How to calculate this improper integral $\int_0^{+\infty} e^{-(ax+\frac{b}{x})^2}\mathrm{d}x$? How to calculate this improper integral 
$$
\int_{0}^{\infty}{\rm e}^{-\left(ax\ +\ b/x\right)^2}\,{\rm d}x\ {\large ?}
$$
 A: Note that
$$ \int_{0}^{\infty} e^{-\left( ax + \frac{b}{x} \right)^{2}} \, dx = e^{-4ab} \int_{0}^{\infty} e^{-\left( ax - \frac{b}{x} \right)^{2}} \, dx. $$
This shows that it suffices to consider the integral on the right-hand side. Associated to this we consider a more general situation. Let assume $a > 0, b > 0$ and $f$ is an integrable even function. With the substitution
$$ x = \frac{b}{at} \quad \Longrightarrow \quad dx = -\frac{b}{at^2} \, dt, $$
we obtain
$$ \int_{0}^{\infty} f\left( ax - \frac{b}{x} \right) \, dx = \int_{0}^{\infty} \frac{b}{at^2} f\left( at - \frac{b}{t} \right) \, dt. $$
Thus if we denote this common value by $I$, then
\begin{align*}
2aI
= \int_{0}^{\infty} \left( a + \frac{b}{x^2} \right) f\left( ax - \frac{b}{x} \right) \, dx 
= \int_{-\infty}^{\infty} f (u) \, du,
\end{align*}
where we used the substitution
$$ u = ax - \frac{b}{x}, \quad du = \left( a + \frac{b}{x^2} \right) \, dx. $$
Therefore we obtain the following identity.
$$ \int_{0}^{\infty} f\left( ax - \frac{b}{x} \right) \, dx = \frac{1}{2a} \int_{-\infty}^{\infty} f (x) \, dx $$
This gives us
$$ \int_{0}^{\infty} e^{-\left( ax + \frac{b}{x} \right)^{2}} \, dx = \frac{\sqrt{\pi}}{2a}  e^{-4ab}. $$
A: $\newcommand{\+}{^{\dagger}}
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$\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}
