TITLE: Simplifying integral:$\int_{0}^{\infty}{\exp\left(-\left(u^2+{ {\alpha^2}\over {16u^2t}}\right)\right)}~\mathrm{d}u$
QUESTION [5 upvotes]: $$I(t)=\int_{0}^{\infty}{\exp\left(-\left(u^2+{ {\alpha^2}\over {16u^2t}}\right)\right)}~\mathrm{d}u $$
where $\alpha$ and $t$ are positive constant.  
P.S.I would like to edit this problem, because I found my mistake about inner sign.
Thanks!

REPLY [3 votes]: I assume that the integral of interest $I$ is 
$$I=\int_0^{\infty}e^{-\left(u^2+\frac{\alpha^2}{16tu^2}\right)}du$$
First, we let $a^2=\frac{\alpha^2}{16t}$ and write
$$\begin{align}
I&=\int_0^{\infty} e^{-a\left(\frac{u^2}{a}+\frac{a}{u^2}\right)}du\\\\
&=e^{-2a}\int_0^{\infty} e^{-a\left(\frac{u}{\sqrt{a}}-\frac{\sqrt{a}}{u}\right)^2}du\tag 1
\end{align}$$
Now, enforcing the substitution $u/\sqrt{a}\to -\sqrt{a}/u$, we find 
$$I=e^{-2a}\int_0^{\infty}e^{-a\left(\frac{u}{\sqrt{a}}-\frac{\sqrt{a}}{u}\right)^2} \frac{a}{u^2}\,du \tag 2$$
Adding $(1)$ and $(2)$ and dividing by $2$  yields
$$I=e^{-2a}\int_0^{\infty}e^{-a\left(\frac{u}{\sqrt{a}}-\frac{\sqrt{a}}{u}\right)^2} \left(1+\frac{a}{u^2}\right)\,du $$
Making one final substitution $\frac{u}{\sqrt{a}}-\frac{\sqrt{a}}{u}\,\to\,u$ and we obtain
$$\bbox[5px,border:2px solid #C0A000]{I=e^{2a}\int_{-\infty}^{\infty}e^{-au^2}\sqrt{a}du=\sqrt{\pi}e^{-2a}=\sqrt{\pi}e^{-\alpha/2\sqrt{t}}}$$