Let $F(a)$ be given by the integral
$$F(a)=\int_{-\infty}^\infty \frac{e^{-ax^2}}{1+x^2}\,dx$$
for $a>0$.
Then the derivative $F'(a)$ of $F(a)$ is
$$\begin{align}
F'(a)&=-\int_{-\infty}^\infty \frac{x^2e^{-ax^2}}{1+x^2}\,dx\\\\
&=F(a)-\int_{-\infty}^\infty e^{-ax^2}\,dx\\\\
&=F(a)-\sqrt{\frac{\pi}{a}}
\end{align}$$
Hence, $F(a)$ satisfies the ODE $F'(a)-F(a)=-\sqrt{\frac{\pi}{a}}$ subject to $F(0)=\pi$. Solution to that ODE can be written
$$\begin{align}
F(a)&=e^a \left(\pi -\sqrt \pi \int_0^a \frac{e^{-x}}{\sqrt x}\,dx\right)\\\\
&=e^a\left(\pi -\pi \frac{2}{\sqrt \pi}\int_0^{\sqrt a}e^{-x^2}\,dx\right)\\\\
&=\pi e^a \left(1-\text{erf}(\sqrt a)\right)\\\\
&=\pi e^a \text{erfc}(\sqrt a)
\end{align}$$
where $\text{erf(x)}=\frac2{\sqrt \pi}\int_0^x e^{-t^2}\,dt$ is the error function and $\text{erfc}(x)=\frac2{\sqrt \pi}\int_x^\infty e^{-t^2}\,dt=1-\text{erf}(x)$ is the complementary error function.