Infinite product of sinc function Is there a closed form for this infinite product? :

$$\prod_{n=1}^{\infty}\operatorname{sinc}\left(\frac{\alpha}{n}\right)$$

where $\operatorname{sinc}(x)=\sin(x)/x$ is the familiar "sinc" function and $\alpha$ is some real parameter.

I would be rather surprised there is, but nevertheless ...

SIDE NOTE : I perform a numerical analysis and my guess for its taylor series in terms of $\alpha$ upto second power may be : $1-\alpha^2\zeta(2)/6$
Since 
$$\frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{\pi^2 n^2}\right) $$
we have:
$$ \log\text{sinc}(x) = -\sum_{n\geq 1}\sum_{m\geq 1}\frac{x^{2m}}{m\pi^{2m} n^{2m}}=-\sum_{m\geq 1}\frac{\zeta(2m)}{m\pi^{2m}}x^{2m}$$
and


  $$ \sum_{n\geq 1}\log\text{sinc}\left(\frac{\alpha}{n}\right) = -\sum_{m\geq 1}\frac{\color{red}{\zeta(2m)^2}}{m\pi^{2m}}\alpha^{2m}.$$
If you are familiar with Fourier transforms, you can interpret this product as the iterated convolution of the fourier transform of the sinc function, which is a constant function on an interval, and 0 outside of that interval. Furthermore if we interpret this function as a probability density function of a random variable (uniform probability distribution), and that convolution of density functions correspond to the density function of the sum of the random variables and also remember https://en.wikipedia.org/wiki/Law_of_large_numbers this will approach a normal distribution for any function that is "nice enough", and certainly a uniform distribution is nice enough for that.