If $c >0$ and $d >0$, show that
$$\int\limits_{0}^\infty e^{-cx} \frac{d}{x^{\frac{3}{2}} \sqrt{2 \pi}} e^{-\frac{d^2}{2x}} \, \mathrm dx = e^{-d\sqrt{2c}}.$$
Obviously we have that $$\int\limits_{0}^\infty e^{-cx} \frac{d}{x^{\frac{3}{2}} \sqrt{2 \pi}} e^{-\frac{d^2}{2x}} \, \mathrm dx = \frac{d}{\sqrt{2\pi}} \int\limits_0^\infty \frac{e^{-cx-\frac{d^2}{2x}}}{x^{\frac{3}{2}}} \, \mathrm dx.$$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}\expo{-cx}\, {d \over x^{3/2}\root{2\pi}}\,\expo{-d^{2}/\pars{2x}}\,\dd x} \\[5mm] = &\ {d \over \root{2\pi}}\int_{0}^{\infty}x^{-3/2} \exp\pars{-\root{c \over 2}d \bracks{{\root{2c} \over d}x + {d \over \root{2c}}{1 \over x} }} \,\dd x \end{align}
\begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}\expo{-cx}\, {d \over x^{3/2}\root{2\pi}}\,\expo{-d^{2}/\pars{2x}}\,\dd x} \\[5mm] = &\ {d \over \root{2\pi}}\int_{0}^{\infty} \bracks{2^{3/4}d^{-3/2}c^{3/4}t^{-3}} \exp\pars{-\root{c \over 2}d\bracks{t^{2} + {1 \over t^{2}}}}\,2^{1/2}dc^{-1/2}t\,\dd t \\[5mm] = &\ {2^{5/4}c^{1/4}d^{1/2} \over \root{2\pi}} \int_{0}^{\infty} \exp\pars{-\root{c \over 2}d \braces{\bracks{t - {1 \over t}}^{2} + 2}}\,{\dd t \over t^{2}} \\[1cm] = &\ {2^{1/4}c^{1/4}d^{1/2} \over \root{2\pi}} \exp\pars{-d\root{2c}}\ \times \\[2mm] & \left(\int_{0}^{\infty}\exp\pars{-\root{c \over 2}d \bracks{t - {1 \over t}}^{2}} \,{\dd t \over t^{2}}\right. \\[2mm] & \ +\left. \int_{\infty}^{0}\exp\pars{-\root{c \over 2}d \bracks{{1 \over t} - t}^{2}} \pars{-1}\,\dd t\right) \\[1cm] = &\ {2^{1/4}c^{1/4}d^{1/2} \over \root{2\pi}}\exp\pars{-d\root{2c}}\ \times \\[2mm] &\ \int_{0}^{\infty}\exp\pars{-\root{c \over 2}d \braces{\bracks{t - {1 \over t}}^{2} + 2}} \pars{1 + {1 \over t^{2}}}\,\dd t \\[1cm] & \stackrel{1/t\ -\ t\ \mapsto\ t}{=}\,\,\, {2^{1/4}c^{1/4}d^{1/2} \over \root{2\pi}}\exp\pars{-d\root{2c}}\ \underbrace{\int_{-\infty}^{\infty}\exp\pars{-\root{c \over 2}dt^{2}}\,\dd t} _{\ds{2^{1/4}\root{\pi} \over c^{1/4}d^{1/2}}} \\[5mm] & = \bbx{\exp\pars{-d\root{2c}}} \end{align}