Krdytkyu

The problem

Can be $zeta(3)$ written as $alphapi^beta$, 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:

$$zeta(3)=alphapi^beta$$
$$alpha,beta in mathbb{C}$$
$$beta ne 0$$
$$alpha text{ not dependent of } pi$$

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.

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).

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.

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).

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).