Integral involving exponential function

Question

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…

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\infty}\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.

---

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

Answer 2

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

$$ \forall \alpha>0,\qquad \int_{0}^{+\infty}\exp\left(-\alpha\left(w+\frac{1}{w}\right)\right)\,dw = 2\cdot K_1(2\alpha).$$