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