<s>[INST] 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!
 [/INST] 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}
</s>