Showing $\int_{-\infty}^{\infty} \exp\left({-a x^2 -\frac{b}{x^2}}\right) \,dx = \sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab} }$ How to calculate the following integral:
$$\int_{-\infty}^{\infty} \exp{\left({-a x^2 -\frac{b}{x^2}}\right)} \,dx = \sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}.$$

$$\left({-a x^2 -\frac{b}{x^2}}\right) = -\left({a x^2 -2\sqrt{ab}+\frac{b}{x^2}}\right)-2\sqrt{ab} =-a\left({x -\frac{\sqrt{b/a}}{x}}\right)^2-2\sqrt{ab}  $$
Then using that 
$$\int_{-\infty}^{+\infty}f\left(x-\frac{c}x\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x$$
With $f(x) = e^{-ax^2}$ and $c=\sqrt{b/a}$
$$I =\int_{-\infty}^{\infty} \exp{\left({-a x^2 -\frac{b}{x^2}}\right)} \,dx =e^{-2\sqrt{ab}}  \int_{-\infty}^{\infty}e^{-ax^2}\,dx = \sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}$$

Reference 
How to prove $\int_{-\infty}^{+\infty} f(x)dx = \int_{-\infty}^{+\infty} f\left(x - \frac{1}{x}\right)dx?$
