calculus - Integral of the form $int_0^infty exp{left(-frac{(a+bx)^2}{x}right)}frac{dx}{sqrt{x^3}}$

I am currently working on the following improper integral:

Let $\sigma>0$ and $H>1$. I would like to show that
$$ \int_0^{\infty} \frac{\frac{\ln H}{\sigma}\exp \left( \frac{- \left( \large \frac{\ln H}{\sigma}+ \frac{t \sigma}{2} \right)^2}{2t} \right)}{\sqrt{2 \pi t^3}} \,dt = \frac{1}{H}.$$

This is the result indicated in my book and also on Wolfram Alpha, but I do not know the substitution required to evaluate this. Any ideas?

Answer

$$I(a,b)=\int_0^\infty \exp{\left(-\frac{(a+bx)^2}{x}\right)}\frac{dx}{\sqrt{x^3}}\overset{x\to x^2}=2\int_0^\infty \exp{\left(-\frac{(a+bx^2)^2}{x^2}\right)}\frac{dx}{x^2}$$
$$\overset{\large \frac{1}{x^2}\to x}=\color{blue}{2\int_0^\infty \exp{\left(-{(ax+b/x)^2}\right)}dx}\overset{\large x\to \frac{b}{ax}}=\color{red}{\frac{2b}{a}\int_0^\infty \exp\left(-(ax+b/x)^2\right)\frac{dx}{x^2}}$$
$$\color{purple}{2I(a,b)}=\color{purple}{2\int_0^\infty \exp\left(-(ax+b/x)^2\right)}\left(\color{blue}{1}+\color{red}{\frac{b}{ax^2}}\right)dx$$

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It would have been perfect if we had $(ax-b/x)$ instead of $(ax+b/x)$ since it's derivative would be found in $\left(1+b/(ax^2)\right)$, but we can adjust things:

$$(ax+b/x)^2=a^2 x^2+b^2/x^2+2ab$$

$$ (ax-b/x)^2=a^2x^2+b^2/x^2-2ab$$
$$\Rightarrow (ax+b/x)^2=(ax-b/x)^2+4ab$$

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$$\Rightarrow I(a,b)=\frac{e^{-4ab}}{a}\int_0^\infty \exp\left(-(ax-b/x)^2\right)\left(a+\frac{b}{x^2}\right)dx$$
$$\overset{ax-b/x=t}=\frac{e^{-4ab}}{a}\int_{-\infty}^\infty e^{-t^2}dt=\frac{\sqrt{\pi}e^{-4ab}}{a}$$