I'm asked to evaluate the following integral:
I know the classic gaussian result, but I cannot seem to use this to help. I'm given a hint to try the substitution: , but I cannot make this work.
Here's my stab at it. Denote the integral by $\displaystyle I$. Because of the symmetry this integral is
$\displaystyle I = 2\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\;dx.\;\;\;(1)$
Under the change of variable $\displaystyle x = \sqrt{\dfrac{b}{a}}\dfrac{1}{y}$, the integral becomes
$\displaystyle I =2\sqrt{\dfrac{b}{a}}\int_0^{\infty} e^{-\frac{1}{2}\left(ay^2 + \frac{b}{y^2}\right)}\dfrac{dy}{y^2}.$.
Now replace the dummy variable with $\displaystyle x$, so
$\displaystyle I =2\sqrt{\dfrac{b}{a}}\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\dfrac{dx}{x^2},\;\;\;(2)$.
Now add $\displaystyle \sqrt{a} \;(1) + \sqrt{a} \;(2)$ giving
$\displaystyle 2 \sqrt{a} I = 2\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\;\left(\sqrt{a}+ \dfrac{\sqrt{b}}{x^2}\right)\;dx\;\;\;(3)$
Now introduce your hint (sort of)
$\displaystyle y = \sqrt{a} x - \dfrac{\sqrt{b}}{x}$
so
$\displaystyle dy = \left(\sqrt{a} + \dfrac{\sqrt{b}}{x^2}\right)\,dx$.
Note that $\displaystyle x \to 0$ gives $\displaystyle y \to - \infty $ and $\displaystyle x \to \infty $ gives $\displaystyle y \to \infty$ so with that and noting that
$\displaystyle y^2 = ax^2 - 2 \sqrt{ab} + \dfrac{b}{x^2}$ gives $\displaystyle ax^2 + \dfrac{b}{x^2} = y^2 + 2 \sqrt{ab} $.
Thus, (3) becomes
$\displaystyle 2 \sqrt{a} I = 2\int_{-\infty}^{\infty} e^{-\frac{1}{2}\left(y^2 + 2 \sqrt{ab}\right)}\;\;dy = 2e^{-\sqrt{ab}}\int_{-\infty}^{\infty} e^{-\frac{1}{2}y^2}\;\;dy$
and solving for $\displaystyle I $ gives
$\displaystyle I = \sqrt{\dfrac{2\pi}{a}} e^{-\sqrt{ab}}$.
Whew! I think I need to lie down.
In that change of variable, there is indeed a minus sign attached to the dy. But the limits of integration get switched: instead of going from 0 to ∞, the integral goes from ∞ to 0. When you switch the limits back again, that introduces a change of sign, which cancels with the other one.
When you make the substitution $\displaystyle x = \phi(y)$ in the definite integral $\displaystyle \int_a^b f(x)\, dx,$ it becomes $\displaystyle \int_{\phi^{-1}(a)}^{\phi^{-1}(b)}f(\phi(y))\phi'(y)\,dy.$ In this case, $\displaystyle \textstyle x = \sqrt{\frac ba}\frac1y,$ and the limits of integration are that x goes from 0 to ∞. When x = 0, y = ∞; and when x = ∞, y = 0. So the dy-integral, instead of going from 0 to ∞, goes from ∞ to 0.