# Multiple integrals involving product of gamma functions

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:

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

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!

• Some similar integrals ( $\sf\mbox{Watson/Van Peype Triple Integrals}$ ) are in the Najin book. – Felix Marin Dec 21 '14 at 0:49

• Some similar integrals ( $\sf\mbox{Watson/Van Peype Triple Integrals}$ ) are in the Najin book. – Felix Marin Dec 21 '14 at 0:45
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.