An improper integrals related to probability, $\int_0^\infty\frac1y \exp(\frac{-x_0}y-y)\,dy$ How can I calculate the integral
$$\int_0^\infty{\frac1y e^{\frac{-x_0}y-y}}dy$$
in terms of well-known constants and functions?
I used some fundamental techniques of integration but got nothing.
 A: Let $u = y+x_0/y$, then 
$$y=\frac12 \left (u \pm \sqrt{u^2-4 x_0} \right ) $$
$$dy=\frac12 \left (1 \pm \frac{u}{\sqrt{u^2-4 x_0}} \right ) du $$
Then the integral may be rewritten as (see this answer)
$$ \int_{\infty}^{2 \sqrt{x_0}} du \left (1 - \frac{u}{\sqrt{u^2-4 x_0}} \right ) \frac{e^{-u}}{\left (u - \sqrt{u^2-4 x_0} \right )} + \int_{2 \sqrt{x_0}}^{\infty} du \left (1 + \frac{u}{\sqrt{u^2-4 x_0}} \right ) \frac{e^{-u}}{\left (u + \sqrt{u^2-4 x_0} \right )} $$
which simplifies to
$$\begin{align}2 \int_{2 \sqrt{x_0}}^{\infty} du \, \left (u^2-4 x_0 \right )^{-1/2} e^{-u} &= 2 \int_1^{\infty} dv \, (v^2-1)^{-1/2} e^{-2 \sqrt{x_0} v}\\ &= 2 \int_0^{\infty} dt \, e^{-2 \sqrt{x_0} \cosh{t}} \\ &= 2 K_0 \left ( 2 \sqrt{x_0}\right )  \end{align}$$
where $K_0$ is the modified Bessel function of the second kind of zeroth order.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\int_{0}^{\infty}{1 \over y}\,\exp\pars{-\,{x_{0} \over y} - y}\,\dd y}\ =\
\overbrace{\int_{0}^{\infty}{1 \over y}\,\exp\pars{-\root{x_{0}}\,
\bracks{{\root{x_{0}} \over y} + {y \over \root{x_{0}}}}}\,\dd y}
^{\ds{\dsc{y}=\dsc{\root{x_{0}}\exp\pars{\theta}}}}
\\[5mm]&=\int_{-\infty}^{\infty}{1 \over \root{x_{0}}\expo{\theta}}\,
\exp\pars{-\root{x_{0}}\bracks{\expo{-\theta} + \expo{\theta}}}\,\root{x_{0}}\expo{\theta}\,\dd\theta
\\[5mm]&=2\int_{0}^{\infty}\exp\pars{-2\root{x_{0}}\cosh\pars{\theta}}\,\dd\theta
=\color{#66f}{\large 2\,{\rm K}_{0}\pars{2\root{x_{0}}}}
\end{align}

$\ds{\,{\rm K}_{\nu}}$ is a Bessel Function. See $\ds{\bf 9.6.24}$
  in this link.

