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?
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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.
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@athos Thanks for the edit!! – Robert Z Dec 06 '18 at 12:43
1
Let $u=x-\frac1x$. Then $\mathrm{d}u=\left(1+\frac1{x^2}\right)\mathrm{d}x$
$$
\begin{align}
\int_0^\infty\frac{e^{-x-1/x}}{x\sqrt{x}}\,\mathrm{d}x
&=2\int_0^\infty\frac{e^{-x^2-1/x^2}}{x^2}\,\mathrm{d}x\tag1\\
&=2\int_0^\infty e^{-x^2-1/x^2}\,\mathrm{d}x\tag2\\
&=\int_{-\infty}^\infty e^{-u^2-2}\,\mathrm{d}u\tag3\\
&=\frac{\sqrt\pi}{e^2}\tag4
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto x^2$
$(2)$: substitute $x\mapsto 1/x$
$(3)$: average $(1)$ and $(2)$ and substitute $u=x-1/x$
$(4)$: $\int_{-\infty}^\infty e^{-u^2}\,\mathrm{d}u=\sqrt\pi$
robjohn
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