Calculate integral with two exponential functions 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 
$$\int\limits_{0}^\infty e^{-cx} \frac{d}{x^{\frac{3}{2}} \sqrt{2 \pi}} e^{-\frac{d^2}{2x}} \, \mathrm dx = \frac{d}{\sqrt{2\pi}} \int\limits_0^\infty \frac{e^{-cx-\frac{d^2}{2x}}}{x^{\frac{3}{2}}} \, \mathrm dx.$$
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\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}

Set $\ds{x \equiv {d \over \root{2c}}t^{2} =
2^{-1/2}dc^{-1/2}\, t^{2} \implies
\dd x = d\root{2 \over c}t\,\dd t =
2^{1/2}dc^{-1/2}t\,\dd t}$:

\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}
