Does the following infinite product have a "nice" closed form?
$$ P = \prod_{k=2}^{\infty} \left(\left(1 - \frac{1}{k^2}\right)^\dfrac{(-1)^k}{k}\right) $$
I know that without the power one could easily use the infinite sine product formula or if the kth root was the $k$th power, there wouldn't be too much difficulty, but I could not find anything on this stated product.
Edit:
Finding the natural log of this is easily shown to be equivalent to finding the sum:
$$ \ln(P) = S = \sum_{k=2}^{\infty} \left(\frac{(-1)^k}{k}\ln\left(1 - \frac{1}{k^2}\right)\right) $$
Edit
By expanding the logarithm in the sum, one obtains:
$$ S = \sum_{k=2}^{\infty} \left(\frac{(-1)^k}{k}\left(\ln\left(1 - \frac{1}{k}\right) + \ln\left(1 + \frac{1}{k}\right)\right)\right) $$
Using the Taylor series for the natural logarithm, switching the order of summation, and then re-summing gives that the sum must be:
$$ S = \sum_{k=2}^{\infty} \left(\frac{\eta \ (k+1)-1}{k}\right)+\sum_{k=2}^{\infty} \left(\frac{(-1)^k(\eta \ (k+1)-1)}{k}\right) $$
Which may then be expanded to the sums:
$$ S = \sum_{k=2}^{\infty} \left(\frac{\zeta \ (k+1)-1}{k}\right)+\sum_{k=2}^{\infty} \left(\frac{(-1)^k(\zeta \ (k+1)-1)}{k}\right)+\sum_{k=2}^{\infty} \left(\frac{\zeta \ (k+1)}{k \ 2^k}\right)+\sum_{k=2}^{\infty} \left(\frac{(-1)^k\zeta \ (k+1)}{k \ 2^k}\right) $$
The first sum is the Lüroth or Alladi-Grinstead constant, and the rest don't seem to have well behaved closed forms. I don't know if things might cancel nicely and leave a decent solution. I doubt they leave a closed form.