i want to understand the solution of the following integral: $$\int_{0}^{+\infty} x^{-1/2}\;\exp({-x-\frac{a}{x}})dx=\sqrt\pi\; \exp(-2\sqrt a),\, a>0$$ The solution was generated by Mathematica, but i can't find a way to solve it my self. I have tried to substitute $\exp(-\frac{a}{x})=\sum_{n=0}^\infty\frac{(-\frac{a}{x})^n}{n!}$ and switching integration and summation (i now know this isn't legal in this case), which gave: $$\sum_{n=0}^\infty\frac{(-a)^n}{n!}\int_0^{+\infty}x^{\frac 1 2 -n+1}\;e^{-x}dx$$ I also now know that these integrals do not converge, but i just set them to $\Gamma(\frac 1 2-n)=\sqrt \pi \frac{n!(-4^n)}{(2n)!}$, which yields: $$ \sum_{n=0}^\infty\frac{(2\sqrt a)^n}{n!}\sqrt \pi= \sqrt\pi \;\exp(2\sqrt a)$$ It seems using $|\Gamma(\frac 1 2-n)|$ instead of $\Gamma(\frac 1 2-n)$ gives the right results. Observing that $$x^{\frac 1 2 -n+1}\;e^{-x}>0\,\forall x\in(0,+\infty)$$ using the absolute value seems somehow justified.

My questions are:
1. How do you evaluate the original integral?
2. Could the methods i used be modified so that they were legal and yielded the right result?

Thank you all very much in advance :)

  • 2
    $\begingroup$ Very good first question :) By the way for formatting you can use "\exp" instead of "Exp". $\endgroup$ – John Doe Jun 1 '17 at 12:41
  • $\begingroup$ Thank you, i've tried my best. And this will be edited right away :) $\endgroup$ – Leonard Jun 1 '17 at 12:42

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\infty}x^{-1/2}\exp\pars{-x - {a \over x}}\,\dd x \\[5mm] \stackrel{x\ =\ \root{a}\exp\pars{2\theta}}{=}\,\,\,& \int_{0}^{\infty}a^{-1/4}\expo{-\theta}\exp\pars{-\root{a}\expo{2\theta} - {a \over \root{a}\expo{2\theta}}}\,2a^{1/2}\expo{2\theta}\,\dd \theta \\[5mm] = &\ 2a^{1/4}\int_{-\infty}^{\infty} \expo{\theta}\exp\pars{-2\root{a}\cosh\pars{2\theta}}\,\dd \theta \\[5mm] = &\ 2a^{1/4}\int_{-\infty}^{\infty} \bracks{\cosh\pars{\theta} + \sinh\pars{\theta}}\exp\pars{-2\root{a}\bracks{2\sinh^{2}\pars{\theta} + 1}}\,\dd \theta \\[5mm] = &\ 4a^{1/4}\exp\pars{-2\root{a}}\int_{0}^{\infty} \exp\pars{-4\root{a}\sinh^{2}{\theta}}\cosh{\theta}\,\dd \theta \\[5mm] \stackrel{\sinh\pars{\theta}\ =\ t}{=}\,\,\,& 4a^{1/4}\exp\pars{-2\root{a}} \int_{0}^{\infty}\exp\pars{-4\root{a}t^{2}}\,\dd t = 4a^{1/4}\exp\pars{-2\root{a}}\pars{\root{\pi}/2 \over 2a^{1/4}} \\[5mm] = &\ \bbx{\root{\pi}\exp\pars{-2\root{a}}} \end{align}


Hint. A way to see the evaluation of the given integral is to use the following property

Property. If $f\in L^1(\mathbb{R})$, then $$ \int_{-\infty}^{+\infty}f(u)\,du = \int_{-\infty}^{+\infty}f\left(u-\frac{1}{u}\right)\,du. $$

See many proofs here.

By the change of variable, $$ x=\sqrt{a}\cdot u^2, \qquad du= 2^{-1}a^{-1/4}\cdot x^{-1/2}\:dx, \qquad a>0. $$ One gets $$ \begin{align} \int_0^\infty x^{-1/2}e^{\large -x-\frac ax}dx& =2a^{1/4}\int_0^\infty e^{\large -\sqrt{a}\left(u^2+\frac 1{u^2}\right)}du \\\\& =a^{1/4}e^{-2\sqrt{a}}\int_{-\infty}^\infty e^{\large -\sqrt{a}\left(u-\frac 1{u}\right)^2}du \\\\& =a^{1/4}e^{-2\sqrt{a}}\int_{-\infty}^\infty e^{\large -\sqrt{a}\cdot u^2}du \\\\& =\sqrt{\pi}\cdot e^{-2\sqrt{a}} \end{align} $$ as expected.

  • 1
    $\begingroup$ See also 'Glasser's Master Theorem' $\endgroup$ – Dilemian Jun 1 '17 at 13:03
  • $\begingroup$ Thank you, Olivier! This identity is new to me, there is alway something to learn :) $\endgroup$ – Leonard Jun 2 '17 at 14:41
  • $\begingroup$ @Leonard You are very welcome. $\endgroup$ – Olivier Oloa Jun 2 '17 at 15:17

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