Integral of complex periodic signal Integrating   $e^{j(k-n)\omega t} \, dt$ over the interval $0$ to $T$ where $T$ is the fundamental time period of the sinusoids yields zero when $k$ is not equal to $n$.... ? how?
assume ω is the fundamental frequency equal to $2\pi/T$.

We have
$$
\int_0^{2\pi} e^{j\ell\omega t} \,dt
$$
where $\ell$ is a nonzero integer.  If we let $s=t+c$, where $c$ is some real constant, then we have $ds=dt$, and as $t$ goes from $0$ to $2\pi$, then $s$ goes from $c$ to $2\pi+c$.  But since the function is periodic, we have
$$
\int_c^{2\pi+c} \cdots\cdots\,ds = \int_0^{2\pi} \cdots\cdots\,ds.
$$
So
$$
\int_0^{2\pi} e^{j\ell\omega t} \,dt = \int_c^{2\pi+c} e^{j\ell\omega(s-c)} \,ds = \int_0^{2\pi} e^{j\ell\omega(s-c)} \,ds = e^{-j\ell\omega c} \int_0^{2\pi} e^{j\ell\omega s} \,ds = e^{-j\ell\omega c} \int_0^{2\pi} e^{j\ell\omega t} \,dt.
$$
The last equality holds because we can rename the bound variable.
Now we have
$$
\text{integral} = (e^{-j\ell\omega c}\cdot\text{same integral}).
$$
If the integral is not $0$, then we can divide both sides by it and get
$$
0 = e^{-j\ell\omega c}.
$$
But this is clearly false, so the integral must be $0$.
