How does $$\frac{1}{\tau}\int_0^\tau \mathrm{e}^{\mathrm{i} (n - m) t} \mathrm{d}t = \delta_{nm}$$ where $$ \delta_{nm} = \begin{cases} 1 & \text{if $n = m$}\\ 0 & \text{if $n \neq m$} \end{cases} $$
$t, \tau \in \mathbb{R}$.
When $n \neq m$, \begin{align} \frac{1}{\tau}\int_0^\tau \mathrm{e}^{\mathrm{i} (n - m) t} \mathrm{d}t & = \frac{1}{\tau} \int_0^\tau \cos{((n - m)t)}\mathrm{d}t + \frac{\mathrm{i}}{\tau} \int_0^\tau \sin{((n - m)t)}\mathrm{d}t\\ & = \frac{1}{\tau (n - m)}(\sin{((n - m)\tau) - \sin(0))} - \frac{\mathrm{i}}{\tau (n - m)}(\cos{((n - m)\tau)} -\cos(0))\\ & = \frac{\sin{(\tau(n - m)})}{\tau (n - m)} - \frac{\mathrm{i} \cos{(\tau(n - m))}}{\tau (n - m)} - 1 \end{align}
How does the above expression evaluate to $0$ when $n \neq m$? I know this is very trivial but I'm unable to prove it.
Edit:
I guess this is a simple typo. $\tau$ is equal to $2\pi / \omega$ where $\omega$ is the fundamental angular frequency. There should be an $\omega$ in the exponential term. The integral should be $$\frac{1}{\tau}\int_0^\tau \mathrm{e}^{\mathrm{i} \boldsymbol{\omega} (n - m) t} \mathrm{d}t.$$ Then everything is alright.
It doesn't work, but if you were integrating over the interval $(-\pi,\pi)$, then it would, due to the periodicity of the trig functions. I think this must be what is meant.
Greg