Convergence of the improper integral $\int _{1}^{\infty}{\frac{x}{1-e^{x}}dx}$

Question

Need help determining the convergence/divergence of the following improper integral:

$$\int \limits_{1}^{\infty}{\frac{x}{1-e^{x}}dx}$$

I tried using comparison tests but with no luck.

Answer 1

For sufficiently large $x$, $e^x-1>x^3$. So there exists a constant $C\ge1$ such that $$ \int_C^\infty\frac x{e^x-1}dx<\int_C^\infty\frac1{x^2}dx<\infty. $$ So your integral also converges.

Answer 2

We can enforce the substitution $x\to \log x$ to reveal that

$$\begin{align} \int_1^{\infty}\frac{x}{e^x-1}\,dx&=\int_e^{\infty}\frac{\log x}{x(x-1)}\,dx\\\\ &\le \int_e^{\infty}\frac{\log x}{(x-1)^2}\,dx\\\\ &=\lim_{L\to \infty}\left.\left(\frac{(1-x)\log|1-x|+x\log |x|}{1-x}\right)\right|_{e}^{L}\\\\ &=-\log(e-1)+\frac{e}{e-1}\\\\ &<\infty \end{align}$$

Therefore, the integral converges!