Indefinite Integration of Heaviside function muliplied by a function

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The discussion centers on the indefinite integration of the Heaviside function multiplied by another function. The integral in question is expressed as integral [H(r-a)/r] with respect to r. It is clarified that the integral simplifies to $$\int_a^\infty \frac{1}{r} ~~dr$$, which is an improper integral. Participants express confusion about the integration process and the relevance of the Heaviside function in this context. The conversation emphasizes the need for understanding improper integrals when dealing with the Heaviside function.
MahaRoho
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Homework Statement
integral [H(r-a)/r] with respect to r
Relevant Equations
integral [H(r-a)/r] with respect to r
Will it be [{(r-a)/r}*H(r-a)]
 
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MahaRoho said:
Homework Statement:: integral [H(r-a)/r] with respect to r
Relevant Equations:: integral [H(r-a)/r] with respect to r

Will it be [{(r-a)/r}*H(r-a)]
No.
##H(r - a) = \begin{cases} 0 & r < a\\ 1/2 & r = a ;\\ 1 & r > a \end{cases}##

So your integral looks to me to be $$\int_a^\infty \frac 1 r ~~dr$$
What do you get for that improper integral?
 
Last edited:
Hello,
Thanks! I don't have any answer tho!
Btw, is not integration of H(r-rd)=(r-rd) H(r-rd)?
 
Mark44 said:
So your integral looks to me to be $$\int_a^\infty \frac 1 r ~~dr$$
What do you get for that improper integral?

MahaRoho said:
Thanks! I don't have any answer tho!
If you aren't able to do that integration, I don't understand why you're trying to integrate the Heaviside function.

MahaRoho said:
Btw, is not integration of H(r-rd)=(r-rd) H(r-rd)?
Do a web search for "indefinite integral of Heaviside function".
 

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