# Is this equality $\lim_{x \to \infty} \int_0^x \frac{t^2}{2(e^t-1)}\mathrm{d}t=\lim_{n \to \infty}\sum_{k=1}^n \frac{1}{k^3}$ true?

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.

• Yes, it's true. See e.g. this answer for a way to derive it. The gist of the derivation: expand $\frac{1}{e^t-1} = \frac{1}{e^t} \frac{1}{1-e^{-t}}$ in a geometrical series $\sum_{n=1}^\infty e^{-nt}$ and integrate term by term using the definition of the Gamma function to evaluate each integral. The general result is $\int_0^\infty \frac{t^{s-1}}{e^t-1}{\rm d}t = \zeta(s)\Gamma(s)$ where $\zeta(s)$ is the Riemann zeta function. – Winther Dec 31 '15 at 1:49
• OK, i see, it looks a very funny way, i will do it. I never use the gamma function in my life but it looks in my way. Thank you. – Hexacoordinate-C Dec 31 '15 at 1:54
• See this MO question for a, imo, particular nice derivation (just take $s=3$ to get your integral). – Winther Dec 31 '15 at 2:07

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 $\zeta (3)=\sum_{k=1}^\infty \frac{1}{k^3}$.