Integral using the Substitution y = $\frac{1}{2}(\sqrt{a}x^2+\frac{\sqrt{b}}{x^2})$ Given that $\int_{-\infty}^{\infty} e^{(-\frac{1}{2}x^2)}dx = \sqrt{2\pi}$,
Find $I(a,b) = \int_{-\infty}^{\infty} e^{-\frac{1}{2}(ax^2+\frac{b}{x^2})}dx$.
The question suggests using the substitution: $$y = \frac{1}{2}(\sqrt{a}x-\frac{\sqrt{b}}{x})$$ if careful attention is paid to the limits. I'm not sure how the limits of integral changes as I can't see why the limits wouldn't remain the same using the substitution.
 A: The key here is to note that this integral is improper at zero, so before performing the substitution notice the integrand is even so we may rewrite the integral as
$$I(a,b)=2\int_{0+}^\infty e^{-\frac{1}{2}(ax^2+\frac{b}{x^2})}dx$$
Let $y=\sqrt{a}x-\frac{\sqrt{b}}{x}$. Note the bounds become $-\infty$ to $\infty$, since as $x\rightarrow 0^+$, $y\rightarrow -\infty$. Also notice $y^2=ax^2+\frac{b}{x^2}-2\sqrt{ab}$, and 
$$dy=\sqrt{a}+\frac{\sqrt{b}}{x^2}=\frac{\sqrt{y^2+4\sqrt{ab}}-y}{2\sqrt{b}}\sqrt{y^2+4\sqrt{ab}}\ dx$$ so the integral is
$$I(a,b)=2\int_{-\infty}^\infty e^{-\frac{1}{2}(y^2+2\sqrt{ab})}\frac{2\sqrt{b}}{\sqrt{y^2+4\sqrt{ab}}}\frac{1}{\sqrt{y^2+4\sqrt{ab}}-y}dy$$
Combining the positive and negative sides of the axis gives
$$I(a,b)=2\int_{0}^\infty e^{-\frac{1}{2}(y^2+2\sqrt{ab})}\frac{2\sqrt{b}}{\sqrt{y^2+4\sqrt{ab}}}\left(\frac{1}{\sqrt{y^2+4\sqrt{ab}}-y}+\frac{1}{\sqrt{y^2+4\sqrt{ab}}+y}\right)dy$$
$$=\frac{2}{\sqrt{a}}\int_0^\infty e^{-\frac{1}{2}(y^2+2\sqrt{ab})}dy$$
And from there you can use your identity.
A: The integrand is even, so restrict the integration range to $x\ge 0$ and introduce a factor of $2$. The only discontinuity is at $x=0$, and to the right of that $dy/dx>0$. If you're still stuck, see here.
