Multiple integrals involving product of gamma functions

Question

The following integral was posted a few days back on Integrals and Series forum: $$\int_0^{2\pi} \int_0^{2\pi} \int_0^{2\pi} \frac{dk_1\,dk_2\,dk_3}{1-\frac{1}{3}\left(\cos k_1+\cos k_2+ \cos k_3\right)}=\frac{\sqrt{6}}{4}\Gamma\left(\frac{1}{24}\right)\Gamma\left(\frac{5}{24}\right)\Gamma\left(\frac{7}{24}\right)\Gamma\left(\frac{11}{24}\right)$$

I am curious if there is a closed form solution for:

$$\int_{\large[0,2\pi]^n} \frac{dk_1\,dk_2\,dk_3\,\cdots \,dk_n}{1-\frac{1}{n}\left(\cos k_1+\cos k_2+\cos k_3+\cdots +\cos k_n\right)}$$

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Since $\left|\dfrac{\cos k_1 + \cos k_2 + \cos k_3}{3}\right|<1$,

$$\int_0^{2\pi} \int_0^{2\pi} \int_0^{2\pi} \frac{dk_1\,dk_2\,dk_3}{1 - \frac 1 3 \left( \cos k_1 + \cos k_2 + \cos k_3 \right)}$$ $$=8\int_0^{\pi} \int_0^{\pi} \int_0^{\pi} \frac{dk_1\,dk_2\,dk_3}{1 - \frac 1 3 \left( \cos k_1 + \cos k_2 + \cos k_3 \right)}$$ $$=8\sum_{n=0}^{\infty} \frac{1}{3^n} \int_0^{\pi} \int_0^{\pi} \int_0^{\pi} \left( \cos k_1 + \cos k_2 + \cos k_3 \right)^n\,dk_1\,dk_2\,dk_3 $$

We can ignore the odd values of $n$ as the integral is zero for them. Also, for even values of $n$, the exponents of cosines in the expansion of $\left( \cos k_1 + \cos k_2 + \cos k_3 \right)^{2n}$ must be even. Hence, from multinomial therem, we can write:

$$8\sum_{n=0}^{\infty}\,\,\sum_{m_1+m_2+m_3=n} \frac{1}{3^{2n}}\frac{(2n)!}{(2m_1)! (2m_2)! (2m_3)!} \int_0^{\pi} \int_0^{\pi} \int_0^{\pi} \cos^{2m_1}k_1\cos^{2m_2}k_2 \cos^{2m_3}k_3\,dk_1\,dk_2\,dk_3$$

$$ = 16\sum_{n=0}^{\infty}\,\,\sum_{m_1+m_2+m_3=n} \frac{1}{3^{2n}}\frac{(2n)!}{(2m_1)! (2m_2)! (2m_3)!} \int_0^{\pi/2} \int_0^{\pi/2} \int_0^{\pi/2} \cos^{2m_1}k_1\cos^{2m_2}k_2 \cos^{2m_3}k_3\,dk_1\,dk_2\,dk_3$$

Using the result: $\int_0^{\pi/2} \cos^{2k}x\,dx=\frac{(2k)!}{4^k (k!)^2}\frac{\pi}{2}$, the integral is,

$$2\pi^3 \sum_{n=0}^{\infty}\,\,\sum_{m_1+m_2+m_3=n} \frac{1}{36^n}\frac{(2n)!}{(m_1!)^2 (m_2!)^2 (m_3!)^2}$$

I am stuck here.

Any help is appreciated. Thanks!

Answer 1

This is not an answer to the question, but it is too long for a comment. Here is a piece of information that I know about the integral that might help you to evaluate it. The integral above is known as one of the three of the Watson/van Peype Triple Integrals: \begin{align} I_1&=\frac{1}{\pi^3}\int_0^\pi\int_0^\pi\int_0^\pi\frac{\mathrm dx\;\mathrm dy\;\mathrm dz\;}{1-\cos x \cos y \cos z}\\[7pt] I_2&=\frac{1}{\pi^3}\int_0^\pi\int_0^\pi\int_0^\pi\frac{\mathrm dx\;\mathrm dy\;\mathrm dz\;}{3-\cos x \cos y -\cos x \cos z -\cos y \cos z }\\[7pt] I_3&=\frac{1}{\pi^3}\int_0^\pi\int_0^\pi\int_0^\pi\frac{\mathrm dx\;\mathrm dy\;\mathrm dz\;}{3-\cos x -\cos y -\cos z } \end{align} These three integrals first appeared in a paper published by W. F. van Peype, “Zur Theorie der Magnetischen Anisotropic Kubischer Kristalle Beim Absoluten Nullpunkt,” Physica, June 1938, pp. 465–482. That is, van Peype was studying magnetic behavior in certain cubic crystalline lattice structures at very low temperatures (low means near absolute zero). The Watson/van Peype integrals turn-up not only in the physics of frozen magnetic crystals, but also in the pure mathematics of random walks. You can find a complete discussion of both the history and the mathematics of the integrals in I. J. Zucker, “70+ Years of the Watson Integrals,” Journal of Statistical Physics, November 2011, pp. 591–612.

The evaluation of these three integrals can be found in a paper published by G. N. Watson, “Three Triple Integrals,” Quarterly Journal of Mathematics, 1939, pp. 266–276. Here are the closed forms of the three integrals: \begin{align} I_1&=\frac{\Gamma^4\!\left(\frac{1}{4}\right)}{4\pi^3}\\[7pt] I_2&=\frac{3\Gamma^6\!\left(\frac{1}{3}\right)}{2^{14/3}\pi^4}\\[7pt] I_3&=\frac{\Gamma\!\left(\frac{1}{24}\right)\Gamma\!\left(\frac{5}{24}\right)\Gamma\!\left(\frac{7}{24}\right)\Gamma\!\left(\frac{11}{24}\right)}{16\sqrt{6}\,\pi^3} \end{align} The evaluation of these three integrals can also be found in Watson's Triple Integrals - Wolfram MathWorld. The evaluation of general form of Watson's integrals \begin{align} W_1(n_1)&=\frac{1}{\pi^3}\int_0^\pi\int_0^\pi\int_0^\pi\frac{\mathrm dx\;\mathrm dy\;\mathrm dz\;}{1-\frac{n_1}{3}(\cos x \cos y +\cos x \cos z +\cos y \cos z )}\\[7pt] W_2(n_2)&=\frac{1}{\pi^3}\int_0^\pi\int_0^\pi\int_0^\pi\frac{\mathrm dx\;\mathrm dy\;\mathrm dz\;}{1-\frac{n_2}{3}(\cos x +\cos y +\cos z )} \end{align} can be found in a paper published by G. S. Joyce and I. J. Zucker, “On The Evaluation of Generalized Watson Integrals”.

Answer 2

I have found the following result: Take the inverse Laplace transform of $$ I_n(a)=\int_{\large[0,2\pi]^n} \frac{dk_1\,dk_2\,dk_3\,\cdots \,dk_n}{a-\frac{1}{n}\left(\cos k_1+\cos k_2+\cos k_3+\cdots +\cos k_n\right)} $$ with respect to $a$. This gives $$ \mathcal{L}^{-1}_{a \to s}[I_n(a)] = \int_{\large[0,2\pi]^n} \exp\left(\frac{s}{n}\left(\sum_{l=1}^n \cos(k_l)\right)\right) \;dk_1 \cdots d k_n$$ which is also $$ \mathcal{L}^{-1}_{a \to s}[I_n(a)] = \prod_{l=1}^n \int_0^{2\pi}\exp\left(\frac{s \cos(k_l)}{n}\right) \; dk_l $$ So the integrals are seperable $$ \mathcal{L}^{-1}_{a \to s}[I_n(a)]= \left(\int_0^{2\pi} e^{s\cos(x)/n}\; dx\right)^n $$ $$ \mathcal{L}^{-1}_{a \to s}[I_n(a)]= \left(2\pi I_0\left(\frac{s}{n}\right)\right)^n $$ and taking the Laplace transform of both sides gives $$ I_n(a) = \int_0^\infty \left(2\pi I_0\left(\frac{s}{n}\right)\right)^n e^{-s a} \; ds $$ where $I_0(x)$ is a Bessel function, so $$ I_n = \int_0^\infty \left(2\pi I_0\left(\frac{s}{n}\right)\right)^n e^{-s} \; ds $$

it indeed seems that $$ \int_0^\infty \left(2\pi I_0\left(\frac{s}{3}\right)\right)^3 e^{-s} \; ds = \frac{\sqrt{6}}{4}\Gamma\left(\frac{1}{24}\right)\Gamma\left(\frac{5}{24}\right)\Gamma\left(\frac{7}{24}\right)\Gamma\left(\frac{11}{24}\right) $$

this solution seems to relate to the elliptic integral singular value $K(k_6)$ in the link. It seems that $I_1$ and $I_2$ don't converge, but $I_4$ and $I_5$ and higher seem to numerically.