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.
$\endgroup$ 86 Answers
$\begingroup$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).
$\endgroup$ 9 $\begingroup$Any complex number can be written as $\alpha \pi^\beta$ if you make no assumptions on $\alpha$ and $\beta$. You can even take $\beta = 0$.
It is however conjectured that $\zeta(2n+1)/\pi^{2n+1}$ is irrational for all $n \geq 1$. This conjecture is wide open.
$\endgroup$ 5 $\begingroup$See this question at MO. The short answer is that we don't know whether $\zeta(3)/\pi^3$ is rational (I assume that that's what you meant), but nobody seriously believes it is. Indeed, it shouldn't be, instead it should be connected to a certain higher regulator (google for Borel regulator and/or Lichtenbaum conjecture if you want to know more, but beware that it's pretty technical stuff).
$\endgroup$ $\begingroup$I'll answer your modified question, as explained in your comment:
I am not searching for a rational number, maybe could be a irrational...a strange irrational. Maybe α uses γ (constant).
It's known through LLL testing that there is no simple form of $\zeta(3)$ involving low powers of $\pi$, certain other constants like $\gamma,$ and simple (small denominator) rationals. It's not expected that such a form exists at all, but of course this is not known (and probably won't be any time soon).
$\endgroup$ $\begingroup$Other than the answers presented in this post, I would like to mention several formulae expressing $\zeta(3)$ (and other odd zeta values) in terms of powers of $\pi$. The most well-known ones are due to Plouffe and Borwein & Bradley:
$$ \begin{aligned} \zeta(3)&=\frac{7\pi^3}{180}-2\sum_{n=1}^\infty \frac{1}{n^3(e^{2\pi n}-1)},\\ \sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} &= -\frac{4}{3}\,\zeta(3)+\frac{\pi\sqrt{3}}{2\cdot 3^2}\,\left(\zeta(2, \tfrac{1}{3})-\zeta(2,\tfrac{2}{3}) \right). \end{aligned} $$
Moreover, in this Math.SE post we have:
$$ \frac{3}{2}\,\zeta(3) = \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(e^{2\pi k\sqrt{2}}-1)}. $$
You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.
$\endgroup$ $\begingroup$I'm hoping it'll turn out to be related somehow to OEIS A104007
$\zeta(3)=c \cdot \pi^3/60$
$\zeta(5)=c \cdot \pi^5/378$
$\zeta(7)=c \cdot \pi^7/2700$
where $c$ is either a constant or a function of $n$. The sequence 6,60,90,378,945,2700 are the denominators of coefficients in the expansion of $\frac{x^2}{(1-e^{-2x})^{2}}$.
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