Prove $\int_{-\pi}^\pi F_n(y) \, dy=1$

Question

Prove $\int_{-\pi}^\pi F_n(y)\,dy=1$, with $$F_n(y)=\frac{1}{2\pi (n+1)}\frac{\sin^2 \left( \frac{(n+1)y}{2} \right)}{\sin^2(\frac{y}{2})}$$

I tried a similar question but there I had given a serie of the function. This time I don't know if there is a serie that can help me with this. I tried it without a serie:

\begin{align} \int_{-\pi}^{\pi}F_n(y) \, dy &= \int_{-\pi}^\pi \frac{1}{2\pi (n+1)} \frac{\sin^2\left(\frac{(n+1)y}{2}\right)}{\sin^2(\frac{y}{2})} \, dy\\ &=\int_{-\pi}^\pi \frac{1}{2\pi (n+1)}\frac{(e^{\frac{i(n+1)y}{2}}-e^{\frac{-i(n+1)y}{2}})^2}{(e^\frac{iy}{2}-e^\frac{-iy}{2})^2} \, dy\\ &=\int_{-\pi}^\pi \frac{1}{2\pi (n+1)}\frac{(e^{i(n+1)y}+e^{-i(n+1)y}-2)}{(e^{iy}+e^{-iy}-2)} \, dy \end{align}

But now I'm stuck again. I think there needs to be an easier way to prove this. Can someone help me out?

Answer 1

Hint: Let $$D_n(x)= \sum_{k=-n}^n e^{ikx},$$ and let $$F_N(x) = \sum_{n=0}^{N-1} D_n(x).$$ Prove that $$ F_N(x) = \frac{1}{N}\frac{\sin^2 (Nx/2)}{\sin^2 (x/2)}.$$

$D_n$ is known as the Dirichlet kernel and $F_N$ is known as the Fejér Kernel.

Answer 2

You have

$$F_n(x) = \frac{2\pi}{n+1}D_n^2(x)=\frac{1}{2\pi (n+1)}\left(\sum_{k=-n}^n e^{ikx} \right)^2= \frac{1}{2\pi (n+1)}\frac{\sin^2 \left(\frac{(n+1)x}{2}\right)}{\sin^2 \left(\frac{x}{2}\right)}.$$

from which follows by switching $\int$ and $\sum$ the equalities

$$\begin{aligned}\int_{-\pi}^{\pi}F_n(x) \ dx&= \frac{1}{2\pi (n+1)}\int_{-\pi}^{\pi}\left(\sum_{k=-n}^n e^{ikx}\right)^2 \ dx\\ &= \frac{1}{2\pi (n+1)}\int_{-\pi}^{\pi}\sum_{k=-n}^n \sum_{l=-n}^ne^{i(k+l)x}\ dx\\ &= \frac{1}{2\pi (n+1)}\sum_{k=-n}^n \sum_{l=-n}^n \int_{-\pi}^{\pi}e^{i(k+l)x}\ dx\\ &=1 \end{aligned}$$

as in the double sum, the only non vanishing terms are for $k=-l$ and there are $n+1$ such terms.

Answer 3

Here is to integrate without resorting to series

\begin{align} \int_{-\pi}^{\pi}F_{n}(y)dy & = \frac1{\pi(n+1)}\int_{0}^{\pi} \frac{\sin^{2}\frac{(n+1)y}{2}}{\sin^{2}\frac{y}{2}}dy\\ & = \frac1{\pi(n+1)}\int_{0}^{\pi} \frac{ \cos(n+1)y-1}{\cos y-1} dy \\ & = \frac1{\pi(n+1)}\cdot \lim_{a\to 0}\int_{0}^{\pi} \frac{ \cos(n+1)y-\cos(n+1)a}{\cos y-\cos a} dy\\ & = \frac1{\pi(n+1)}\cdot \lim_{a\to 0}\frac{\pi\sin(n+1)a}{\sin a}\\ &=1 \end{align} where the result $$\int_{0}^{\pi}{\frac{\cos(nx)-\cos(na)}{\cos x-\cos a}}dx = \frac{\pi \sin(na )}{\sin a}$$ derived in Parametric trigonometric integral $\int_{0}^{\pi}{\frac{\cos(nx)-\cos(na)}{\cos x-\cos a}}dx$ is used.