Jtdylktuy

If $c >0$ and $d >0$, show that

$$intlimits_{0}^infty e^{-cx} frac{d}{x^{frac{3}{2}} sqrt{2 pi}} e^{-frac{d^2}{2x}} , mathrm dx = e^{-dsqrt{2c}}.$$

Obviously we have that
$$intlimits_{0}^infty e^{-cx} frac{d}{x^{frac{3}{2}} sqrt{2 pi}} e^{-frac{d^2}{2x}} , mathrm dx = frac{d}{sqrt{2pi}} intlimits_0^infty frac{e^{-cx-frac{d^2}{2x}}}{x^{frac{3}{2}}} , mathrm dx.$$

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newcommand{ds}[1]{displaystyle{#1}}
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newcommand{mc}[1]{mathcal{#1}}
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{2pi}},expo{-d^{2}/pars{2x}},dd x}
\[5mm] = &
{d over root{2pi}}int_{0}^{infty}x^{-3/2}
exppars{-root{c over 2}d
bracks{{root{2c} over d}x + {d over root{2c}}{1 over x} }}
,dd x
end{align}

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

begin{align}
&bbox[10px,#ffd]{int_{0}^{infty}expo{-cx}, {d over x^{3/2}root{2pi}},expo{-d^{2}/pars{2x}},dd x}
\[5mm] = &
{d over root{2pi}}int_{0}^{infty}
bracks{2^{3/4}d^{-3/2}c^{3/4}t^{-3}}
exppars{-root{c over 2}dbracks{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{2pi}}
int_{0}^{infty}
exppars{-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{2pi}}
exppars{-droot{2c}} times
\[2mm] &
left(int_{0}^{infty}exppars{-root{c over 2}d
bracks{t - {1 over t}}^{2}}
,{dd t over t^{2}}right.
\[2mm] & +left.
int_{infty}^{0}exppars{-root{c over 2}d
bracks{{1 over t} - t}^{2}}
pars{-1},dd tright)
\[1cm] = &
{2^{1/4}c^{1/4}d^{1/2} over root{2pi}}exppars{-droot{2c}}
times
\[2mm] &
int_{0}^{infty}exppars{-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{2pi}}exppars{-droot{2c}}
underbrace{int_{-infty}^{infty}exppars{-root{c over 2}dt^{2}},dd t}
_{ds{2^{1/4}root{pi} over c^{1/4}d^{1/2}}}
\[5mm] & = bbx{exppars{-droot{2c}}}
end{align}