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I have added to step function statement and proof of its representation as
$\begin{aligned} \Theta(x) & = \frac{1}{2\pi i} \int_{-\infty}^\infty \frac{e^{i \omega x}}{\omega - i 0^+} \\ & \coloneqq \underset{ \epsilon \to 0^+}{\lim} \frac{1}{2 \pi i} \int_{-\infty}^\infty \frac{e^{i \omega x}}{\omega - i \epsilon} d\omega \,, \end{aligned}$(here).
While I was it, I also added statement and proof that $\partial \Theta = \delta$ (copied that also over to the Properties-section of delta distribution and to the Examples-section at derivative of distributions).
step function (typo in your link)
Thanks, fixed now.
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