# 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 $$y$$).

• You are more likely to get an answer if you present your attempt at the problem. Aug 15, 2019 at 2:07
• Start by enforcing the substitution $\omega \mapsto \omega^2$. Aug 15, 2019 at 2:25

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

$$\int_0^\infty \frac{e^{-y^2/4\omega-t^2\omega}}{\sqrt {\pi\omega}}\,d\omega=\frac{e^{-ty}}{t}$$
