Prove $\int_0^\infty \frac{e^{-x-1/x}}{x \sqrt{x}} dx = \frac{\sqrt{\pi}}{e^2}$

Question

I went into this integration,
$$\int_0^\infty \frac{e^{-x-1/x}}{x \sqrt{x}} dx = \frac{\sqrt{\pi}}{e^2}$$
Wolfram Online Integral Calculator gives the result above without process. May I know how to derive the result?

Answer 1

After letting $x=1/s^2$ we get $$\begin{align}\int_0^\infty \frac{e^{-x-1/x}}{x \sqrt{x}} dx&=2\int_0^\infty e^{-s^2-1/s^2} ds=\frac{2}{e^2}\int_0^\infty e^{-(s-1/s)^2} ds\\ &=\frac{1}{e^2}\int_{-\infty}^\infty e^{-(s-1/s)^2} ds=\frac{1}{e^2}\int_{-\infty}^\infty e^{-s^2} ds=\frac{\sqrt{\pi}}{e^2}\end{align}$$ where at the last two steps we used How to prove $\int_{-\infty}^{+\infty} f(x)dx = \int_{-\infty}^{+\infty} f\left(x - \frac{1}{x}\right)dx?$ and then the Gaussian integral.