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?
 A: Notice that your final line can be rewritten as 
$$F'(a) = -2b^2 F(a)$$
And $F(0)$ is a usual Gaussian integral. Can you take it from here?
A: Since you had to prove an identity which had an exponential at the right hand side and you needed to use the Leibnitz integral rule, it should have been natural to think about  reducing the problem to a differential equation whose solution would involve a logarithm.From what you have worked out, it gives $dF=-2b^2Fda$.Since $F$ is never zero, you can write it as $\frac{dF}{F}=-2b^2 da$.Integrating would give you the general solution.In addition to that you need to use the condition $F(0)=\int_{0}^{\infty} e^{-b^2x^2} dx$.The evaluation of this Gaussian integral is pretty standard, you can use properties of the gamma function or polar coordinates to find it out.Once you get this, you have your solution.
A: 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
$$F(a)= -\int_a^\infty F’(s)ds=  \sqrt{\pi}b \int_a^\infty e^{-2sb^2}ds 
= \frac {\sqrt {\pi}}{2b} e^{-2ab^2} $$
