0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I am not sure that an antiderivative could be found (even with special functions). Effectively $$\int_0^\infty \frac{e^{-\frac{(a+b t)^2}{t}}}{t^{3/2}}\,dt=\frac{\sqrt{\pi } e^{-4 a b}}{a}$$ $\endgroup$ – Claude Leibovici Sep 29 at 6:37
  • $\begingroup$ @ClaudeLeibovici Thanks. May you please explain how to get this formula? Contour integration? I still cannot derive this.... $\endgroup$ – Richard Sep 29 at 6:41
  • $\begingroup$ No idea and I am sorry for that. This is a result given by a CAS. $\endgroup$ – Claude Leibovici Sep 29 at 6:44
  • $\begingroup$ I would say that the original integral is quite scary at first sight, but if you set $a=\frac{\ln H}{\sqrt 2 \sigma}$ and $b=\frac{\sigma}{2\sqrt 2}$, as Claude did, then things become much friendlier. $\endgroup$ – Nyssa Sep 29 at 8:57
6
$\begingroup$

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


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:

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


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

$\endgroup$
  • 1
    $\begingroup$ This is really nice ! Thanks for posting such a beautiful anser and $+1$ for sure. $\endgroup$ – Claude Leibovici Sep 29 at 8:57
  • $\begingroup$ Thank you Claude! Regarding the comments from bellow the question, wolfram can produce a closed form in terms of the error function for $\int \exp{\left(-{(ax+b/x)^2}\right)}dx$ so the original integral must have an antiderivative too. Not sure why it didn't spit one out initially. $\endgroup$ – Nyssa Sep 29 at 9:02
  • 1
    $\begingroup$ I got it thanks to your work. Cheers :-) $\endgroup$ – Claude Leibovici Sep 29 at 9:24
  • $\begingroup$ At the price of a bunch of $\sqrt{a^2}$ and $\sqrt{b^2}$ we can cover all situations but $(a,b>0)$ give the simplest formula. $\endgroup$ – Claude Leibovici Sep 29 at 12:49
1
$\begingroup$

This is not an answer.

After カカロット's elegant solution, assuming $a>0$ and $b>0$, we can find $$\int \frac{e^{-\frac{(a+b x)^2}{x}}}{x^{3/2}} \, dx=\frac{\sqrt{\pi } e^{-4 a b}}{2 a} \left(\text{erfc}\left(\frac{a-b x}{\sqrt{x}}\right)+e^{4 a b} \text{erfc}\left(\frac{a+b x}{\sqrt{x}}\right)-2 \right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.