Evaluating $\pi^3+8\pi^3\sum_{k\geq1}\frac{\Gamma(2k)^3}{k^32^{6k}\Gamma(k)^6}$, the result of $\iiint_{(0,\pi)^3}\frac{dxdydz}{1-\cos x\cos y\cos z}$

Question

So I was attempting to solve an integral given to me by a friend: $$I = \iiint_{(0, \pi)^3} \frac{dxdydz}{1 - \cos (x)\cos (y)\cos (z)}$$ I started out with a simple series expansion: $$\sum_{k \geq 0} \iiint_{(0, \pi)^3} \cos^k(x)\cos^k(y)\cos^k(z)\, dxdydz$$ However since all variables are seperable, the integral can now be written as: $$\sum_{k \geq 0} \left ( \int_0^{\pi} \cos^k(x) \,dx \right )^3$$ Notice that for odd values of $k$, the integrand is the negative of itself, meaning that for odd $k$, the integral must be $0$. Calculating the sum for $k = 0$ (this will be important later), and reindexing the sum we have: $$\pi^3 + \sum_{k \geq 1} \left (\int_0^{\pi} (\cos x)^{2k}\, dx \right )^3$$ The integrand is also symmetric about $\pi/2$, which yields: $$\pi^3 + 8\sum_{k \geq 1}\left ( \int_0^{\frac{\pi}{2}} (\cos x)^{2k} \, dx \right )^3$$ Using the Beta function and rewriting it in terms of the composite Gamma functions we have: $$\pi^3 + \sum_{k \geq 1} \left ( \frac{\Gamma (1/2)\Gamma(k + 1/2)}{\Gamma (k + 1)} \right )^3$$ After some rewriting and using Legendre's duplication formula we have the monstrous: $$\pi^3 + 8\pi^3\sum_{k \geq 1} \frac{\Gamma (2k)^3}{k^32^{6k}\Gamma (k)^6}$$ The evaluation of this sum is what I'm stuck with and Wolfram Alpha recovers a beautiful result, how would I proceed? Thank you in advance.

Answer 1

There is a way to find this sum using some properties of generalised hypergeometric functions -- definitely not an obvious one, but it does the job.

$$ \sum_{k = 0}^{\infty} \left(\frac{\Gamma (1/2)\Gamma(k + 1/2)}{\Gamma (k + 1)} \right)^3 = \pi^3 \sum_{k = 0}^{\infty} \left(\frac{(2k-1)!!}{2^{k}k!}\right)^3 = \\ = \pi^3 \sum_{k = 0}^{\infty} \frac{ \left(\frac{1}{2}\frac{3}{2}\cdots\frac{2k-1}{2}\right)^3 }{(k!)^2}\frac{1}{k!} = \pi^3 \sum_{k = 0}^{\infty} \frac{ \left(\frac{1}{2}\left(1+\frac{1}{2}\right)\cdots\left(\frac{1}{2}+k-1\right)\right)^3 }{(k!)^2}\frac{1}{k!} = \\ = \pi^3 \cdot {}_3F_2\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}; 1, 1; 1\right). $$ Surprisingly, this value can be expressed in terms of the gamma function by using the Dixon's identity for $a = b = c = \frac{1}{2}$: $$ {}_3F_2\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}; 1, 1; 1\right) = \frac{\Gamma(1+1/4)\Gamma(1/4)}{\Gamma(1+1/2)\Gamma(1/2) (\Gamma(3/4))^2}. $$ From the properties of the gamma function we have $$ \Gamma(1+1/4)\Gamma(1/4) = \frac{1}{4}(\Gamma(1/4))^2, \\ \Gamma(1+1/2)\Gamma(1/2) = \frac{1}{2}(\Gamma(1/2))^2 = \frac{1}{2}\pi, \\ \Gamma(1/4)\Gamma(3/4) = \frac{\pi}{\sin(\pi/4)} = \sqrt{2}\pi, $$ which miraculously gives us $$ {}_3F_2\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}; 1, 1; 1\right) = \frac{1}{2\pi} \left(\frac{\Gamma(1/4)}{\Gamma(3/4)}\right)^2 = \frac{1}{2\pi} \left(\frac{(\Gamma(1/4))^2}{\sqrt{2}\pi}\right)^2 = \frac{(\Gamma(1/4))^4}{4\pi^3}. $$ So, finally, $I = \frac{1}{4} (\Gamma(1/4))^4$.

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P.S.: just an interesting fact, $1 - \pi^3 I^{-1} \approx 0.28223$ is the probability of returning to the origin for a random walk on $\mathbb{Z}^3$ with transition probabilities in all eight directions, namely $(i, j, k) \to (i\pm 1, j\pm 1, k \pm 1)$, equal to $1/8$.