Definite integrals by parts

Question

If $\int_{0}^{\infty} e^{-(a^{2}x^{2} + \frac{b^{2}}{x^2})} dx \ = \frac{\sqrt\pi}{2a} e^{-2ab}$, then the value of $\int_{0}^{\infty} x^{-2}e^{-(a^{2}x^{2} + \frac{b^{2}}{x^2})} dx $ is equal to?

To solve this problem I started by integrating by parts.

For ease of writing I kept $ \int_{0}^{\infty} e^{-(a^{2}x^{2} + \frac{b^{2}}{x^2})} dx = \frac{\sqrt\pi}{2a} e^{-2ab} = C $

Integrating by parts:

$$\int_{0}^{\infty} x^{-2}e^{-(a^{2}x^{2} + \frac{b^{2}}{x^2})} = x^{-2}C - \int_{0}^{\infty} -2x^{-3} Cdx $$

$$= x^{-2}C + C $$ $$ = \frac{\sqrt\pi}{2a} e^{-2ab}(x^{-2} + 1)$$

Is this the answer?

Answer 1

Let $b^2=c$ then $$I=\int_{0}^{\infty} e^{-a^2x^2-c/x^2} dx=\frac{\sqrt{\pi}}{2a} e^{-2a \sqrt{c}}$$ D.w.r.t. $c$ on both sides, to get $$J=\int_{0}^{\infty}\frac{1}{x^2} e^{-a^2x^2-c/x^2} dx=\frac{1}{2}\sqrt{\pi/c}~e^{-2a\pi \sqrt{c}}.$$ Put $c=b^2$ to get the value of the improper integral that converges.

Answer 2

Denote $I=\int_{0}^{\infty} e^{-(a^{2}x^{2} + \frac{b^{2}}{x^2})} dx ,\>J=\int_{0}^{\infty} {x^{-2}}e^{-(a^{2}x^{2} + \frac{b^{2}}{x^2})} dx $ and note that

\begin{align} aI-bJ=e^{2ab}\int_{0}^{\infty}\left(a-\frac b{x^2}\right)\> e^{-(ax + \frac{b}{x})^2}dx\overset{ ax+\frac bx\to x} =0 \end{align} Thus, $J= \frac ab I = \frac{\sqrt\pi}{2b} e^{-2ab} $.