= Solution
For $x>0$, the <geometric series> gives
$$
\frac{xe^{-\alpha x}}{1-e^{-x}}=\sum_{n=0}^{\infty}xe^{-(n+\alpha)x}.
$$
Every summand is nonnegative and has <integral> $(n+\alpha)^{-2}$ by part (a). This numerical series converges by comparison with $\sum_{n\geq1}n^{-2}$. The series-integration rule proved above therefore yields
$$
\boxed{\int_0^\infty\frac{xe^{-\alpha x}}{1-e^{-x}}\,dx
=\sum_{n=0}^{\infty}\frac1{(n+\alpha)^2}<\infty.}
$$
The value at the single endpoint zero does not affect this Lebesgue <integral>.
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