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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?
$$I(a,b)=\int_0^\infty \exp{\left(-\frac{(a+bx)^2}{x}\right)}\frac{dx}{\sqrt{x^3}}\overset{x\to x^2}=2\int_0^\infty \exp{\left(-\frac{(a+bx^2)^2}{x^2}\right)}\frac{dx}{x^2}$$ $$\overset{\large \frac{1}{x}\to x}=\color{blue}{2\int_0^\infty \exp{\left(-{(ax+b/x)^2}\right)}dx}\overset{\large x\to \frac{b}{ax}}=\color{red}{\frac{2b}{a}\int_0^\infty \exp\left(-(ax+b/x)^2\right)\frac{dx}{x^2}}$$ $$\color{purple}{2I(a,b)}=\color{purple}{2\int_0^\infty \exp\left(-(ax+b/x)^2\right)}\left(\color{blue}{1}+\color{red}{\frac{b}{ax^2}}\right)dx$$
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:
$$(ax+b/x)^2=a^2 x^2+b^2/x^2+2ab$$ $$ (ax-b/x)^2=a^2x^2+b^2/x^2-2ab$$ $$\Rightarrow (ax+b/x)^2=(ax-b/x)^2+4ab$$
$$\Rightarrow I(a,b)=\frac{e^{-4ab}}{a}\int_0^\infty \exp\left(-(ax-b/x)^2\right)\left(a+\frac{b}{x^2}\right)dx$$ $$\overset{ax-b/x=t}=\frac{e^{-4ab}}{a}\int_{-\infty}^\infty e^{-t^2}dt=\frac{\sqrt{\pi}e^{-4ab}}{a}$$
This is not an answer.
After カカロット's elegant solution, assuming $a>0$ and $b>0$, we can find $$\int \frac{e^{-\frac{(a+b x)^2}{x}}}{x^{3/2}} \, dx=\frac{\sqrt{\pi } e^{-4 a b}}{2 a} \left(\text{erfc}\left(\frac{a-b x}{\sqrt{x}}\right)+e^{4 a b} \text{erfc}\left(\frac{a+b x}{\sqrt{x}}\right)-2 \right)$$
