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I am unsure how to solve an integral equation. As you know the orthogonality relation for cylindrical harmonics is: $$ \int_0^{2\pi}e^{in\phi'}e^{-im\phi'}d\phi'=2\pi\delta_{m,n}\ $$

The problem I have comes when you add a unit vector inside the integral:

$$ \int_0^{2\pi}\hat{\phi'}e^{in\phi'}e^{-im\phi'}d\phi' $$

How can this integral be solved?

Tommy
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  • Of the top of my head I'd say residues, altho Mathematica gives $\frac{-1+e^{-2 i \pi (m-n)} (1+2 i \pi (m-n))}{(m-n)^2}$ – dingo_d Feb 21 '14 at 15:28
  • If you decompose the vector into its Cartesian coordinates, Cartesian unit vectors are constants and can be pulled outside the integral. – David H Feb 21 '14 at 15:36
  • This is the solution if you don't consider $\hat{\phi'}$ as a unitary vector, else a variable $\phi'$. –  Feb 21 '14 at 15:41
  • David H, Is the only way to solve that? I did it and for cylindrical coordinates is not so easy. The thing is that I have to solve this: $$ \int_0^{2\pi}e^{in\phi'}e^{-im\phi'}d\phi'=2\pi\delta_{m,n}
    $$     –  Feb 21 '14 at 15:49
  • Sorry the time pass to edit the comments: I would like to say this: David H, Is the only way to solve that? I did it and for cylindrical coordinates is not so easy. The thing is that I have to solve this: $$ \int \hat{\phi}Y^{m*}nY^m_nd\Sigma=\delta{m,n}
    $$ And this is more difficult to solve     –  Feb 21 '14 at 15:57
  • @Tommy I don't know another way to do it, but I'd be willing to bet one exists. I have a hunch that maybe you could solve it by rewriting the integral as a line-ntegral and applying Stoke's Theorem, but I'd have to check and see if that pans out. – David H Feb 21 '14 at 15:57
  • Ok, thank you, @David H –  Feb 21 '14 at 16:00
  • @Tommy Alright, I did a little bit of work and it looks like my hunch was right. If you are still interested, I'll write up a solution. – David H Feb 21 '14 at 16:59
  • We are interested David. – Physics_maths Feb 22 '14 at 10:30
  • Sure @David H, I would appeciate it. I am eager to know it ;) – Tommy Feb 23 '14 at 14:49

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