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.

The question is whether or not $\zeta(3)$ is connected with $\pi$. The answer is yes. Moreover, $\zeta(3) = \beta \pi^{\alpha}$ for some complex $\alpha, \beta$. Take $\alpha = 3$ and $\beta = 0.0387682...$. It is not known whether $\beta = 0.0387682...$ is algebraic or transcendental. It is known, however, that $\zeta(3)$ is irrational as shown by Apery.

Ramanujan conjectured and Grosswald proved that the following holds. If $\alpha, \beta > 0$ such that $\alpha \beta = \pi^{2}$, then for each non-negative integer $n$,
\begin{align}
 \alpha^{-n} \left( \frac{\zeta(2n+1)}{2} + \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 k \alpha} - 1} \right) & = (- \beta)^{-n} \left( \frac{\zeta(2n+1)}{2} + \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 k \beta} - 1} \right) - 
\end{align}
\begin{align}
\qquad 2^{2n} \sum_{k = 0}^{n+1} (-1)^{k} \frac{B_{2k} \ B_{2n- 2k + 2}}{(2k)! \ (2n  - 2k + 2)!} \alpha^{n - k + 1} \beta^{k}.
\end{align}
where $B_n$ is the $n^{\text{th}}$-Bernoulli number.

For odd positive integer $n$, we take $\alpha = \beta = \pi$,
\begin{align}
\zeta(2n+1) = -2^{2n} \left( \sum_{k = 0}^{n+1} (-1)^{k} \frac{B_{2k} \ B_{2n- 2k + 2}}{(2k)! \ (2n  - 2k + 2)!} \right) \pi^{2n+1} - 2 \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 \pi k} - 1}.
\end{align}

In particular, for $n = 1$,
\begin{align}
\zeta(3) = -4 \left( \sum_{k = 0}^{2} (-1)^{k} \frac{B_{2k} \ B_{2- 2k + 2}}{(2k)! \ (2  - 2k + 2)!} \right) \pi^{3} - 2 \sum_{k \geq 1} \frac{k^{-3}}{e^{2 \pi k} - 1}.
\end{align}

Observe that the coefficient of $\pi^{3}$ is rational, however, nothing is known about the algebraic nature of the infinite sum. This is a current topic of research. Indeed, it is conjectured that $\frac{\zeta(3)}{\pi^{3}}$ is transcendental. 

Update: Recently, Takaaki Musha claims to have proved that $\frac{\zeta(2n+1)}{(2 \pi)^{2n+1}}$ is irrational for positive $n \geq 1$. However, some objection has since been raised (read comments below).
