<s>[INST] Use differentiation under integral sign to prove $\int_{0}^{\infty} e^{-(x^2+\frac {a^2}{x^2})b^2} dx=\frac {\sqrt {\pi}}{2b} e^{-2ab^2}$ Using differentiation under integral sign, prove that $$\int_{0}^{\infty} e^{-\Big(x^2+\frac {a^2}{x^2}\Big)b^2} dx=\frac {\sqrt {\pi}}{2b}e^{-2ab^2}$$
My Attempt:
Let,
$$F(a)= \int_{0}^{\infty} e^{-\Big(x^2+\frac {a^2}{x^2}\Big)b^2} dx$$
Differentiate with respect to $a$
$$\frac {dF(a)}{da}=\int_{0}^{\infty} e^{-\Big(x^2+\frac {a^2}{x^2}\Big)b^2} \Bigg(-\frac {2ab^2}{x^2}\Bigg) dx$$
Put $\frac {a}{x}=y$
$$\frac {dF(a)}{da}=-2b^2 \int_{0}^{\infty} e^{-\Big(\frac {a^2}{y^2}+y^2\Big)b^2} dy$$
How to proceed after this?
 [/INST] Proceed by averaging the two expressions as follows
\begin{align}
F’(a)&= -b^2 \int^{\infty}_{0} e^{-\Big(x^2+\frac {a^2}{x^2} \Big)b^2} \left(1+\frac a{x^2}\right)dy\\
&=- b^2 e^{-2ab^2}\int^{\infty}_{0} e^{-\left(x-\frac {a}{x} \right)^2 b^2} d\left(x-\frac a{x}\right)\\
 &=- b \>e^{-2ab^2}\int^{\infty}_{-\infty} e^{-u^2} du
= -\sqrt{\pi}b \>e^{-2ab^2}
\end{align}
Then
$$F(a)= -\int_a^\infty F’(s)ds=  \sqrt{\pi}b \int_a^\infty e^{-2sb^2}ds 
= \frac {\sqrt {\pi}}{2b} e^{-2ab^2} $$
</s>