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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeSep 6th 2017
• (edited Sep 6th 2017)

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).

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeSep 6th 2017

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).

• CommentRowNumber3.
• CommentAuthorDavidRoberts
• CommentTimeSep 6th 2017

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeSep 6th 2017

Thanks, fixed now.

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