Computing the integral $\int_0^\infty e^{ -t- \frac{1}{t}} \frac{dt}{\sqrt{t}}$ I am trying to Compute the integral, 
$$I=\int_0^\infty  e^{ -t- \frac{1}{t}} \frac{dt}{\sqrt{t}}$$
My attempt. 
Enforcing the change of variables $x= \sqrt{t}$ it  becomes 
$$I=2\int_0^\infty  e^{-t^2- \frac{1}{t^2}} dt\\
=2e^{±2}\int_0^\infty  e^{ -\left(t\pm\frac{1}{t}\right)^2} dt$$
How do I move from here?
 A: There is a formula which says that for $f\in L^{1}(-\infty,\infty)$,
\begin{align*}
\int_{-\infty}^{\infty}f\left(x-\dfrac{1}{x}\right)dx=\int_{-\infty}^{\infty}f(u)du.
\end{align*}
This can be done by substituting $u=x-1/x$. So the integral in question is somehow integrating with $e^{-(\cdot)^{2}}$ along the line.
A: You're on the right track.  We have
$$\begin{align}
\int_0^\infty \frac{e^{-(t+1/t)}}{\sqrt t}\,dt&\overbrace{=}^{t\mapsto t^2}\\\\
&=2\int_0^\infty e^{-(t^2+1/t^2)}\,dt\\\\
&=2e^{-2}\int_{0}^\infty e^{-(t-1/t)^2}\,dt\\\\
&\overbrace{=}^{t\mapsto 1/t}2e^{-2}\int_0^\infty e^{-(t-1/t)^2}\frac1{t^2}\,dt\\\\
&=2e^{-2}\int_0^\infty e^{-(t-1/t)^2} \left(1+\frac1{t^2}\right)\,dt\\\\
&\overbrace{=}^{t-1/t\mapsto t}2e^{-2}\int_0^\infty e^{-t^2}\,dt\\\\
&= e^{-2}\sqrt \pi
\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
I & \equiv \int_{0}^{\infty}\expo{-t - 1/t}{\dd t \over \root{t}}
\,\,\,\stackrel{t\ =\ \exp\pars{\theta}}{=}\,\,\,
\int_{-\infty}^{\infty}\expo{-2\cosh\pars{\theta}}\expo{\theta/2}\dd\theta =
2\int_{0}^{\infty}\expo{-2\cosh\pars{\theta}}\cosh\pars{\theta \over 2}
\,\dd\theta
\\[5mm] & =
2\int_{0}^{\infty}\expo{-4\sinh^{2}\pars{\theta/2} - 2}
\cosh\pars{\theta \over 2}\,\dd\theta =
4\expo{-2}\ \underbrace{\int_{\theta\ =\ 0}^{\theta\ \to\ \infty}\expo{-4\sinh^{2}\pars{\theta/2}}\,\dd\sinh\pars{\theta \over 2}}
_{\ds{\root{\pi} \over 4}}
\\ & =
\bbx{\root{\pi} \over \expo{2}} \approx 0.2399
\end{align}
