how to derive the steps of this intergral for the product of two exponential terms? For homework, I was looking for closed form solutions of exponential functions from this wiki site.
https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions
But I was not sure how to derive the steps for this particular integral,
$\int_{-\infty}^{\infty}e^{-ax^{2}}e^{-\frac{b}{x^{2}}}dx={\sqrt {\frac {\pi }{a}}}e^{-2{\sqrt {ab}}}\quad (a,b>0)$
I am fairly new to normal distribution but I assume this has to do with Taylor series..
Any pointers would be greatly appreciated!

Start completing the square
$$ax^2+\frac b {x^2}=a\left(x+\sqrt{\frac{b}{a}}\frac{1}{x}\right)^2-2\sqrt{ab}$$
$$x+\sqrt{\frac{b}{a}}\frac{1}{x}=t \implies x=\frac{1}{2} \left(t\pm\sqrt{t^2+4 \sqrt{\frac{b}{a}}}\right)$$
Making the calculation for the $\color{red}{-}$ branch.
This makes now the integrand to be
$$\frac{1}{2} e^{2 a \sqrt{\frac{b}{a}}} \left(e^{-a t^2}-\frac{t e^{-a t^2}}{\sqrt{t^2-4 \sqrt{\frac{b}{a}}}}\right)$$ Now, we face two "simple" integrals
$$\int e^{-a t^2}\,dt=\frac{\sqrt{\pi } \text{erf}\left(\sqrt{a} t\right)}{2 \sqrt{a}}$$
$$\int \frac{t e^{-a t^2}}{\sqrt{t^2-4 \sqrt{\frac{b}{a}}}}\,dt=\frac{\sqrt{\pi } e^{-4 a \sqrt{\frac{b}{a}}} \text{erf}\left(\sqrt{a} \sqrt{t^2-4
   \sqrt{\frac{b}{a}}}\right)}{2 \sqrt{a}}$$
Back to $x$
$$\frac{\sqrt{\pi } e^{-2 \sqrt{a} \sqrt{b}}
   \left(\text{erfc}\left(\frac{\sqrt{b}}{x}-\sqrt{a} x\right)-e^{4 \sqrt{a}
   \sqrt{b}} \text{erfc}\left(\sqrt{a} x+\frac{\sqrt{b}}{x}\right)\right)}{4
   \sqrt{a}}$$ Integrateing between $-p$ and $+p$
$$\frac{\sqrt{\pi } e^{-2 \sqrt{a} \sqrt{b}} \left(e^{4 \sqrt{a} \sqrt{b}}
   \text{erf}\left(\sqrt{a}
   p+\frac{\sqrt{b}}{p}\right)-\text{erf}\left(\frac{\sqrt{b}}{p}-\sqrt{a}
   p\right)\right)}{2 \sqrt{a}}$$ and the limit is
$$\frac{\sqrt{\pi } e^{-2 \sqrt{a b}}}{\sqrt{a}}$$
