Inverse fourier transform - Where did the Heaviside function come from?

Question

I asked this question on another forum but no answers so I'm copy/pasting it here in hopes that someone can help out

Answer 1

Try in the opposite direction: \begin{align} \mathcal{F}(e^{-ax}H(x)) &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ax}H(x)e^{-isx}dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-ax}e^{-isx}dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-x(a+is)}dx \\ & = \frac{1}{\sqrt{2\pi}}\left.\frac{e^{-x(a+is)}}{a+is}\right|_{x=0}^{\infty} \\ & = \frac{1}{\sqrt{2\pi}}\frac{1}{a+is} \end{align} Because $\mathcal{F}^{-1}(\mathcal{F}(e^{-ax}H(x))=e^{-ax}H(x)$ (except at the discontinuity): $$ \mathcal{F}^{-1}\left(\frac{1}{\sqrt{2\pi}}\frac{1}{a+is}\right)=e^{-ax}H(x). $$ To see where $H(x)$ comes from using more general principles requires some background in Complex Analysis. If you like Fourier and Laplace transform analysis, you'd enjoy such a class. Complex Analysis is one of the most elegant subjects in Mathematics.