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

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.

• Linear term in the top, an e-power in the bottom, well.....If you use taylor expansion of the e-power, what are the first couple of terms? – imranfat Oct 13 '15 at 21:45
• @imranfat, the first few terms are not going to help, as the problem is at $\infty$... – Mariano Suárez-Álvarez Oct 13 '15 at 21:45
• Can you tell us what you tried? With what did you try to compare? This integral is so convergent that it is difficult to come up with a comparison which does not work! :-) – Mariano Suárez-Álvarez Oct 13 '15 at 21:46
• The denominator is growing exponentially while the numerator grows linearly, as $x\to\infty$. That's most of the answer. ${}\qquad{}$ – Michael Hardy Oct 13 '15 at 21:55
• @MarianoSuárez-Alvarez The first 4 would do, wouldn't it? A linear divided by a cubic? Easy... – imranfat Oct 13 '15 at 22:38

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.
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}
• I feel like this would be clearer if you made the substitution $x = \log u$ to make it clearer what is happening. Good solution though. – Brevan Ellefsen Oct 14 '15 at 0:00