Improper integral $\int_{0}^{\infty} \frac{x}{e^{x}-1} \ dx$ 
$$\int_{0}^{\infty} \frac{x}{e^{x}-1}\ dx$$
  Is there a way to evaluate this integral without using the zeta function?


$$\text{ if } L(f) = F(s) \implies \int_0^{\infty} \frac f{e^t-1}dt = \sum_{s\ge1} F(s)$$
$$L(t) =\frac 1{s^2}\implies\int_0^{\infty} \frac t{e^t-1}dt = \sum_{s\ge1}\frac 1{s^2}  = \zeta(2) $$
