Derivation of Heaviside step function in its integral form

Question

I'm studying the Heaviside step function: $$ \theta(x) = \begin{cases} 1, & \text{if}& x>0\\ 0,&\text{if} &x<0 \end{cases} $$ I have problem to prove that this function can be represented as: $$ \theta(\tau) = \frac{-1}{2\pi i}\lim_{\varepsilon \rightarrow 0}\int_{-\infty}^\infty\frac{d\omega}{\omega + i\varepsilon}e^{-i\omega\tau} $$ I was thinking about integrating this expression using residue theorem, but I haven't done it for a long time, and it didn't work out and now I don't think it is the right way. So my question is how you derive such representation?

Answer 1

I thought it might be instructive to present an approch that circumvents contour integration. Proceeding, we note that we have

$$\begin{align} \int_{-\infty}^\infty \frac{1}{\omega+i\varepsilon}e^{-i\omega \tau}\,d\omega&=\int_{-\infty}^\infty \frac{\omega -i\varepsilon}{\omega^2+\varepsilon^2}e^{-i\omega \tau}\,d\omega\\\\ &=-i\varepsilon \int_{-\infty}^\infty \frac{\cos(\omega \tau)}{\omega^2+\varepsilon^2}\,d\omega -i \int_{-\infty}^\infty\frac{\omega \sin(\omega\tau)}{\omega^2+\varepsilon^2}\,d\omega\\\\ &=i\left( \frac{d}{d\tau}-\varepsilon \right)\int_{-\infty}^\infty \frac{\cos(\omega \tau)}{\omega^2+\varepsilon^2}\,d\omega \end{align}$$

Therefore, we need only evaluate the integral, for $\varepsilon>0$, $\int_{-\infty}^\infty \frac{\cos(\omega \tau)}{\omega^2+\varepsilon^2}\,d\omega=\frac1\varepsilon \int_{-\infty}^\infty \frac{\cos(\omega \epsilon \tau)}{\omega^2+1}\,d\omega=\frac\pi{\varepsilon} e^{-|\varepsilon \tau|}$, which can be accomplished by a number of real analytical methodologies.

Thus, for $\tau\ne 0$, we find that

$$\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{1}{\omega+i\varepsilon}e^{-i\omega \tau}\,d\omega=-2\pi i \theta(\tau)$$

as was to be shown.