Integral of the form $\int_{-\infty}^{\infty} \frac{e^{-(ax)^2}}{1 + x^2}dx$ I was reading a paper on nuclear physics when I came across the following definite integral:
$$\int_{-\infty}^{\infty}\frac{\zeta}{2\sqrt\pi} \frac{e^{-\frac{\zeta^2}{4} y^2}}{1 + y^2}\mathrm dy$$
The paper gives the expression of the above integral as:
$$\int_{-\infty}^{\infty}\frac{\zeta}{2\sqrt\pi} \frac{e^{-\frac{\zeta^2}{4} y^2}}{1 + y^2}\mathrm dy = \frac{\zeta \sqrt\pi}{2} e^{\frac{\zeta^2}{4}}\left(1-\operatorname{erf}\left (\frac{\zeta}{2}\right )\right)$$
Basically, I have no clue where this result comes from. I have tried the substitution $u = \tan^{-1}y$ so that $\mathrm du = \frac{1}{1 + y^2}\mathrm dy$, but I get the following expression:
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\zeta}{2\sqrt\pi} e^{-\frac{\zeta^2}{4} \tan^2(u)}\mathrm du$$
Which I do not know how to evaluate. Any help on the above integral would be greatly appreciated. Even a hint as to how to proceed further is welcomed. Thank you so much in advance!
PS: This is my first question so I hope the formatting/question wording is not too confusing.
Best,
Nathan
 A: Consider the following integral:
$$I(a)=\int_{-\infty}^\infty \frac{e^{-a^2(1+x^2)}}{1+x^2}dx$$
Note that initially the constant $e^{-a^2}$ wasn't there, but bringing it helps to simplify the denominator when we take a derivative with respect to $a$. Afterwards we just mutiply by $e^{a^2}$ and everything is unchanged, but let's take a derivative:
$$ I'(a)=-2a\int_{-\infty}^\infty e^{-a^2(1+x^2)}dx=-2\sqrt \pi e^{-a^2}$$
Now notice that $I(\infty)=0$ and we're after $I\left(\frac{\zeta }{2}\right)$.
$$I\left(\frac{\zeta }{2}\right)=-\left(I(\infty)-I\left(\frac{\zeta}{2}\right)\right)=2\sqrt \pi \int_{\frac{\zeta}{2}}^\infty e^{-a^2}da=\pi\operatorname{erfc}\left(\frac{\zeta }{2}\right)$$
Finally we just need to multiply by $\frac{\zeta }{2\sqrt \pi}e^{\zeta^2/4}$ and the result follows.
A: Define
$$
f(a)=\int_{-\infty}^\infty\frac{e^{-ax^2}}{1+x^2}\,\mathrm{d}x\tag1
$$
then
$$
\begin{align}
f(a)-f'(a)
&=\int_{-\infty}^\infty\frac{e^{-ax^2}}{1+x^2}\,\mathrm{d}x-\frac{\mathrm{d}}{\mathrm{d}a}\int_{-\infty}^\infty\frac{e^{-ax^2}}{1+x^2}\,\mathrm{d}x\\
&=\int_{-\infty}^\infty e^{-ax^2}\,\mathrm{d}x\\
&=\sqrt{\frac\pi{a}}\tag2
\end{align}
$$
We can solve $(2)$ using an integrating factor. Note that
$$
\begin{align}
\left(e^{-a}f(a)\right)'
&=-e^{-a}f(a)+e^{-a}f'(a)\\[3pt]
&=-e^{-a}(f(a)-f'(a))\\
&=-e^{-a}\sqrt{\frac\pi{a}}\tag3
\end{align}
$$
Therefore, using the complementary error function,
$$\newcommand{\erfc}{\operatorname{erfc}}
\begin{align}
f(a)
&=e^a\int_a^\infty e^{-t}\sqrt{\frac\pi{t}}\,\mathrm{d}t\\
&=2\sqrt\pi e^a\int_{\sqrt{a}}^\infty e^{-t^2}\,\mathrm{d}t\\
&=\pi e^a\erfc\left(\sqrt{a}\right)\tag4
\end{align}
$$
Thus,
$$
\begin{align}
\int_{-\infty}^\infty\frac{e^{-a^2x^2}}{1+x^2}\,\mathrm{d}x
&=f\!\left(a^2\right)\\
&=\pi e^{a^2}\erfc(a)\tag5
\end{align}
$$
A: Starting with 
$$\Re\int_0^\infty e^{-(1+ix)y}\ dy=\Re\frac1{1+ix}=\frac1{1+x^2}\tag{1}$$
$$\int_0^\infty e^{-(ax^2+bx+c)}\ dx=\frac12\sqrt{\frac{\pi}{a}}\ e^{\frac{b^2}{4a}-c}\ \text{erfc}\left(\frac{b}{2\sqrt{a}}\right)\tag{2}$$
Multiply both sides of (1) by $e^{-a^2x^2}$ then integrate from $x=0$ to $\infty$ we have
\begin{align}
\int_0^\infty\frac{e^{-a^2x^2}}{1+x^2}\ dx&=\int_0^\infty e^{-y}\left(\Re \int_0^\infty e^{-(a^2x^2+iyx)}\ dx\right)\ dy\\
&\overset{\text{use (2)}}{=}\int_0^\infty e^{-y}\left(\frac{\sqrt{\pi}}{2a}e^{-\frac{y^2}{4a^2}}\right)\ dy\\
&=\frac{\sqrt{\pi}}{2a}\int_0^\infty e^{-(\frac{y^2}{4a^2}+y)}\ dy\\
&\overset{\text{use (2)}}{=}\frac{\sqrt{\pi}}{2a}\left(a\sqrt{\pi}e^{a^2}\text{erfc}(a)\right)\\
&=\frac{\pi}{2}\ e^{a^2}\text{erfc}(a)
\end{align}
and since the integrand is even function, then 

$$\int_{-\infty}^\infty\frac{e^{-a^2x^2}}{1+x^2}\ dx=2\int_0^\infty\frac{e^{-a^2x^2}}{1+x^2}\ dx=\pi\ e^{a^2}\text{erfc}(a)$$

