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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

    Θ(x) =12πi e iωxωi0 + limε0 +12πi e iωxωiεdω, \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 Θ=δ\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

    step function (typo in your link)

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeSep 6th 2017

    Thanks, fixed now.

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