A curious infinite product of factorials

Question

I found the following infinite product of factorials without proof: $$\small\prod_{n=1}^\infty\frac{{(2 n)!}^{20}\,{(8 n)!}^{32}\,{(32 n)!}^2}{{n!}^4\,{(4 n)!}^{37}\,{(16 n)!}^{13}}=\\\frac{\sin ^{14}\!\frac{\pi }{8}\cdot\sin \frac{\pi }{32} \cdot\sin \frac{3 \pi }{32} \cdot\sin \frac{5 \pi }{32} \cdot\sin \frac{7 \pi }{32}}{\sin ^6\!\frac{\pi }{16} \cdot\sin ^6\!\frac{3\pi }{16}}\cdot \frac{2^{1283/64}\,\pi^{14}\,\Gamma^{10} \!\left(\frac{1}{8}\right) \Gamma^2\! \left(\frac{5}{32}\right) \Gamma^2 \!\left(\frac{7}{32}\right)}{\Gamma^{18} \!\left(\frac{5}{8}\right) \Gamma^{10} \!\left(\frac{1}{16}\right) \Gamma^{10} \!\left(\frac{3}{16}\right) \Gamma^2 \!\left(\frac{17}{32}\right) \Gamma^2 \!\left(\frac{19}{32}\right)}.$$ We can verify that $$\small\frac{{(2 n)!}^{20}\,{(8 n)!}^{32}\,{(32 n)!}^2}{{n!}^4\,{(4 n)!}^{37}\,{(16 n)!}^{13}}=1+\mathcal O\left(n^{-3}\right),$$ so the product indeed converges.

• Can you suggest how to prove its closed form on the right?
• Is it possible to further simplify it?
• Is it possible to find a simpler convergent infinite product of this form (involving only integer powers of factorials of integer multiples of $n$)?

Answer 1

A summary that I have now is as follows.

Let $f(x)=\sum_{k=1}^d c_k x^k$ with $\sum_{k=1}^d c_k=\sum_{k=1}^d kc_k=0$. Then $$\int_0^\infty\frac{f(e^{at})\,dt}{t(e^{bt}-1)}=\sum_{k=1}^d c_k\log\Gamma\left(1-\frac{ak}{b}\right).\qquad(b>0,a<b/d)$$ If also $\sum_{k=1}^d c_k/k=\sum_{k=1}^d c_k\log k=\sum_{k=1}^d kc_k\log k=0$, then $$\sum_{n=1}^\infty\sum_{k=1}^d c_k\log[(kn)!]=\int_0^\infty\frac{g(e^t)\,dt}{t(e^t-1)},\qquad g(x)=\sum_{k=1}^d\frac{c_k}{x^k-1}.$$

The first part is obtained using the integral representation $$\log\Gamma(1+z)=\int_0^\infty\left(\color{blue}{z}-\frac{\color{blue}{1}-e^{-zt}}{1-e^{-t}}\right)\frac{e^{-t}}{t}\,dt\qquad(\Re z>-1)$$ (attributed to Malmsten; follows from this one by Gauss). The conditions on $c_k$ (which are basically $f(1)=f'(1)=0$) ensure that the integrand has no singularity at $t=0$, and that the "blue" parts of the last integral vanish when summed.

The second part would follow from the first one if we could justify $\sum\int\mapsto\int\sum$ (and then the additional conditions on $c_k$ - which are essential - would have nowhere to come from). Let $$r_N(t)=\sum_{n=N}^\infty\sum_{k=1}^d c_k e^{-nkt};$$ then $r_N(t/N)/N$ and $r_N'(t/N)/N^2$ have finite limits as $N\to\infty$, i.e. $\sup_{t>0}|r_N(t)|$ behaves like $O(N)$; we can't justify $\int_0^\infty\frac{r_N(t)}{t(e^t-1)}\,dt\to 0$ as $N\to\infty$. A workaround is to consider $$\lambda(z):=\log\Gamma(1+z)-\left(z+\frac12\right)\log z+z-\frac12\log 2\pi-\frac1{12z};$$ then $\sum_{k=1}^d c_k\log[(kn)!]=\sum_{k=1}^d c_k\lambda(kn)$ and, thanks to $$\lambda(z)=\int_0^\infty\rho(t)e^{-zt}\,dt,\qquad\rho(t):=\frac1t\left(\frac12-\frac1t+\frac1{e^t-1}\right)-\frac1{12},$$ we're able to show $\int_0^\infty\rho(t)r_N(t)\,dt\to 0$ (as $N\to\infty$) since now $\rho(t)=O(t^2)$ as $t\to 0$.

In our case, $f(x)=2x^{32}-13x^{16}+32x^8-37x^4+20x^2-4x$ meets all the conditions, and we have $g(x)/(x-1)=p(x)/(x^{32}-1)$ with $p(x)$ a polynomial of degree $30$. This gives an expression for the given product in terms of $\Gamma(\cdot/32)$, and it remains to reduce the basis of these values using the multiplication/reflection formulae for $\Gamma$.

(Still editing the answer, to include a computation, and possibly a simpler product.)