Simplifying integral:$\int_{0}^{\infty}{\exp\left(-\left(u^2+{ {\alpha^2}\over {16u^2t}}\right)\right)}~\mathrm{d}u$ $$I(t)=\int_{0}^{\infty}{\exp\left(-\left(u^2+{ {\alpha^2}\over {16u^2t}}\right)\right)}~\mathrm{d}u $$
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!
 A: 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
$$\bbox[5px,border:2px solid #C0A000]{I=e^{2a}\int_{-\infty}^{\infty}e^{-au^2}\sqrt{a}du=\sqrt{\pi}e^{-2a}=\sqrt{\pi}e^{-\alpha/2\sqrt{t}}}$$
A: Hint. 
For any $f$ such that $\displaystyle \int_{-\infty}^{+\infty} f(u)\: \mathrm{d}x<\infty$, we have (see a proof here)

$$
\int_{-\infty}^{+\infty}f\left(u-\frac{s}{u}\right)\mathrm{d}u=\int_{-\infty}^{+\infty} f(u)\: \mathrm{d}u, \quad s>0. \tag1
$$

Apply it to $f(u)=e^{-u^2}$, you get
$$
\int_{-\infty}^{+\infty}e^{-(u-s/u)^2}\mathrm{d}u=\int_{-\infty}^{+\infty} e^{-u^2} \mathrm{d}u=\sqrt{\pi}, \quad s>0. \tag2
$$
or
$$
\int_{-\infty}^{+\infty}e^{-u^2-s^2/u^{2}}\mathrm{d}u=\sqrt{\pi}\:e^{-2s}\tag3
$$ then put $s=\dfrac{\alpha}{4\sqrt{t}}$ to obtain the value of your integral, noticing that $f$ is even here.
A: Using 
$$x^{2} - \frac{a^{2}}{x^{2}} = \left(x - \frac{i a}{x} \right)^{2} + 2 i a $$
then
\begin{align}
I &= \int_{0}^{\infty} e^{- \left(u^{2} - \frac{\alpha^{2}}{16 \, t \, u^{2}}\right)} \, du 
= e^{-2 i a} \, \int_{0}^{\infty} e^{- \left(u - \frac{i a}{u}\right)^{2}} \, du
\end{align}
where $4 \sqrt{t} \, a = \alpha$. Let $u = \sqrt{i a} x$ to obtain
\begin{align} \tag{1}
e^{2 i a} \, I = \sqrt{i a} \, \int_{0}^{\infty} e^{- i a \, \left(x - \frac{1}{x}\right)^{2}} \, dx.
\end{align}
Now let $x \to \frac{1}{x}$ to obtain
\begin{align}\tag{2}
e^{2 i a} \, I &= \sqrt{i a} \, \int_{0}^{\infty} e^{-i a \, \left( x - \frac{1}{x} \right)^{2}} \, \frac{dx}{x^{2}}. 
\end{align}
Adding equations (1) and (2) lead to
\begin{align}
\frac{2 \, e^{2 i a}}{\sqrt{i a}} \, I = \int_{0}^{\infty} e^{-i a \, \left(x - \frac{1}{x}\right)^{2}} \, \left(1 + \frac{1}{x^{2}}\right) \, dx. 
\end{align}
Let $y = x + \frac{1}{x}$ to obtain
\begin{align}
\frac{2 \, e^{2 i a}}{\sqrt{i a}} \, I = \int_{0}^{\infty} e^{-i a \, y^{2}} \, dy 
\end{align}
Set $y = \sqrt{\frac{x}{i a}}$ to obtain
\begin{align}
\frac{2 \, e^{2 i a}}{\sqrt{i a}} \, I = \frac{1}{2} \, \sqrt{\frac{\pi}{i a}}
\end{align}
From this the integral is given by
\begin{align}
\int_{0}^{\infty} e^{- \left(u^{2} - \frac{a^{2}}{u^{2}}\right)} \, du = \frac{\sqrt{\pi}}{4} \, e^{-2 i a}.
\end{align}
When $4 \sqrt{t} a = \alpha$ then
\begin{align}
\int_{0}^{\infty} e^{- \left(u^{2} - \frac{\alpha^{2}}{16 \, t \, u^{2}}\right)} \, du = \frac{\sqrt{\pi}}{4} \, e^{- \frac{i \, \alpha}{2 \sqrt{t}}}.
\end{align}
