Evaluate the Integral $$\int_{0}^{\infty} \frac{\mathrm{e}^{-x}}{x^{1/2}+ax^{3/2}} \mathrm{d}x$$

How to evaluate

\int\limits_{0}^{\infty} \frac{\mathrm{e}^{-x}}{x^{1/2}+ax^{3/2}} \mathrm{d}x

was a question posed at Mathematics Stack Exchange. Here is my solution.

\begin{align}
\tag{a}
\int\limits_{0}^{\infty} \frac{\mathrm{e}^{-x}}{x^{1/2}+ax^{3/2}} \mathrm{d}x &=
2\int\limits_{0}^{\infty} \frac{z\mathrm{e}^{-z^{2}}}{z+az^{3}} \mathrm{d}z \\
\tag{b}
&= \frac{2}{\sqrt{a}} \int\limits_{0}^{\infty} \mathrm{e}^{-y^{2}/a} \frac{1}{1+y^{2}} \mathrm{d}y \\
\tag{c}
&= \frac{\pi}{\sqrt{a}} \mathrm{e}^{1/a} \,\mathrm{erfc}\left(\frac{1}{\sqrt{a}}\right)
\end{align}

Notes:

a. $$x=z^{2}$$
b. $$y^{2}=az^{2}$$
c. From DLMF, we have the following integral definition of the complementary error function

\mathrm{erfc}(z) = \frac{2}{\pi} \mathrm{e}^{-z^{2}} \int\limits_{0}^{\infty} \mathrm{e}^{-z^{2}t^{2}}
\frac{1}{t^{2} + 1} \mathrm{d}t

To derive the complementary error function integral above, we begin with

\frac{2}{\pi} \mathrm{e}^{-z^{2}} \int\limits_{0}^{\infty} \frac{\mathrm{e}^{-z^{2}x^{2}}}{x^{2} + 1} \mathrm{d}x =
\frac{2z}{\pi} \mathrm{e}^{-z^{2}} \int\limits_{0}^{\infty} \frac{\mathrm{e}^{-t^{2}}}{z^{2} + t^{2}} \mathrm{d}t
= \frac{z}{\pi} \mathrm{e}^{-z^{2}} \int\limits_{-\infty}^{\infty} \frac{\mathrm{e}^{-t^{2}}}{z^{2} + t^{2}} \mathrm{d}t

We used the substitution $$x=t/z$$.

For the integral

\int\limits_{-\infty}^{\infty} \frac{\mathrm{e}^{-t^{2}}}{z^{2} + t^{2}} \mathrm{d}t

we let $$f(t) = \mathrm{e}^{-t^{2}}$$ and $$g(t) = 1/(z^{2} + t^{2})$$ and take Fourier transforms of each,

\mathrm{F}(s) = \mathcal{F}[f(t)] = \frac{\mathrm{e}^{-s^{2}/4}}{\sqrt{2}}

and

\mathrm{G}(s) = \mathcal{F}[g(t)] = \frac{1}{z}\sqrt{\frac{\pi}{2}} \mathrm{e}^{-z|s|}

then invoke Parseval’s theorem

\int\limits_{-\infty}^{\infty} f(t)\overline{g(t)} \mathrm{d}t
= \int\limits_{-\infty}^{\infty} \mathrm{F}(s)\overline{\mathrm{G}(s)} \mathrm{d}s

Dropping constants, the integral becomes
\begin{align}
\int\limits_{-\infty}^{\infty} \mathrm{e}^{-s^{2}/4} \mathrm{e}^{-z|s|} \mathrm{d}s
&= 2\int\limits_{0}^{\infty} \mathrm{e}^{-s^{2}/4} \mathrm{e}^{-z|s|} \mathrm{d}s \\
&= 2\mathrm{e}^{z^{2}} \int\limits_{0}^{\infty} \mathrm{e}^{-(s+2z)^{2}/4} \mathrm{d}s \\
&= 4\mathrm{e}^{z^{2}} \int\limits_{0}^{\infty} \mathrm{e}^{-y^{2}} \mathrm{d}y \\
&= 2\sqrt{\pi}\mathrm{e}^{z^{2}} \mathrm{erfc}(z)
\end{align}
We completed the square in the exponent and used the substitution $$y=z+s/2$$.

Putting the pieces together yields our desired result
\begin{align}
\frac{2}{\pi} \mathrm{e}^{-z^{2}} \int\limits_{0}^{\infty} \frac{\mathrm{e}^{-z^{2}x^{2}}}{x^{2} + 1} \mathrm{d}x &=
\frac{z}{\pi} \mathrm{e}^{-z^{2}} \int\limits_{-\infty}^{\infty} \frac{\mathrm{e}^{-t^{2}}}{z^{2} + t^{2}} \mathrm{d}t \\
&= \frac{z}{\pi} \mathrm{e}^{-z^{2}} \frac{1}{\sqrt{2}} \frac{1}{z} \sqrt{\frac{\pi}{2}} 2\sqrt{\pi} \mathrm{e}^{z^{2}} \mathrm{erfc}(z) \\
&= \mathrm{erfc}(z)
\end{align}

Thanks to Jack D’Aurizio for outlining this derivation.