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I found this text online:

"In general, integrating the $\delta$ function or one of its integrals makes it smoother. Differentiating it increases the discontinuities. For example $\int\delta $ is discontinuous itself. $\int \int \delta $ is continuous but with a discontinuous first derivative. $\int \int \int \delta$ is continuous, but with a discontinuous second derivative, etc..."

I agree that $\int\delta $ is certainly discontinuous since it equates to the Heaviside function. However I do not agree that $\int \int \delta $ is continuous since wouldn't this simply be the Ramp function ($\int H(x)$) And isn't the Ramp function also discontinuous at $x = 0$?

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  • $\begingroup$ What is according to you the limit of the ramp function approaching it from negative and positive zero? $\endgroup$ Commented Oct 24, 2015 at 22:45
  • $\begingroup$ Zero, I would say $\endgroup$
    – user32882
    Commented Oct 24, 2015 at 22:47
  • $\begingroup$ So what makes you say that it is discontinues? $\endgroup$ Commented Oct 24, 2015 at 22:48
  • $\begingroup$ Because there is a kink there... $\endgroup$
    – user32882
    Commented Oct 24, 2015 at 22:48
  • $\begingroup$ I guess discontinuous necessarily means there is a jump with infinite slope? $\endgroup$
    – user32882
    Commented Oct 24, 2015 at 22:49

1 Answer 1

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The first integral is the Heaviside function. It's discontinuous at $0$ because the limits from the left and from the right are different ($0$ and $1$). One might explain the discontinuity by saying that one can't trace the graph without lifting the pencil.

The second integral is continuous: you can trace this graph without lifting your pencil. The limit as $x\to 0$ is $0$, and it agrees with the value of the function there.

ramp

The fact that there is a corner at $0$ means something different: this function is not differentiable at $0$. (The slopes from the left and from the right don't match.)

And the third integral is differentiable:

parabola

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