Indefinite Integration of Heaviside function muliplied by a 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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