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 $y$).

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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}
