Is it possible to find some kind of closed form for the following function ?
$$L(x) = \prod_{k=1}^\infty \frac{\sin \pi \frac{x}{k}}{\pi \frac{x}{k}} = \prod_{k=1}^\infty \prod_{n=1}^\infty (1-\frac{x^2}{k^2 n^2})\\ = \exp(-\sum_{k=1}^\infty \sum_{n=1}^\infty\sum_{m=1}^\infty \frac{x^{2m}}{m k^{2m} n^{2m}})=\exp(-\sum_{m=1}^\infty \frac{x^{2m}}{m}\zeta(2m)^2)$$
It converges for $0<x<1$ and this is the interval I'm interested in.
We have the following identities:
$$\sum_{m=1}^\infty \frac{\zeta(2m)}{m} x^{2m} =\log \frac{\pi x}{\sin \pi x}$$
$$\sum_{m=1}^{\infty} \frac{\zeta(2m)}{2m+1} x^{2m}= \frac{1}{2} \left(1-\log 2-\log (\sin \pi x)-\frac{1}{2 \pi x}\operatorname {Cl} _{2}(2\pi x) \right)$$
$$\sum_{m=1}^{\infty} \frac{\zeta(2m)}{m(2m+1)} x^{2m}= \log (2 \pi x)-1+\frac{1}{2 \pi x}\operatorname {Cl} _{2}(2\pi x)$$
See https://en.wikipedia.org/wiki/Clausen_function.
However, I don't know a closed form when zeta is squared.
Using one of the integrals for the zeta function, I have also derived:
$$\frac{1}{L(x)}= \exp \left[ \frac{1}{2} \log \frac{\pi x}{\sin \pi x}+x \int_0^x \log \frac{\pi u}{\sin \pi u} \frac{du}{u^2}+ \\ + 2 \int_0^\infty \frac{\arctan t dt}{e^{2 \pi t}-1}+ \frac{1}{i} \int_0^\infty \frac{dt}{e^{2 \pi t}-1} \left( \log \sin \frac{\pi x}{1+ i t}- \log \sin \frac{\pi x}{1- i t} \right) \right]$$
Or, using automaticallyGenerated's comment, we can simplify:
$$L(x)= \frac{\sqrt{2 \pi}}{e} \sqrt{\frac{\sin \pi x}{\pi x}} e^{-I_1(x)-I_2(x)}$$
Where:
$$I_1(x)=x \int_0^x \log \frac{\pi u}{\sin \pi u} \frac{du}{u^2}$$
$$I_2(x)=\frac{1}{i} \int_0^\infty \frac{dt}{e^{2 \pi t}-1} \left( \log \sin \frac{\pi x}{1+ i t}- \log \sin \frac{\pi x}{1- i t} \right) $$
We can derive the following identity:
$$\log \sin (a+i b)= \frac{1}{2} \log ( \cosh^2 b- \cos^2 a)+ i \arctan \frac{\tanh b}{\tan a}$$
Which leads to:
$$I_2(x)=2 \int_0^\infty \frac{dt}{e^{2 \pi t}-1} \arctan \frac{\tanh \frac{\pi x t}{1+ t^2} }{\tan \frac{\pi x}{1+ t^2}} $$
Or:
$$I_2(x)=2 \int_0^{\pi/2} \frac{du}{(e^{2 \pi \tan u}-1) \cos^2 u} \arctan \frac{\tanh \frac{\pi x}{2} \sin 2 u }{\tan \frac{\pi x}{2} (1+ \cos 2u)} $$
$I_2$ seems hopeless, but I'd love a partial answer for $I_1$ at least.
This can be considered a generalization of my old question, though I just remembered it.