Closed form for $\zeta(3)$

Question

I was reading about Euler's approach for the closed form of $\zeta(2)$. On the same lines, if we instead consider the factorization

$x(1-\frac{x}{\pi})(1+\frac{x}{\pi})(1-\frac{x}{2\sqrt2\pi})(1+\frac{x}{2\sqrt2\pi})(1-\frac{x}{3\sqrt3\pi})(1+\frac{x}{3\sqrt3\pi})(1-\frac{x}{4\sqrt4\pi})(1+\frac{x}{4\sqrt4\pi})...$,

i.e.,

$x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{2^3\pi^2})(1-\frac{x^2}{3^3\pi^2})(1-\frac{x^2}{4^3\pi^2})...$,

will it just be a matter of finding a trigonometric function representation of this polynomial ($\sin(x)$ in case of Euler's method), that when expanded as a Taylor series, and its co-efficients compared with the polynomial, that we shall be able to find a closed form for

$1 + \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + ...$ ?

Further, if it can be proved that no trigonometric function can express the above polynomial, will that mean that no closed form for $\zeta(3)$ exists?

Please forgive my limited knowledge on this topic, lack of research, and only quoting Wikipedia.

Answer 1

I've seen this closed form over internet a long time ago, wolframalpha confirm that it is correct, but I've also a very limited math knowledge, I have no idea if it can be considered as closed form since it involve the third-order derivative of Gamma or polyGamma functions

I think it's coming from Plouffe's inverter

$\zeta(3) = -{1\over2}\biggl(\gamma^3+{1\over2}(\gamma\pi^2)+\Gamma'''(1)\biggl)$

wolframalpha computation result

Proposed as alternate form by wolframalpha :

$\zeta(3) = -{{\psi^{(2)}(1)}\over2}$

wolframalpha computation result

Answer 2

Not an answer.

I'm not sure if Ch'nycos's answer will stay, but to support its claims a bit, by rearranging the Polygamma function's recurrence relation we obtain, $$\frac{1}{n^3}=\frac{\psi^{(2)}(n+1)-\psi^{(2)}(n)}{2}$$ summing over $n\in\mathbb{N}$, $$\zeta(3)=\sum_{n=1}^\infty\frac{1}{n^3}=\frac{1}{2}\sum_{n=1}^\infty\psi^{(2)}(n+1)-\psi^{(2)}(n)=-\frac{\psi^{(2)}(1)}{2}$$ after telescoping the series.