## Remarks on ζ(2n+1)

[Posted on december 15, 2016.]

Having for a few days stupidly and incorrectly claimed a proof of the irrationality of all $\zeta(2n+1)$ after getting confused with a definition (sic) for which I express my deepest apologies for any confusion this may have caused, a new version of the preprint has withdrawn that claim.

The little that survives then is the formula $\displaystyle \frac{\zeta(2n+1)}{r^{2n}}=(-1)^{n+1} \int\limits_{[0;1]^n} \left (\prod_{i=1}^{n}\frac{\log(x_i)}{x_i}\right ) \log\left (1-\left(\prod_{i=1}^nx_i\right)^r \right ) dx_1\cdots dx_n$ which for $r=1$, due to its product form, I thought might have some uses that the classical $\displaystyle \zeta(k)=\int\limits_{[0;1]^k} \frac{dx_1\cdots dx_k}{1-x_1\cdots x_k}$ might not have.

Related expressions that I then obtained, namely $\displaystyle \int\limits_{[0;1]^2} \frac{\log(x)}{x}\frac{\log(y)}{y}\log(1-(xy)^1) \log(x)\log(y) (xy)^{2k+1} dxdy =$
$\displaystyle \frac{4\zeta(6)}{(2k+1)} + \frac{4\zeta(5)}{(2k+1)^2} + \frac{4\zeta(4)}{(2k+1)^3} + \frac{4\zeta(3)}{(2k+1)^4} + \frac{4\zeta(2)}{(2k+1)^5} -\frac{4}{(2k+1)^6}\sum_{n=1}^{2k+1}\frac{1}{n}-4\sum_{j=1}^{6}\left (\frac{1}{(2k+1)^j} \sum_{i=1}^{2k+1}\frac{1}{i^{7-j}}\right )$

and

$\displaystyle \int\limits_{[0;1]^2} \frac{\log(x)}{x}\frac{\log(y)}{y}\log(1-(xy)^2) \log(x)\log(y) (xy)^{2k+1} dxdy =$
$\displaystyle \frac{63\zeta(6)}{8(2k+1)} + \frac{31\zeta(5)}{4(2k+1)^2} + \frac{15\zeta(4)}{2(2k+1)^3} + \frac{7\zeta(3)}{(2k+1)^4} + \frac{6\zeta(2)}{(2k+1)^5}$
$\displaystyle +\frac{8\log(2)}{(2k+1)^6}-\frac{4}{(2k+1)^6}\sum_{i=1}^{k}\frac{1}{2i+1}-8\sum_{j=1}^{6}\left (\frac{1}{(2k+1)^j}\sum_{i=0}^{k}\frac{1}{(2i+1)^{7-j}}\right )$

do not seem to be of any help towards a proof of irrationality of at least $\zeta(5)$.  A combination of those two expressions, perhaps with other expressions such as those of Vasilyev, might allow to improve matters, but I was not able to find how.