This might not be a difficult integral but it made me come up with a new method to solve it so I think it's quite exotic.
Let’s do the general integral
$\displaystyle I(a,b)=\int_{0}^{\infty}e^{-(ax^{-2}+bx^{2})}dx$
Differentiate with respect to a
$\displaystyle \frac{\partial I}{\partial a}=\int_{0}^{\infty}x^{-2}e^{-(ax^{-2}+bx^{2})}dx$
Now differentiate with respect to b
$\displaystyle \frac{\partial^2 I}{\partial a \partial b}=\int_{0}^{\infty}x^{-2}x^{2}e^{-(ax^{-2}+bx^{2})}dx$
$\displaystyle \frac{\partial^2 I}{\partial a \partial b}=\int_{0}^{\infty}e^{-(ax^{-2}+bx^{2})}dx$
$\displaystyle \frac{\partial^2 I}{\partial a \partial b}=I$
Thus our integral satisfies this PDE.This is a hyperbolic homogenous PDE. It is a second order PDE but it is first order with respect to each of the variables so we’ll need two boundary conditions to determine a unique solution.(In this case two asympotic BCs and one Drichlet boundary condition will be used).Keep this in mind we’ll need it later.
Let’s complete the square of expression in the exponential.
$\displaystyle I(a,b)=\int_{0}^{\infty}e^{-(ax^{-2}+bx^{2}-2\sqrt{ab}+2\sqrt{ab})}dx$
$\displaystyle I(a,b)=\int_{0}^{\infty}e^{-(\sqrt{a}x^{-1}-\sqrt{b}x)^{2}-2\sqrt{ab}}dx$
$\displaystyle I(a,b)=e^{-2\sqrt{ab}}\int_{0}^{\infty}e^{-(\sqrt{a}x^{-1}-\sqrt{b}x)^{2}}dx$
Now let’s explore more of it’s properties.One thing to note is that this integral diverges(blows up) at b=0 but at a=0 it has a well known value. It is the Gaussian integral so
$\displaystyle I(0,b)=\int_{0}^{\infty}e^{-(bx^{2})}dx=\frac{1}{2}\sqrt{\frac{\pi}{b}}$
The negative exponential was extracted from the integral rather than the positive one beacause
$\displaystyle \lim_{a\to\infty}\int_{0}^{\infty}e^{-(ax^{-2}+bx^{2})}dx=0$
and
$\displaystyle \lim_{a\to\infty}e^{-2\sqrt{ab}}=0$
So let’s assume that we assume that the solution to our PDE is of the form
$\displaystyle I(a,b)=e^{-2\sqrt{ab}}K(b)$
where K is a function of b(and diverges at b=0)
Let’s put this in the PDE
$\displaystyle \frac{\partial I}{\partial a}=-\sqrt{\frac{b}{a}}e^{-2\sqrt{ab}}K(b)$
$\displaystyle \frac{\partial^2 I}{\partial a \partial b}=-\sqrt{\frac{b}{a}}e^{-2\sqrt{ab}}K^{'}(b)-\frac{1}{2\sqrt{ab}}e^{-2\sqrt{ab}}K(b)+\sqrt{\frac{b}{a}}\sqrt{\frac{a}{b}}e^{-2\sqrt{ab}}K(b)$
$\displaystyle \frac{\partial^2 I}{\partial a \partial b}=e^{-2\sqrt{ab}}(-\sqrt{\frac{b}{a}}K^{'}(b)-\frac{K(b)}{2\sqrt{ab}}+K(b))$
As
$\displaystyle \frac{\partial^2 I}{\partial a \partial b}=I$
So
$\displaystyle e^{-2\sqrt{ab}}(-\sqrt{\frac{b}{a}}K^{'}(b)-\frac{K(b)}{2\sqrt{ab}}+K(b))=e^{-2\sqrt{ab}}K(b)$
$\displaystyle -\sqrt{\frac{b}{a}}K^{'}(b)-\frac{K(b)}{2\sqrt{ab}}+K(b)=K(b)$
$\displaystyle -\sqrt{\frac{b}{a}}K^{'}(a)=\frac{K(b)}{2\sqrt{ab}}$
$\displaystyle K^{'}(b)=-\frac{K(b)}{2b}$
This is a separable ODE.Let’s solve it
$\displaystyle \frac{1}{K}dK=-\frac{1}{2}\frac{1}{b}db$
Let’s integrate
$\displaystyle \int \frac{1}{K}dK=-\frac{1}{2}\int \frac{1}{b}db$
$\displaystyle \ln(K)=-\frac{1}{2}\ln(b)+C$
$\displaystyle \ln(K)=\ln(b^{-\frac{1}{2}})+C$
$\displaystyle K=e^{C}b^{-\frac{1}{2}}$
Let
$\displaystyle v=e^{C}$
So
$\displaystyle K(b)=vb^{-\frac{1}{2}}$
Thus the solution is
$\displaystyle I(a,b)=ve^{-2\sqrt{ab}}b^{-\frac{1}{2}}$
This expression diverges at b=0 which is exactly what we wanted. Now let’s determine the constant v. As
$\displaystyle I(0,b)=\frac{1}{2}\sqrt{\frac{\pi}{b}}$
So
$\displaystyle \frac{1}{2}\sqrt{\frac{\pi}{b}}=vb^{-\frac{1}{2}}e^{0}$
$v=\frac{\sqrt{\pi}}{2}$
Thus the integral is
$\displaystyle \boxed{I(a,b)=\frac{1}{2}\sqrt{\frac{\pi}{b}}e^{-2\sqrt{ab}}} (0\leqslant a,b)$