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$\displaystyle \int_0^{2\pi} K_N^4(s)\ ds = \frac{c_{N-1}}{8 \pi^3 N^4}$ 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 $$ {\frac {45+49 {N}^{2}+ 70 N^4 + 151{N}^{6}}{2520 {N }^{3}{\pi }^{3}}} $$ |
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$\displaystyle \int_0^{2\pi} K_N^4(s)\ ds = \frac{c_{N-1}}{8 \pi^3 N^4}$ 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. |
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