Using a little program in Python, it looks true for at least two hundred digits after the comma, but I have absolutely no idea, how to begin. Any hint sould be appreciate.
$$\lim_{x \to \infty} \int_0^x \frac{t^2}{2(e^t-1)}\mathrm{d}t=\lim_{n \to \infty}\sum_{k=1}^n \frac{1}{k^3}$$
It looks not very difficult, I tried a integration by part but it looks not to be the better way to compute it. I'm stuck there.
I should be glad, thank you in advance!
This may be more a long comment than an answer.
In a spirit similar to Winther's comments, it could be of interest to you to know that, using polylogarithms and their properties, $$\int \frac{t^2}{2(e^t-1)}\,{dt}=t \text{Li}_2\left(e^t\right)-\text{Li}_3\left(e^t\right)-\frac{1}{6} t^2 \left(t-3 \log \left(1-e^t\right)\right)$$ So,$$\int_0^x \frac{t^2}{2(e^t-1)}\,{dt}=x \text{Li}_2\left(e^x\right)-\text{Li}_3\left(e^x\right)-\frac{1}{6} x^2 \left(x-3 \log \left(1-e^x\right)\right)+\zeta (3)$$ and, if $x\to \infty$, the only term left is $\zeta (3)=\sum_{k=1}^\infty \frac{1}{k^3}$.