)da.$$Getting the integration constant right and substituting $a=b^2c^2$, $$f=\frac{\pi}{b}-\frac{2\sqrt{\pi}}{b}\exp (ab^2)\int_0^{b\sqrt{a}} \exp (-c^2)dc.$$As Dr. Sonnhard Graubner noted, we can rewrite this in terms of the error function.
A: Differentiating under the integral sign:
Clearly the integral is just
$$ \DeclareMathOperator{\diff}{\,d\!}
2 \int_0^{+\infty} \frac {\exp(-ax^2)}{x^2+b^2} \diff x. 
$$
Let the integral be $I(a)$ and we regard $b$ as a constant. We want to differentiate w.r.t. $a$. For each $c > 0$, on $J_c=[c, +\infty)$, the integral converges for $a \in J_c$, and 
$$\newcommand{\Abs}[1]{{\left|#1\right|}}
\Abs{\partial_a \frac {\exp(-ax^2)}{x^2+b^2}} \leqslant \exp(-cx^2),
$$
where $\int_0^{+\infty} \exp(-cx^2) \diff x$ converges, so by Weierstrass M-test, 
$$
\int_0^{+\infty} \frac {\exp(-ax^2)(-x^2)}{x^2+b^2}\diff x 
$$
converges uniformly for $a\in J_c$.  Therefore $I(a)$ could be differentiated under the integral symbol. Now
\begin{align*}
I'(a) &= \int_0^{+\infty} \frac {\exp(-ax^2) (-x^2)}{x^2+b^2}\diff x\\
&= \int_0^{+\infty} \exp(-ax^2) \left( \frac {b^2}{x^2+b^2}-1 \right)\diff x\\
&= b^2 I(a) - \int_0^{+\infty} \exp(-ax^2)\diff x\\
&= b^2 I(a) - \frac {\sqrt \pi} {2\sqrt a}. 
\end{align*}
Now solve this differential equations under the initial value condition 
$$
I(0) = \int_0^{+\infty}\frac {\diff x}{x^2+b^2} = \frac \pi{2b}.
$$
By the formula for 1st-order ODE, we have 
$$
I(a) = \frac{\exp(b^2a)}b \left(-\int_0^{b\sqrt a} \exp(-u^2)\diff u + \frac \pi {2b}\right),
$$
and the original is just $2I(a)$. 
