<s>[INST] Integral of an exponential 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.
 [/INST] 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
$$ \int_{-\infty}^\infty e^{-\frac{1}{2}\left(a x^2 + \frac{b}{x^2}\right)} \, dx = \frac{1}{\sqrt{a}} I(a b), $$
where 
$$ I(\omega) = \int_{-\infty}^\infty e^{-\frac{1}{2}\left(t^2 + \frac{\omega}{t^2}\right)} = 2\int_{0}^\infty e^{-\frac{1}{2}\left(t^2 + \frac{\omega}{t^2}\right)} \, dt. $$
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 
$$ \frac{1}{\sqrt{a}} I(a b) = \sqrt{\frac{2\pi}{a}} e^{-\sqrt{a b}}. $$
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