$$\int_0^\infty\frac{\exp{\left(-\frac {y^2}{4w}-t^2w\right)}}{\sqrt {\pi w}}dw=\frac{\exp(-ty)}t$$ 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 $y$).
1
$\begingroup$
$\endgroup$
-
3$\begingroup$ You are more likely to get an answer if you present your attempt at the problem. $\endgroup$ – JG123 Aug 15 at 2:07
-
$\begingroup$ Start by enforcing the substitution $\omega \mapsto \omega^2$. $\endgroup$ – Mark Viola Aug 15 at 2:25
2
$\begingroup$
$\endgroup$
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
$$\int_0^\infty \frac{e^{-y^2/4\omega-t^2\omega}}{\sqrt {\pi\omega}}\,d\omega=\frac{e^{-ty}}{t}$$
