Integral involving exponential function Can you give any hints how to solve the integral
$$
\int\limits_{0}^{\infty} {\exp \left( \frac{-w (z-u)^2}{2u^2 z}  \right) dz}
$$
for $u>0, w>0$, or does no close form exist? Substitution seems to be difficult…
 A: By expanding the square and performing a substitution we have that the given integral depends on a modified Bessel function of the second kind, since
$$ \forall \alpha>0,\qquad \int_{0}^{+\infty}\exp\left(-\alpha\left(w+\frac{1}{w}\right)\right)\,dw = 2\cdot K_1(2\alpha).$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&\int_{0}^{\infty}\exp\pars{-\,{w\bracks{z - u}^{2} \over 2u^{2}z}}\,\dd z =
\int_{0}^{\infty}\exp\pars{-\,{w \over 2u^{2}}\,
\bracks{\root{z} - {u \over \root{z}}}^{2}}\,\dd z
\\[5mm] \stackrel{\root{z}\ =\ \root{u}\expo{\theta}}{=}\,\,\,&
\int_{-\infty}^{\infty}\exp\pars{-\,{2w \over u}\,
\sinh^{2}\pars{\theta}}\pars{2u\expo{2\theta}}\,\dd\theta
\\[5mm] = &\
u\int_{-\infty}^{\infty}\exp\pars{-\,{2w \over u}\,
{\cosh\pars{2\theta} - 1 \over 2}}\bracks{\sinh\pars{2\theta} + \cosh\pars{2\theta}}\,2\,\dd\theta
\\[5mm] = &\
2u\expo{w/u}\int_{0}^{\infty}\exp\pars{-\,{w \over u}\,
\cosh\pars{\theta}}\cosh\pars{\theta}\,\dd\theta
\\[5mm] = &\
\left.-2u\expo{w/u}\partiald{}{z}\int_{0}^{\infty}
\exp\pars{-\,z\cosh\pars{\theta}}\,\dd\theta\,\right\vert_{\ z\ =\ w/u}
\\[5mm] = &\
-2u\expo{w/u}\,\underbrace{\mrm{K}_{0}'\pars{w \over \mu}}
_{\ds{-\,\mrm{K}_{1}\pars{w/u}}}\,,\qquad\qquad\qquad\qquad\qquad\qquad
\verts{\mrm{arg}\pars{w \over u}} < {\pi \over 2}
\end{align}

$\ds{\,\mrm{K}_{\nu}}$ is a
  Modified Bessel Function. See
  A & S Table.


$$
\bbx{\int_{0}^{\infty}\exp\pars{-\,{w\bracks{z - u}^{2} \over 2u^{2}z}}\,\dd z =
2u\expo{w/u}\,\mrm{K}_{1}\pars{w \over u}}
\qquad\qquad
\verts{\mrm{arg}\pars{w \over u}} < {\pi \over 2}
$$
