$f: \mathbb{R} \to \mathbb{R}$, $$f (x) = \left\{\begin{array}{ll} \frac{x^2}{(e^x-1)^2} & x> 0\\ 0 & x \leq 0\end{array}\right.$$
Can the function be integrated Lebesgue in the measure space $(\mathbb{R}, M, m)$? Also write the $\int f dm $ integral as a function containing the expression $\sum_{n=0}^∞ 1/n^3$. (Guidance: $1 / (1-t) ^ 2 = \sum_{n=0}^∞ 1/(nt^{n-1})$, use the series expansion $| t | <1$.) $(\mathbb{R}, M, m)$ is a set of real numbers and a set of measurable functions, and m is a measure function.
The formula for $\frac 1 {(1-t)^{2}}$ given in the hint is false. The series there does not even converge. The correct formula is $\frac 1 {(1-t)^{2}} = \sum\limits_{n=1}^{\infty} n t^{n-1}$
For $x>0$ we can write $f(x)=\frac {x^{2}e^{-2x}} {(1-e^{-x})^{2}}$. Using the formula $\frac 1 {(1-t)^{2}} = \sum\limits_{n=1}^{\infty} n t^{n-1}$ this becomes $f(x)=x^{2}e^{-2x} \sum\limits_{n=1}^{\infty} n e^{-(n-1)x}=\sum\limits_{n=1}^{\infty} x^{2} e^{-(n+1)x} $. I will let you finish by integrating term by term and using integration by parts for $\int_0^{\infty} x^{2}e^{-(n+1)x}dx$.