MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

2 added 104 characters in body

$\displaystyle \int_0^{2\pi} K_N^4(s)\ ds = \frac{c_{N-1}}{8 \pi^3 N^4}$ 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

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

1

$\displaystyle \int_0^{2\pi} K_N^4(s)\ ds = \frac{c_{N-1}}{8 \pi^3 N^4}$ 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.