The problem
Can be $\zeta(3)$ written as $\alpha\pi^\beta$, where ($\alpha,\beta \in \mathbb{C}$), $\beta \ne 0$ and $\alpha$ doesn't depend of $\pi$ (like $\sqrt2$, for example)?
Details
Several $\zeta$ values are connected with $\pi$, like:
$\zeta$(2)=$\pi^2/6$
$\zeta$(4)=$\pi^4/90$
$\zeta$(6)=$\pi^6/945$
...
and so on for all even numbers.
See this mathworld link to more details: Riemann Zeta Function
So the question is, could $\zeta(3)$ be written as:
$$\zeta(3)=\alpha\pi^\beta$$ $$\alpha,\beta \in \mathbb{C}$$ $$\beta \ne 0$$ $$\alpha \text{ not dependent of } \pi$$
See $\alpha$ not essencially belongs $\mathbb{Q}$ and $\alpha,\beta$ could be real numbers too.
When I wrote $\alpha$ is not dependent of $\pi$ it's a strange and a hard thing to be defined, but maybe $\alpha$ can be written using $e$ or $\gamma$ or $\sqrt2$ or some other constant.
Edit:
Maybe this still a open question. If
$ \sum_{k = 0}^{2} (-1)^{k} \frac{B_{2k} \ B_{2- 2k + 2}}{(2k)! \ (2 - 2k + 2)!}$
in $-4 \sum_{k = 0}^{2} (-1)^{k} \frac{B_{2k} \ B_{2- 2k + 2}}{(2k)! \ (2 - 2k + 2)!}\pi^3$ be of the form $\frac{\delta}{\pi^3}$ with $\delta$ not dependent of $\pi$
and $- 2 \sum_{k \geq 1} \frac{k^{-3}}{e^{2 \pi k} - 1}$ not dependent of $\pi$ too, this question still hard and open.
Edit 2:
I discovered a result, but later I've seen that this is something already known, either way, it is an interesting one to have it here.
$$\zeta(3)=-4\pi^2\zeta'(-2)$$
but, if $\zeta'(-2)$ is of the form $\frac{\alpha}{\pi^2}$, with $\alpha$ not dependent of $\pi$, then this still remains as a hard and an open question.
I have a conjecture that $\zeta'(-2)$ will not cancel the $\pi^2$ term, but since I wasn't able to prove it and I can't use it here.
We can express $\zeta$ of odd numbers with $\zeta'$ in a easy way, with a "closed" form like this one.