Integrals of the form $\int \log(2+2\cos x)^ndx$

Question

$\log$ will be the natural logarithm and $\zeta$ the Riemann zeta function. I'm interested in the following family of integrals: $$ I_n = \int_0^\pi(\log(2+2\cos x))^n\mathrm{d}x $$ Some of the values are:
$$\begin{array}{c|c|c|} n& \text{$I_n$}\\ \hline 1 & 0 \\ \hline 2 & 2\pi\zeta(2) \\ \hline 3 & -12\pi\zeta(3) \\ \hline 4 & 114\pi\zeta(4) \\ \hline 5 & -1177.6529...\cdot\pi\zeta(5) \\ \hline 6 & 14420.2439...\cdot\pi\zeta(6) \\ \hline 7 & -203649.3734...\cdot\pi\zeta(7) \\ \hline \end{array}$$ Now the values for $n=2,3,4$ are particularly nice when written in this form, and I managed to calculate those three integrals using dubiously rigorous contour integral methods which are far too unwieldy for anything more. The half-angle substitution leads to various ugly nested sums which I don't know how to evaluate. However (according to WolframAlpha) the patter apparently breaks at $n=5$, even though it was going so well! Does anyone know if/how it is possible to obtain a closed form solution?

Answer 1

One way to go about this would be by defining $$E(s)=\int_0^\pi(2+2\cos x)^sdx$$ Then your integral is given by, from the Leibniz rule, $$I_n=E^{(n)}(0)$$ Anyway, we have $$E(s)=2^s\int_0^\pi(1+\cos x)^s dx$$ We have the identity $1+\cos x=2\cos^2(x/2)$ so $$E(s)=4^s\int_0^\pi \cos^{2s}(x/2)dx$$ $t=\frac{x}2$: $$E(s)=2^{2s-1}\int_0^{\pi/2}\cos^{2s}(x)dx$$ We have that $$\int_0^{\pi/2}\sin^a(x)\cos^b(x)dx=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}{2}+1)}\qquad \text{Re }a,\text{Re }b>-1$$ Hence $$E(s)=2^{2s-2}\sqrt{\pi}\frac{\Gamma(s+1/2)}{\Gamma(s+1)}$$ And $$I_n=\frac{\sqrt\pi}{4}\left(\frac{\partial}{\partial s}\right)^n2^{2s}\frac{\Gamma(s+1/2)}{\Gamma(s+1)}\Bigg|_{s=0}$$