I would like to show that
$$\frac{1}{n\pi}\int_{-\pi}^\pi \frac{\sin^2(nx/2)}{2\sin^2(x/2)} dx = 1$$
My attempt is very similar to the accepted answer to this question.
$$\int_{-\pi}^\pi \frac{\sin^2(nx/2)}{2\sin^2(x/2)} dx = \frac{1}{2} \int_{-\pi}^\pi {1-\cos(n x)\over 1-\cos(x)}dx$$
I have $\int_{-\pi}^{\pi}{1-\cos(n x)\over 1-\cos(x)}dx$, then
$$\cos\bigl((n+1)x\bigr)+\cos\bigl((n-1)x\bigr)=2\cos x\, \cos(nx)\ ,$$
that gives
$$1-\cos\bigl((n+1)x\bigr)=2(1-\cos x)\cos(nx)+ 2\bigl(1-\cos(nx)\bigr)-\bigl(1-\cos((n-1)x\bigr)\ .$$
Except $$\int_{-\pi}^{\pi}\cos(nx)\ dx= \frac{1}{n} \sin(n\pi) \ne 0$$ does not go away.
So I do not get the right conclusion in the recursion at the end. Where is my mistake?
Best Answer
We have: $$ \frac{\sin(nz)}{\sin z} = \frac{e^{inz}-e^{-inz}}{e^{iz}-e^{-iz}}=\frac{e^{iz}}{e^{inz}}\cdot\frac{e^{2niz}-1}{e^{2iz}-1}=e^{-(n-1)iz}\sum_{k=0}^{n-1}e^{2kiz}$$ so: $$\frac{\sin(nx/2)}{\sin(x/2)}=e^{-\frac{n-1}{2}iz}\sum_{k=0}^{n-1}e^{kix}$$ and: $$\left(\frac{\sin(nx/2)}{\sin(x/2)}\right)^2 = e^{-(n-1)iz}\sum_{j,k=0}^{n-1}e^{(k+j)ix}$$ so: $$\int_{-\pi}^{\pi}\left(\frac{\sin(nx/2)}{\sin(x/2)}\right)^2\,dx = 2\pi\cdot\#\{(j,k)\in[0,n-1]^2: j+k=n-1\}$$ and the claim:
follows.