Fourier Question: Why is $\int^\pi_{-\pi} e^{i(n+m)x }dx = 0$ when $n \neq m$

[Mechdude]

Homework Statement

i was wondering why this is equal to zero : \int^{\pi}_{-\pi} e^{i(n+m)x }dx= 0 when n\neq m

Homework Equations

e^{inx} = cos(nx) + isin(nx)

 

[Dick]

m and n are integers, right? And I think you want (n-m) in the integral if your condition is n not equal to m. Just work out the integral to see why it's zero.

 

[Mechdude]

Yes they are integers, and thanks for the correction it should be n-m , thanks, now anyone with links to a tutorial on why, \left[ \frac {e^{2nix} } {2ni} \right]^{\pi}_{-\pi} = \pi

 

[Dick]

e^(i*n*pi) (where n is an integer) is +1 if n is even and -1 if n is odd. e^(-i*n*pi) is that same value. I don't think you need a tutorial to prove the difference is 0.

 

[Mechdude]

Ok, very nice thanks