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 :)
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\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}
