Integrate[e^(ix(m+n)),{x,0,2pi}] = 2pi*delta(m+n)
[tex]\int_0^{2\pi} e^{ix(m+n)} dx= 2\pi \delta_{m+n}[/tex]
(click on the equation to see the code)
If m+n is not 0, then the integral is
[tex]-\frac{i}{m+n}e^{ix(m+n)}[/tex]
evaluated from 0 to [itex]2\pi[/itex]. But [itex]e^{ix(m+n)}[/itex] is 0 at both 0 and [itex]2\pi[/itex] so the integral is 0.
If m+n= 0 then the integral is
[tex]\int_0^{2\pi}dx= 2\pi[tex]
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Yep, it looks like that "delta" should be "1 if m+n= 0, 0 otherwise".[/tex][/tex]