Computing the normalization of Jackson Kernel?

Question

The Jackson Kernel is $$ K_n(x) = C_n\frac{\sin^4(nx/2)}{\sin^4(x/2)} $$

Where $C_n = \frac{3}{2\pi n(2n^2+1)}$ is chosen to make sure that $\int_{-\pi}^\pi K_n(x) dx =1.$

How could I evaluate this integral? I have not found a reference for this.

Answer 1

Following the logic of this answer for the integral $$\int_{-\pi}^{\pi}\frac{\sin^2(nx/2)}{\sin^2(nx)}$$

we can write $$\frac{\sin^4(nx/2)}{\sin^4(nx)}=e^{-2(n-1)}\sum_{j,k,l,m=0}^{n-1}e^{i(j+k+l+m)x}$$

and so the integral is equal to $$2\pi\cdot \#\{j,k,l,m\in[n-1]:j+k+l+m=2(n-1)\}$$

From there it's as simple as counting the number of integer non-negative solutions of the equation $x_1+x_2+x_3+x_4=2n-2$ under the constraints $x_i \leq n-1$. This number can be easily determined by the inclusion-exclusion principle:

• There are $\binom{2n-2+4-1}{4-1}=\binom{2n+1}{3}$ non-negative integer solutions (without any constraints)
• From that, subtract the number of solutions with $x_1 > n-1$. This is equal to the number of non-negative integer solutions of $y_1+x_2+x_3+x_4=2n-2-n=n-2$. There are $\binom{n+1}{3}$ such solutions.
• Do the same for the number of solutions with $x_2>n-1$, $x_3>n-1$, and finally, $x_4>n-1$.

There are no more solutions to take into account (i.e. no solutions such that $x_i > n-1$ and $x_j>n-1$ that we would have subtracted twice) so the total number of such solutions is $$\binom{2n+1}{3}-4\binom{n+1}{3}=\frac{n(2n^2+1)}{3}$$

Putting everything together, the integral is equal to $\frac{2\pi n(2n^2+1)}{3}$ and so $C_n=\frac{3}{2\pi n(2n^2+1)}$, as desired.