Integral Operation: Why Does nηIJ Vanish?

[moriheru]

My question concerns a integral and why it vanishes:
-nη^IJ 1/2π ∫₀^2π dσ e^i(m+n)σ=-nη^IJ delta_m+n=0Just to justify why this should be on the beyond the standard model forum, this is part of a calulation concerning the comutators of the alpha modes.

 

[RUber]

I assume m and n are integers?
If $m+n \neq 0$ then, $m+n = a \in \mathbb{Z}$ and
$\int_0^{2\pi} e^{i (m+n) \sigma} d\sigma = \frac{1}{i(m+n)} e^{i (m+n) \sigma} |_0^{2\pi} = \frac{1}{i(m+n)} - \frac{1}{i(m+n)} = 0.$
If $m+n=0$ then $\int_0^{2\pi} e^{i (m+n) \sigma} d\sigma = \int_0^{2\pi} 1 d\sigma= 2\pi.$

 

[moriheru]

Thankyou.

 

[RUber]

No problem--this idea is fundamental to Fourier transforms and a host of other applications requiring orthogonality of sines and cosines.

 

[moriheru]

Oh... that's an FT I missed that :) would've made that easyer,thanks.