I have the following: $$ I(a,b) \equiv\int_{-\infty}^\infty e^{\frac{-1}{2}\left(ax^2+\frac{b}{x^2}\right)}dx$$ where $a,b>0$. And I have the following substitution as a hint: $$y=\frac{1}{2}\left(\sqrt{a}x-\frac{\sqrt{b}}{x}\right)$$ And the integral must be evaluated. But I seem to be having trouble going through it. Many thanks in advance.

  • $\begingroup$ Where do you get stuck when using the hint? $\endgroup$
    – Huy
    Dec 31, 2014 at 20:29
  • $\begingroup$ Rearranging the integrand in a form where the substitution can be implemented really. $\endgroup$
    – user211337
    Dec 31, 2014 at 20:31
  • $\begingroup$ @bigman Instead of rearranging the integrand, why not rearrange the substitution relation instead so that $x$ is solved for as a function of $y$? $\endgroup$
    – David H
    Dec 31, 2014 at 21:15
  • $\begingroup$ After substitution, the power becomes $-2y^2+\sqrt{ab}$, and the latter power of $e$ is easily removed. Replacing the $dx$ makes it look terrible though. $\endgroup$
    – David P
    Dec 31, 2014 at 21:44
  • $\begingroup$ Interestingly enough, this happened to me too after a longer moments thought. Odd, this 'hint' appears to be making matters worse! $\endgroup$
    – user211337
    Dec 31, 2014 at 22:14

2 Answers 2


I too got stuck on your hint. Here's how I did your integral.

First letting $x = \frac{t}{\sqrt{a}}, \; dx = \frac{dt}{\sqrt{a}}$, we get

$$ \int_{-\infty}^\infty e^{-\frac{1}{2}\left(a x^2 + \frac{b}{x^2}\right)} \, dx = \frac{1}{\sqrt{a}} I(a b), $$


$$ I(\omega) = \int_{-\infty}^\infty e^{-\frac{1}{2}\left(t^2 + \frac{\omega}{t^2}\right)} = 2\int_{0}^\infty e^{-\frac{1}{2}\left(t^2 + \frac{\omega}{t^2}\right)} \, dt. $$

To find $I(\omega)$, let $t = 1/z, \; dt = -dz/z^2$ to get

$$ I(\omega) = 2\int_{0}^\infty e^{-\frac{1}{2}\left(\frac{1}{z^2} + \omega z^2\right)} \, \frac{dz}{z^2}. $$

Differentiating both sides with respect to $\omega$ gives

$$ I'(\omega) = -\int_{0}^\infty e^{-\frac{1}{2}\left(\frac{1}{z^2} + \omega z^2\right)} \, dz, $$

and letting $z = \frac{y}{\sqrt{\omega}}, \; dz = \frac{dy}{\sqrt{\omega}}$,

$$ I'(\omega) = -\frac{1}{\sqrt{\omega}}\int_{0}^\infty e^{-\frac{1}{2}\left(\frac{\omega}{y^2} + y^2\right)} \, dy = -\frac{1}{2\sqrt{\omega}}I(\omega). $$

Using $I(0) = \sqrt{2\pi}$, solving the separable ODE gives

$$ I(\omega) = \sqrt{2\pi} e^{-\sqrt{\omega}}. $$

Therefore the original integral is

$$ \frac{1}{\sqrt{a}} I(a b) = \sqrt{\frac{2\pi}{a}} e^{-\sqrt{a b}}. $$

  • $\begingroup$ I see! A very interesting approach. Wouldn't have thought of this in a million years! Many thanks! $\endgroup$
    – user211337
    Dec 31, 2014 at 22:15

solving $y=\frac{1}{2}\left(\sqrt{a}x-\frac{\sqrt{b}}{x}\right)$ for $x$ we get $$\left\{\left\{x\to \frac{y}{\sqrt{a}}-\frac{\sqrt{\sqrt{a} \sqrt{b}+y^2}}{\sqrt{a}}\right\},\left\{x\to \frac{\sqrt{\sqrt{a} \sqrt{b}+y^2}}{\sqrt{a}}+\frac{y}{\sqrt{a}}\right\}\right\}$$ and now must be insert $x$ for $x$ in the exponent and after this we get $$dy=\frac{y}{\sqrt{a} \sqrt{\sqrt{a} \sqrt{b}+y^2}}+\frac{1}{\sqrt{a}}dx$$ i hope this will help you

  • $\begingroup$ I see! Odd, how the hint that was given was so obscure! Many thanks for your help! $\endgroup$
    – user211337
    Dec 31, 2014 at 22:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .