How to compute this integral? :
$$\int_0^{\infty}\sqrt x \exp\left(-x-\frac{1}{x}\right) \, dx$$
Wolframalpha gives the answer $\dfrac{3\sqrt{\pi}}{2e^2}$, but how to compute this?
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1$\begingroup$ Break it up into $\int_0^1+\int_1^\infty$ and let $t=\dfrac1x$ in one of them. $\endgroup$ – Lucian Apr 3 '14 at 19:19
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$\begingroup$ @lucian tried, but I couldn't make it work $\endgroup$ – Ziqian Xie Apr 4 '14 at 0:27
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A little roundabout, but here goes. Write
$$\begin{align}I &= \underbrace{\int_0^{\infty} dx \, \sqrt{x} \, e^{-\left (x+\frac1{x} \right )}}_{x=u^2} \\ &= 2 e^2 \underbrace{\int_0^{\infty} du \, u^2 \, e^{-\left (u+\frac1{u} \right )^2}}_{v=u+\frac1{u}}\\ &= e^2 \int_{\infty}^2 dv \, \left (1-\frac{v}{\sqrt{v^2-4}} \right )\left (\frac{v^2}{2}-\frac{v}{2} \sqrt{v^2-4}-1 \right )\, e^{-v^2}\\&+e^2 \int_2^{\infty} dv \, \left (1+\frac{v}{\sqrt{v^2-4}} \right )\left (\frac{v^2}{2}+\frac{v}{2} \sqrt{v^2-4}-1 \right )\, e^{-v^2}\\ &= 2 e^2 \underbrace{\int_2^{\infty} dv \,v \left ( \sqrt{v^2-4}+\frac1{\sqrt{v^2-4}}\right )\, e^{-v^2}}_{y=v^2-4}\\ &= e^{-2} \int_0^{\infty} dy \, \left (\sqrt{y}+\frac1{\sqrt{y}} \right ) e^{-y}\\ &= e^{-2} \left (\frac{\sqrt{\pi}}{2} + \sqrt{\pi} \right ) \\ &= \frac{3 \sqrt{\pi}}{2 e^2}\end{align}$$
For a little more background on how the integral over $v$ gets split, see this answer. Also note that, in the third step, I made use of the fact that $u^2=v u-1$.
answered Apr 3 '14 at 19:49
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\begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}\root{x} \exp\pars{-x - {1 \over x}}\,\dd x} \\[5mm] \stackrel{x\ =\ \expo{2\theta}}{=}\,\,\,& \int_{-\infty}^{\infty}\expo{\theta} \exp\pars{-2\cosh\pars{2\theta}}\pars{2\expo{2\theta}}\,\dd\theta \\[5mm] = &\ 4\int_{0}^{\infty}\cosh\pars{3\theta} \exp\pars{-2\cosh\pars{2\theta}}\,\dd\theta \\[5mm] = &\ 4\int_{0}^{\infty}\ \overbrace{\cosh\pars{\theta}\bracks{4\sinh^{2}\pars{\theta} + 1}}^{\ds{\cosh\pars{3\theta}}}\ \exp\pars{-4\sinh^{2}\pars{\theta} - 2}\,\dd\theta \\[5mm] \stackrel{t\ =\ \sinh{\theta}}{=}\,\,\,\,\,\,& 4\expo{-2}\int_{0}^{\infty}\pars{4t^{2} + 1}\expo{-4t^{2}}\,\dd t = \bbx{3\root{\pi} \over 2\expo{2}} \approx 0.3598 \\ & \end{align}
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$
\begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}\root{x} \exp\pars{-x - {1 \over x}}\,\dd x} \\[5mm] \stackrel{x\ =\ \expo{2\theta}}{=}\,\,\,& \int_{-\infty}^{\infty}\expo{\theta} \exp\pars{-2\cosh\pars{2\theta}}\pars{2\expo{2\theta}}\,\dd\theta \\[5mm] = &\ 4\int_{0}^{\infty}\cosh\pars{3\theta} \exp\pars{-2\cosh\pars{2\theta}}\,\dd\theta \\[5mm] = &\ 4\int_{0}^{\infty}\ \overbrace{\cosh\pars{\theta}\bracks{4\sinh^{2}\pars{\theta} + 1}}^{\ds{\cosh\pars{3\theta}}}\ \exp\pars{-4\sinh^{2}\pars{\theta} - 2}\,\dd\theta \\[5mm] \stackrel{t\ =\ \sinh{\theta}}{=}\,\,\,\,\,\,& 4\expo{-2}\int_{0}^{\infty}\pars{4t^{2} + 1}\expo{-4t^{2}}\,\dd t = \bbx{3\root{\pi} \over 2\expo{2}} \approx 0.3598 \\ & \end{align}
answered Nov 30 '20 at 4:17
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With $t=\sqrt x$
\begin{align} \int_0^{\infty}\sqrt x e^{-x-\frac{1}{x}} \, dx &= \int_{-\infty}^{\infty}t^2e^{-t^2-\frac1{t^2}}dt = \frac1{e^2}\int_{-\infty}^{\infty}t^2e^{-(t-\frac1{t})^2}dt\\ &= \frac1{e^2}\left(\int_{-\infty}^{\infty}\left(t^2 -1\right)e^{-(t-\frac1{t})^2}dt +\int_{-\infty}^{\infty}\underset{t\to \frac1t}{e^{-(t-\frac1{t})^2}}dt\right)\\ &=\frac1{e^2}\int_{-\infty}^{\infty}\left((t-\frac1t)^2 +1 \right)e^{-(t-\frac1{t})^2}dt \\ &= \frac1{e^2}\int_{-\infty}^{\infty}\left(t^2 +1 \right)e^{-t^2}dt =\frac{3\sqrt\pi}{2e^2}\\ \end{align} where $\int_{-\infty}^{\infty}f(t-\frac1t)dt = \int_{-\infty}^{\infty}f(t)dt$ is used.
