Integration techniques: Fourier--Laplace Commutation

Once upon a time, I was studying Laplace transforms as they related to generating functions and other applied-math linear transforms. One of the really fascinating examples that came up was the calculation of \begin{gather} \int_{0}^{\infty} \frac{ e^{-st-\frac{x^2}{2Dt}} }{ \sqrt{2\pi D t} } dt. \end{gather} This transformation comes up when one is working to solve certain examples of the diffusion equation \[ \frac{\partial n}{\partial t} = \frac{D}{2} \frac{\partial^2 n}{\partial x^2} \] What is interesting about this integral is that it provides an example of an unusual integration technique that isn't often taught in calculus classes. It can be solved by using Fourier transforms! Making the problem harder at first, actually eventually makes it easier!

The Fourier transform of a function $f(x)$ is \begin{gather} \mathscr{F}[f] := \int_{-\infty}^{\infty} e^{2 \pi i \omega x} f(x) \; dx \end{gather} The inverse Fourier transform \begin{gather} \mathscr{F}^{-1}[\hat{f}] := \int_{-\infty}^{\infty} e^{-2 \pi i \omega x} \hat{f}(\omega) \; d\omega \end{gather} The closely related Laplace transform \begin{gather} \mathscr{L}[f] := \int_{0}^{\infty} e^{-st} f(t) dt. \end{gather}

If the Laplace and Fourier transforms commute, then \[ \mathscr{L}[f] = \mathscr{F}^{-1}[ \mathscr{L}[ \mathscr{F}[f]]].\] Making this explict, it turns out that we can calculate the Laplace transform AFTER we do the Fourier transform, but not before. We then discover... \begin{gather} \int_{0}^{\infty} \frac{ e^{-st-\frac{x^2}{2Dt}} }{ \sqrt{2\pi D t} } dt = \int_{-\infty}^{\infty} e^{-2 \pi i \omega x} \int_{0}^{\infty} \int_{-\infty}^{\infty} \frac{ e^{-\frac{x^2}{2Dt}-st+2\pi i \omega x} }{\sqrt{2 \pi D t}} dx\; dt \; d\omega \\ = \frac{e^{\sqrt{\frac{2 s x^2}{D}}}}{\sqrt{2Ds}} \end{gather} This Laplace transform can be performed by first calculating the Fourier transform in $x$, then calculating the Laplace transform in $t$, and then inverting the Fourier transform. Residual calculus is helpful when doing the inversion.

This is just an illustrative example. I don't know how often this trick is actually useful. But it's similar in flavor to the old technique of differentiation under the integral sign.