# Evaluation of $\int_{0}^{+\infty} x^{-1/2}e^{-x-\frac{a}{x}}dx, a>0$

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 :)

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


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.

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