TITLE: How to calculate this improper integral $\int_0^{+\infty} e^{-(ax+\frac{b}{x})^2}\mathrm{d}x$?
QUESTION [5 upvotes]: How to calculate this improper integral 
$$
\int_{0}^{\infty}{\rm e}^{-\left(ax\ +\ b/x\right)^2}\,{\rm d}x\ {\large ?}
$$

REPLY [3 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\infty}\expo{-\pars{ax\ +\ b/x}^{2}}\,\dd x:\ {\large ?}}$
\begin{align}
&\color{#66f}{\Large\int_{0}^{\infty}\expo{-\pars{ax\ +\ b/x}^{2}}\,\dd x}\ =\
\overbrace{\int_{0}^{\infty}
\exp\pars{-ab\bracks{\root{a \over b}x\ + \root{b \over a}\,{1 \over x}}^{2}}
\,\dd x}^{\ds{\mbox{Set}\ \root{a \over b}x \equiv \expo{\theta}}}
\\[3mm]&=\int_{-\infty}^{\infty}\exp\pars{-4ab\cosh^{2}\pars{\theta}}
\root{b \over a}\expo{\theta}\,\dd\theta
\\[3mm]&=\root{b \over a}\int_{-\infty}^{\infty}
\exp\pars{-4ab\bracks{\sinh^{2}\pars{\theta} + 1}}
\bracks{\cosh{\theta} + \sin\pars{\theta}}\,\dd\theta
\\[3mm]&=2\root{b \over a}\expo{-4ab}\
\overbrace{\int_{0}^{\infty}
\exp\pars{-4ab\sinh^{2}\pars{\theta}}\cosh{\theta}\,\dd\theta}
^{\ds{2\root{ab}\sinh\pars{\theta} \equiv t}}
\\[3mm]&=2\root{b \over a}\expo{-4ab}\,{1 \over 2\root{ab}}\
\overbrace{\int_{0}^{\infty}\exp\pars{-t^{2}}\,\dd t}^{\ds{\root{\pi} \over 2}}\
=\ \color{#66f}{\Large{\root{\pi} \over 2a}\,\expo{-4ab}}
\end{align}