- #1

binbagsss

- 1,278

- 11

## Homework Statement

I think I am being stupid, I am trying to show that

## \int^{T}_{0} e^i\frac{2\pi(n-m)t}{T} dt = 0 ## [1] if ## n \neq m##

## = T ## if ##n=m##, ##T## the period.

## Homework Equations

[/B]

I am using the following ##cos## and ##sin## orthogonal identities:

1) ##\int^{2 \pi / g}_0 sin ( n g x) sin ( m g x ) dx = \delta_{nm} ##

2) ##\int^{2 \pi / g}_0 cos ( n g x) cos ( m g x ) dx = \delta_{nm} ##

3) ##\int^{2 \pi / g}_0 sin ( n g x) cos ( m g x ) dx = \delta_{nm} ##

## The Attempt at a Solution

## e^i\frac{2\pi(n-m)t}{T} dt = e^i\frac{2\pi(n)t}{T} . e^i\frac{2\pi(n-m)t}{T} ##

(Let ##g=\frac { 2 \pi i}{T} ##)

##= (cos ( g n t) + i sin (gnt) ).(cos ( g m t) - i sin ( g m t))##

So using the orthogonal identities [1] the contribution to [1] from the ##coscos## and ##sinsin## term gives ##2 \delta _{nm} ##, and then from the cross ##sin## and ##cos## terms I am getting

## i \delta_{nm} - i \delta_{nm} = 0 ##.

So I am getting ## 2 \delta _{nm} ##

Which I can see straight off is wrong since I am not dividing by the period to get the coefficient. (independent of it)

Many thanks in advance.