**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.