TITLE: Riemann zeta function at odd positive integers
QUESTION [52 upvotes]: Starting with the famous Basel problem, Euler evaluated the Riemann zeta function for all even positive integers and the result is a compact expression involving Bernoulli numbers. However, the evaluation of the zeta function at odd positive integers (in terms of getting a closed form sum) is still open. There has been some progress in the form of Apery's theorem and other results such as "infinitely many of $\zeta(2n+1)$ are irrational" or "at least one of $\zeta(5),\zeta(7),\zeta(9)$ or $\zeta(11)$ is irrational". 
Question(s): Is there a high level understanding for this disparity between even and odd integers? Is it a case of there being a simple expression for $\zeta(3)$ that is out there waiting for an ingenious attack like Euler did with $\zeta(2)$? Or is the belief that such a closed form summation is unlikely? Where do the many many proofs powerful enough to evaluate $\zeta(2n)$ stumble when it comes to evaluating $\zeta(2n+1)$?
Motivation: The Basel problem and Euler's solution are my all-time favorites for the sheer surprise factor and ingenuity of proof (what do $\pi$ and $\frac{sin(x)}{x}$ have to do with $\zeta(2)$??). However, I currently lack the more advanced analytical tools to appreciate the deeper results of this area. I have wondered for a while about the questions above and Internet search hasn't helped much. I would greatly appreciate any answers/references. Thanks.

REPLY [2 votes]: To piggy-back off of the discussion from @Matt E above, I would like to add that there is probably good reason to believe that all odd integer values for (s) are algebraically independent of each other, but not necessarily of π. Euler evaluated the sinc(x) function to obtain a closed form value for (2). You can also see this function in the product formula:
$f(x) = \prod_{k=1}^\infty 1 -\frac{x^2}{k^2 π^2} = \frac{sin(x)}{x}$
If we look at a similar function for (4), we obtain:
$f(x) = \prod_{k=1}^\infty 1 -\frac{x^4}{k^4 π^4} = \frac{sin(x) sinh(x)}{x^2}$
There are other transcendental functions associated with the higher, even integer values for (s). Given these transcendental functions, i.e. the sinc(x) function and the function involving the hyperbolic sin associated with (4), you can simply take the absolute value of the second term in each of their Taylor series expansions and set x = π. You then obtain the closed form values for (2) and (4). You can similarly perform this for the higher, even values for (s).
The question: is there a transcendental function that we can similarly use to obtain the value for (3)? We do know this, using the gamma function:
$f(x) = \prod_{k=1}^\infty 1 -\frac{x^3}{k^3 π^3} = \frac{1}{\Gamma(1 - \frac{x}{π}) \Gamma(1 + \frac{(-1)^{(1/3)} x}{π}) \Gamma(1 - \frac{(-1)^{(2/3)} x}{π})}$
If you look at the second term of the series expansion for this inverted "multi-gamma" function above, we obtain:
$\frac{x^3 ψ^2(1)}{2 π^3}$
where $ψ^n(x)$ is ${n^t}^h$ derivative of the digamma function.
When you set x = π, you can see that it cancels out with the $π^3$ in the denominator and you're left with the digamma function value divided by 2, which is (3). So it seems that $π^3$ is kind of related to (3), but not in a way that's tangible. My opinion is that there is most likely another transcendental quantity involved along with π, one that may need to be discovered, or one that is in lieu of π. That is, π might be a stepping stone, so to speak, to reach another transcendental which is then involved with a closed form value for (3).