# Simple Fourier coefficient exponential

1. Dec 24, 2016

### binbagsss

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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)

2. Dec 24, 2016

### Ssnow

mmmhh, I think $g=\frac{2\pi}{T}$, using the orthogonal relations is good, in fact another way can be calculating directing the integral treating $i$ as a constant, in this way you can see the rersult that is still immediate ...

Ssnow

3. Dec 24, 2016

### Ray Vickson

If $a \neq 0$ we have
$$\int_0^T e^{i a t/T} \, dt = \frac{1}{a} iT \left(1-e^{ia}\right)$$.
Put $a = 2 \pi (n-m)$ and simplify, assuming $n,m$ are unequal integers.

Of course, when $a=0$ the integral is just $\int_0^T 1 \, dt.$.

4. Dec 24, 2016

### binbagsss

so where have I gone wrong?

5. Dec 24, 2016

### binbagsss

but what is the error in my working anybody please?

6. Dec 24, 2016

### Ray Vickson

Your equations (1), (2), (3) are incorrect. $\int_0^{2\pi/g} \cos(n g t) \cos(m g t) \, dt = c \delta_{m,n} ,$ but $c \neq 1$.