How does one derive the following formula of integration? $$\int_0^\infty\frac{\exp{\left(-\frac {y^2}{4w}-t^2w\right)}}{\sqrt {\pi w}}dw=\frac{\exp(-ty)}t$$ for $t$ and $y$ positive. This integral is useful in the following context: suppose we are given $$\int_0^\infty tf(t){\exp{\left(-t^2w\right)}}dt$$ (a function of $w$) and we want to convert it to the Laplace transform of $f(t)$. Multiplying by $$\frac{\exp{\left(-\frac {y^2}{4w}\right)}}{\sqrt {\pi w}}$$ and integrating over  $w$ from $0$ to $\infty$ will clearly do the trick (resulting in a function of $y$).
 A: Let $F(y,t)$ be given by the integral 
$$F(y,t)=\int_0^\infty \frac{e^{-y^2/4\omega-t^2\omega}}{\sqrt \omega}\,d\omega\tag1$$
First, enforcing the substitution $\omega\mapsto\omega^2$ reveals
$$\begin{align}
F(y,t)&=2\int_0^\infty e^{-y^2/4\omega^2-t^2\omega^2}\,d\omega\tag2
\end{align}$$
Second, making the substitution $x=\sqrt{\frac{2t}{y}}\,\omega$ in $(2)$, we find 
$$\begin{align}
F(y,t)&=\sqrt{\frac{2y}{t}}\int_0^\infty e^{-ty(x^2+1/x^2)/2}\,dx\\\\
&=\sqrt{\frac{2y}{t}}e^{-ty}\int_0^\infty e^{-ty(x-1/x)^2/2}\,dx\tag3
\end{align}$$
Third, enforcing the substitution $x\mapsto1/x$ in $(3)$ yields
$$F(y,t)=\sqrt{\frac{2y}{t}}e^{-ty}\int_0^\infty e^{-ty(x-1/x)^2/2}\,\frac1{x^2}\,dx\tag4$$
Adding $(3)$ and $(4)$ gives
$$\begin{align}
2F(y,t)&=\sqrt{\frac{2y}{t}}e^{-ty}\int_0^\infty e^{-ty(x-1/x)^2/2}\,\left(1+\frac1{x^2}\right)\,dx\\\\
&=\sqrt{\frac{2y}{t}}e^{-ty}\int_0^\infty e^{-ty(x-1/x)^2/2}\,d\left(x-\frac1x\right)\\\\
&=\sqrt{\frac{2y}{t}}e^{-ty} \int_{-\infty}^\infty e^{-tyu^2/2}\,du\\\\
&=2\frac{\sqrt \pi }{t}e^{-ty}\tag5
\end{align}$$
Dividing both sides of $(5)$ by $2\sqrt\pi$ yields the coveted result
$$\int_0^\infty \frac{e^{-y^2/4\omega-t^2\omega}}{\sqrt {\pi\omega}}\,d\omega=\frac{e^{-ty}}{t}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{w \equiv {y \over 2t}\,\expo{2\theta}}$:
\begin{align}
&\bbox[15px,#ffd]{\int_{0}^{\infty}
\exp\pars{-{y^{2} \over 4w} -t^{2}w}{\dd w \over \root{\pi w}}}
\\[5mm] = &\
{1 \over \root{\pi}}\int_{-\infty}^{\infty}
\exp\pars{-{y^{2} \over 4\bracks{y\expo{2\theta}/2t}} -
t^{2}\bracks{{y \over 2t}\,\expo{2\theta}}}
{y \over 2t}\,{2\expo{2\theta}\,\dd\theta \over \root{y\expo{2\theta}/\pars{2t}}}
\\[5mm] = &\
{1 \over \root{\pi}}\root{2y \over t}\int_{-\infty}^{\infty}
\expo{-ty\cosh\pars{2\theta}}\expo{\theta}\dd\theta
\\[5mm] = &\
{2 \over \root{\pi}}\root{2y \over t}\int_{0}^{\infty}
\expo{-ty\cosh\pars{2\theta}}\cosh\pars{\theta}\dd\theta
\\[5mm] = &\
{2 \over \root{\pi}}\root{2y \over t}\int_{0}^{\infty}
\exp\pars{-ty\bracks{2\sinh^{2}\pars{\theta} + 1}}
\cosh\pars{\theta}\dd\theta
\\[5mm] \stackrel{\sinh\pars{\theta}\ =\ x}{=}\,\,\,&
{2 \over \root{\pi}}\root{2y \over t}\expo{-ty}\int_{0}^{\infty}
\exp\pars{-2tyx^{2}}\,\dd x
\\[5mm] = &\
{2 \over \root{\pi}}\root{2y \over t}\expo{-ty}\pars{\root{\pi/2} \over 2\root{ty}} = \bbx{\expo{-ty} \over t}
\end{align}
