I know with the help of Wolfram Alpha how get the closed-form of these variations of a well-known integral representation of the Riemann zeta function $$\int_0^1\int_0^1\frac{1}{1-x\cos\left(\frac{\pi y}{2}\right)}dxdy,\tag{1}$$ and $$\int_0^1\int_0^1\int_0^1\frac{1}{1-xy\cos\left(\frac{\pi z}{2}\right)}dxdydz.\tag{2}$$

I don't know if these are in the literature, I would like to know what about the integral $$\mathcal{J}=\int_0^1\int_0^1\int_0^1\frac{1}{1-x\cos\left(\frac{\pi y}{2}\right)\cos\left(\frac{\pi z}{2}\right)}dxdydz.$$ If it is in the literature feel free to refer the article or exercise, answering this question as a reference request, and I try to search and read such closed-form from the literature.

Question. From my approach I know that $$\mathcal{J}=\frac{2}{\sqrt{\pi}}\sum_{k=0}^\infty\frac{1}{(k+1)^3}\left(\frac{\Gamma\left(\frac{k+3}{2}\right)}{\Gamma\left(\frac{k}{2}+1\right)}\right)^2,\tag{3}$$ where $\Gamma(s)$ denotes the gamma function. Do you know how to calculate $\mathcal{J}$ using integration or well finishing my approach providing the closed-form of previous series? Many thanks.

  • $\begingroup$ I'm sorry for previous edits, there were typos. $\endgroup$ – user243301 Apr 13 '18 at 21:52

By separating even/odd values of $k$, $$\sum_{k\geq 0}\frac{\Gamma\left(\frac{k+3}{2}\right)^2}{(k+1)^3\,\Gamma\left(\frac{k+2}{2}\right)^2}=G+\frac{1}{2\pi}\,\phantom{}_4 F_3\left(1,1,1,1;\tfrac{3}{2},\tfrac{3}{2},2;1\right)\tag{1} $$ and the hypergeometric term can be computed through the Fourier-Legendre series expansions machinery. In particular $$ \sum_{k\geq 0}\frac{\Gamma\left(\frac{k+3}{2}\right)^2}{(k+1)^3\,\Gamma\left(\frac{k+2}{2}\right)^2}= 2G-\frac{7\,\zeta(3)}{4\pi}\tag{2} $$ and $$\mathcal{J} = \frac{4G}{\sqrt{\pi}}-\frac{7\,\zeta(3)}{4\pi\sqrt{\pi}}.\tag{3} $$

  • $\begingroup$ Many thanks for your answer, it is incredible. $\endgroup$ – user243301 Apr 14 '18 at 8:12
  • 1
    $\begingroup$ @user243301: you're welcome. I am happy that some free investigations about FL expansions turn out to be useful in tackling probabilistic problems and multiple integrals. $\endgroup$ – Jack D'Aurizio Apr 14 '18 at 12:05

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