Hi, I need an estimation or an exact closed form expression for the following integral 


$\int\_{0}^{2\pi} K\_N^4(s) ds $


where $K\_N(s)= \frac{1}{N2\pi} (\frac{sin(Ns/2)}{sin(s/2)})^2$, the Fejer kernel.


I don't know how to obtain an estimation better than


$\int\_{0}^{2\pi} K\_N^4(s) ds < N^4$


Does anyone know a better estimation or some trigonometric tricks that can help me to improve my estimation?


Thanks in advance 


Imma



Does anybody know an estimation of L4 norm of fejer kernel ?
$\displaystyle \int\_0^{2\pi} K\_N^4(s)\ ds = \frac{c\_{N-1}}{8 \pi^3 N^4}$ where $c\_n$ is the coefficient of $z^{4n}$ in $(1 + z + \ldots + z^n)^8$. 


$c\_n$ appears to have the closed form
$$ c\_n = \frac{\left( 315+1284 n + 2734 n^2+3300{n}^{3}+2335{n}^{4}+906{n}^{5}
+151{n}^{6} \right) \left( n+1 \right)}{315} 
$$


It doesn't appear to be in the OEIS yet. Thus your integral is


$$ {\frac {45+49 {N}^{2}+ 70 N^4 + 151{N}^{6}}{2520 {N
}^{3}{\pi }^{3}}}
$$


