let θ(x-x') be the function such that θ = 1 when x-x' > 0 and θ = 0 when x-x' < 0. Show that d/dx θ(x-x') = δ(x - x').(adsbygoogle = window.adsbygoogle || []).push({});

it is easy to show that d/dx θ(x-x') is 0 everywhere except at x = x'. To show that d/dx θ(x-x') is the dirac delta function i also need to show that the integral over the real line of d/dx θ(x-x') = 1.

this is what i tried: ∫ d/dx θ(x-x') dx' = d/dx ∫ θ(x-x') dx' = d/dx ∫ 1 dx' but now i am a little stuck and not sure if this is the way to go or not. my plan was to show that the integral equals x so that d/dx (x) = 1 like we wanted. but it seems like the bottom limit of integration will be x and i'll end up getting -1 instead of 1. can anyone give me some advice on this problem? thanks.

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# Homework Help: Prove that derivative of the theta function is the dirac delta function

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