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There is this question that ask me to compute

enter image description here

by performing a change of coordinates to polar coordinates.

My attempts:

I manage to change it to $$I=\int_0^\infty \int_0^{2\pi} re^{-r}δ(r-R)d\theta\,dr$$ but the problem is how do I integrate delta function?

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  • $\begingroup$ Hint: $\int_{-\infty}^{\infty}f(x)\delta(x-a)dx=f(a)$ $\endgroup$
    – aleden
    Commented Nov 10, 2017 at 15:54
  • $\begingroup$ hi @aleden thx for reply so for my integral is from 2pi to 0 for Re^(-R) d(theta) is it correct if I include the part where I use your method to integrate delta function? $\endgroup$
    – kevin
    Commented Nov 10, 2017 at 15:59
  • $\begingroup$ Not quite, notice that the bounds of your integral are $[0,\infty]$ as opposed to those of the dirac functional equation, being $[-\infty,\infty]$. You would have to manipulate the equation in order make the bounds suitable to apply the above formula. $\endgroup$
    – aleden
    Commented Nov 10, 2017 at 16:01
  • $\begingroup$ hi @aleden, So do I multiply 2 to my integral??? but my integrand is still Re^(-R) d(theta)?? $\endgroup$
    – kevin
    Commented Nov 10, 2017 at 16:05
  • $\begingroup$ You cannot just multiply by 2, since the integrand is not even. I would suggest utilizing the Heaviside step function, since it is $0$ for $r\lt 0$. $\endgroup$
    – aleden
    Commented Nov 11, 2017 at 13:19

1 Answer 1

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$$I~=~\int_{[0,2\pi]}\! d\theta \int_{\mathbb{R}_+} \! dr~ re^{-r}~\delta (r-R) ~=~2\pi \int_{\mathbb{R}} \! dr~H(r) re^{-r}~\delta (r-R)~=~2\pi H(R)Re^{-R}, $$ where $H$ denotes the Heaviside step function.

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  • $\begingroup$ Evaluating between $[-\infty,\infty]$ would yield $0$ right? $\endgroup$
    – aleden
    Commented Nov 11, 2017 at 13:20

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