Methods to solve $\int_{0}^{\infty} \frac{e^{-x^2}}{x^2 + 1}\:dx$ I have a feel this will be a duplicate question. I have had a look around and couldn't find it, so please advise if so.
Here I wish to address the definite integral:
\begin{equation}
 I = \int_{0}^{\infty} \frac{e^{-x^2}}{x^2 + 1}\:dx
\end{equation}
I have solved it using Feynman's Trick, however I feel it's limited and am hoping to find other methods to solve. Without using Residues, what are some other approaches to this integral?
My method:
\begin{equation}
 I(t) = \int_{0}^{\infty} \frac{e^{-tx^2}}{x^2 + 1}\:dx
\end{equation}
Here $I = I(1)$ and $I(0) = \frac{\pi}{2}$. Take the derivative under the curve with respect to '$t$' to achieve:
\begin{align}
 I'(t) &= \int_{0}^{\infty} \frac{-x^2e^{-tx^2}}{x^2 + 1}\:dx = -\int_{0}^{\infty} \frac{x^2e^{-tx^2}}{x^2 + 1}\:dx \\
&= -\left[\int_{0}^{\infty} \frac{\left(x^2 + 1 - 1\right)e^{-tx^2}}{x^2 + 1}\:dx \right] \\
&= -\int_{0}^{\infty} e^{-tx^2}\:dx  + \int_{0}^{\infty} \frac{e^{-tx^2}}{x^2 + 1}\:dx \\
&= -\frac{\sqrt{\pi}}{2}\frac{1}{\sqrt{t}} + I(t)
\end{align}
And so we arrive at the differential equation:
\begin{equation}
 I'(t) - I(t) = -\frac{\sqrt{\pi}}{2}\frac{1}{\sqrt{t}}
\end{equation}
Which yields the solution:
\begin{equation}
 I(t) = \frac{\pi}{2}e^t\operatorname{erfc}\left(t\right)
\end{equation}
Thus, 
\begin{equation}
 I = I(1) \int_{0}^{\infty} \frac{e^{-x^2}}{x^2 + 1}\:dx =  \frac{\pi}{2}e\operatorname{erfc}(1)
\end{equation}
Addendum:
Using the exact method I've employed, you can extend the above integral into a more genealised form:
\begin{equation}
 I = \int_{0}^{\infty} \frac{e^{-kx^2}}{x^2 + 1}\:dx =  \frac{\pi}{2}e^k\operatorname{erfc}(\sqrt{k})
\end{equation}
Addendum 2:
Whilst we are genealising:
\begin{equation}
 I = \int_{0}^{\infty} \frac{e^{-kx^2}}{ax^2 + b}\:dx =  \frac{\pi}{2b}e^\Phi\operatorname{erfc}(\sqrt{\Phi})
\end{equation}
Where $\Phi = \frac{kb}{a}$ and $a,b,k \in \mathbb{R}^{+}$
 A: You can use Plancherel's theorem. Note that
$$
2I = \int_{-\infty}^{\infty} \frac{e^{-x^2}}{x^2 + 1}dx.
$$Let $f(x) = e^{-x^2}$ and $g(x) = \frac{1}{1+x^2}$. Then we have
$$
\widehat{f}(\xi) = \sqrt{\pi}e^{-\pi^2\xi^2},
$$ and
$$
\widehat{g}(\xi) = \pi e^{-2\pi|\xi|}.
$$ By Plancherel's theorem, we have
$$\begin{eqnarray}
\int_{-\infty}^{\infty} f(x)g(x)dx&=&\int_{-\infty}^{\infty} \widehat{f}(\xi)\widehat{g}(\xi)d\xi\\&=&\pi^{\frac{3}{2}}\int_{-\infty}^{\infty}e^{-\pi^2\xi^2-2\pi|\xi|}d\xi\\
&=&2\pi^{\frac{3}{2}}\int_{0}^{\infty}e^{-\pi^2\xi^2-2\pi\xi}d\xi\\
&=&2\pi^{\frac{3}{2}}e\int_{\frac{1}{\pi}}^{\infty}e^{-\pi^2\xi^2}d\xi\\
&=&2\pi^{\frac{1}{2}}e\int_{1}^{\infty}e^{-\xi^2}d\xi = \pi e \operatorname{erfc}(1).
\end{eqnarray}$$
This gives $I = \frac{\pi}{2}e \operatorname{erfc}(1).$
A: Here is a method that employs the old trick of converting the integral into a double integral.
Observe that
$$\frac{1}{1 + x^2} = \int_0^\infty e^{-u(1 + x^2)} \, du.$$
So your integral can be rewritten as
$$I = \int_0^\infty e^{-x^2} \int_0^\infty e^{-u(1 + x^2)} \, du \, dx.$$
or
$$I =  \int_0^\infty e^{-u} \int_0^\infty e^{-(1 + u)x^2} \, dx \, du,$$
on changing the order of integration.
Enforcing a substitution of $x \mapsto x/\sqrt{1 + u}$ gives
$$I = \int_0^\infty \frac{e^{-u}}{\sqrt{1 + u}} \int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2} \int_0^\infty \frac{e^{-u}}{\sqrt{1 + u}} \, du.$$
Next exforcing a substitution of $u \mapsto u^2 - 1$ gives
$$I = \sqrt{\pi} e \int_1^\infty e^{-u^2} \, du = \sqrt{\pi} e \cdot \frac{\sqrt{\pi}}{2} \text{erfc} (1) = \frac{\pi e}{2} \text{erfc} (1),$$
as expected.
