Here $t$ is a real positive parameter, and I am trying to solve the following integral:
$$I_t=\int_0^\infty\frac{x-1}{(x+1)^3(x^t-1)}dx $$
This evaluation could lead (maybe) to an explicit form for $\zeta(3)$, given the following relation:
$$\zeta(3)=\frac{2\pi^2}{7}\lim_{t\to 0}t\cdot I_t $$
I tried letting $x\to\frac1{x}$, leading to
$$I_t=\int_0^\infty\frac{x^t(x-1)}{(x+1)^3(x^t-1)}dx$$
but seems to be worse. So I tried evaluating first the integral from $0$ to $1$ with $x$ in the numerator:
\begin{align}
\int_0^1\frac{x}{(x+1)^3(x^t-1)}dx & =-\int_0^1\frac{x}{(x+1)^3}\left(\sum_{k=0}^\infty x^{kt}\right)dx \\
& =-\sum_{k=0}^\infty\int_0^1\frac{x^{kt+1}}{(x+1)^3}dx \\
\end{align}
but now the integrals of the form $\int_0^1\frac{x^{\alpha}}{(x+1)^3}$ are tough themselves, and result in messy digamma functions, together with other terms, so I believe this is not the way.
I should add that I am not looking for a proof of the limit as I already have it. I am looking only for a closed form of the integral.
Any ideas or hints?
Best Answer
Just for the fun of using (once more) my favored $1,400^+$ years old approximation of the sine function.
Starting from @Lai solution $$J=\frac{\pi}{2} \int_0^1 \frac{x-x^2}{\sin (\pi x)}\,dx=\frac 1{2\pi^2}\int_0^\pi \frac{(\pi -t)\, t} { \sin (t)}\,dt$$ Using $$\sin(t) \simeq \frac{16 (\pi -t) t}{5 \pi ^2-4 (\pi -t) t}\qquad \text{for} \qquad 0\leq t\leq\pi$$
$$J\sim \frac 1{32\pi^2}\int_0^\pi \left(4 t^2-4 \pi t+5 \pi ^2\right)\,dt=\frac{13 \pi }{96}$$ (relative error of $0.2$%) which would then give as an approximation $$\zeta(3) \sim \frac{13 \pi ^3}{336}$$ for an absolute error of $0.0024$.
This is slightly worse than the apparoximation $$\zeta(3)\sim \frac{\pi ^3 }{3 \sqrt{105}}\left(\tan \left(\frac{\pi }{18}\right)+\sec \left(\frac{\pi }{18}\right)\right) $$ which was proposed on the site.