In this paper on pages 6 & 7. Page 6 lists the variables used in the equation on page 7.
The author claims a closed form expression of $\zeta(3)$ (he also goes on to claim a closed form expression of $\zeta(2n+1)$ ). I tried calculating it in my calculator and Wolfram Alpha as well, but get an answer of $0$ both times.
I am new to reading math papers, but I believe I interpreted the equation of $\zeta(3)$ correctly. Please let me know where I went wrong.
To save all of us the trouble, I evaluated the formula in the paper and it is incorrect (perhaps not a surprise). In page 7, the author gives it as,
$$\zeta(3)= \frac{8}{3}u\,\big(\beta\,B(\tfrac{1}{10}) + \alpha\,B(\tfrac{2}{10}) + \alpha\,B(\tfrac{3}{10}) + \beta\,B(\tfrac{4}{10})\big) - \frac{16}{3}u\big(\alpha\,B(\tfrac{1}{5}) +\beta\,B(\tfrac{2}{5})\big) + \frac{2}{3}u\big(-\beta\,B(\tfrac{1}{5}) + \alpha\,B(\tfrac{2}{5})\big)\tag{1}$$
where,
$$u = \frac{2\pi^3}{3},\;\;\alpha = \sqrt{\frac{5+\sqrt{5}}{8}},\;\;\beta = \sqrt{\frac{5-\sqrt{5}}{8}},\;\;B(x) = x(x-\tfrac{1}{2})(x-1)$$
As the OP also noticed, the expression (1) in fact, is equal to zero. (Among other things, the $\sqrt{5}$ was already a bad sign.)
