Problem: Let $f_1(x)=\dfrac{x}{e^x-1}, f_2(x)=\dfrac{x}{e^x+1}$. Show that $f_1,f_2$ are Lebesgue-integrable and $\int_{(0,\infty)}f_1 d\lambda=\sum_{n\in\mathbb{N}}\dfrac{1}{n^2}$
$\int_{(0,\infty)}f_2 d\lambda=\sum_{n\in\mathbb{N}}(-1)^{n+1}\dfrac{1}{n^2}$.
My ideas: We were given the hint that $\dfrac{e^{-x}}{1-e^{-x}}=\sum_{n=1}^\infty e^{-nx}$. I see that $f_1(x)=\dfrac{x}{e^x-1}=\dfrac{xe^{-x}}{1-e^{-x}}$ but I don't know how to go from here or how to prove the hint.
(1) Regarding the hint, note that for $x > 0$ we have $\exp(-x) \in (0,1)$ and hence by the geometric series $$ \sum_{n=0}^\infty \exp(-x)^n = \frac 1{1-\exp(-x)} $$ and so $$ \sum_{n=1}^\infty \exp(-nx)=\sum_{n=1}^\infty \exp(-x)^n = \frac 1{1-\exp(-x)} - 1 = \frac{\exp(-x)}{1-\exp(-x)}. $$ (2) Integrating termwise, as $$ \int_0^\infty x\exp(-nx)\,dx = \frac 1{n^2} $$ by partial integration, this gives the first integral.
(3) For the second integral note that $$ \frac 1{1 + \exp(-x)} = \sum_{n=0}^\infty \bigl(-\exp(-x)\bigr)^n = \sum_{n=0}^\infty (-1)^n \exp(-nx) $$ and continue as above.
