Indefinite Integration of Heaviside function muliplied by a function

[MahaRoho]

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)]

 

[Mark44]

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?

 

[MahaRoho]

Hello,
Thanks! I don't have any answer tho!
Btw, is not integration of H(r-rd)=(r-rd) H(r-rd)?

 

[Mark44]

So your integral looks to me to be $$\int_a^\infty \frac 1 r ~~dr$$
What do you get for that improper integral?

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.

Btw, is not integration of H(r-rd)=(r-rd) H(r-rd)?

Do a web search for "indefinite integral of Heaviside function".