Computing $\int_{0}^{\infty} \ln(1 - e^{-\beta \hbar \omega}) \frac{3v}{2\pi^2 c_{s}^{3}} \omega^{2} \mathop{dw}$ I'm having trouble computing the integral 
$$\int_{0}^{\infty} \ln(1 - e^{-\beta \hbar \omega}) \frac{3V}{2\pi^2 c_{s}^{3}} \omega^{2} \mathop{d\omega},$$
where $e$, $\hbar, \omega, \pi, c_{s}, V$ and  $\beta$ are all constants. 
(side-note: this is part of an equation that's used in physics to calculate the partition function for a Debye solid). 
In a book that I am reading, it is said that this integral can be evaluated by using a Taylor series expansion (I think it will be of the $\ln$ term), and then doing integration by parts once. 
I have no clue how to do this, and I've tried to solve this integral for a few hours now. I think that one of my initial steps are incorrect because I'm more familiar with using integration by parts. I also tried to use this relation:
$$\ln(1 - e^{-\beta \hbar\omega}) = \ln\left(1 - \frac{1}{e^{\beta \hbar \omega}}\right) = \ln\left(\frac{e^{\beta \hbar \omega} - 1}{e^{\beta \hbar \omega}}\right) = \ln(e^{\beta\hbar\omega} - 1) - \ln(e^{\beta \hbar \omega}) = \ln(e^{\beta\hbar\omega} - 1) - \beta \hbar \omega,$$
but that also didn't get me anywhere. If anyone has any suggestions, I would much appreciate it.
EDIT: maybe writing $e^{-\beta \hbar \omega} = \cosh(x) - \sinh(x)$ can help? 
 A: \begin{eqnarray*}
&&\int_{0}^{\infty} \ln(1 - e^{-\beta \hbar \omega}) \frac{3v}{2\pi^2 c_{s}^{3}} \omega^{2} \mathop{d\omega}\\
&=&-\frac{3v}{2\pi^2 c_{s}^{3}} \int_{0}^{\infty} \sum_{n=1}^\infty\frac{1}{n}e^{-n\beta \hbar \omega} \omega^{2} \mathop{d\omega}\\
&=&-\frac{3v}{2\pi^2 c_{s}^{3}}\sum_{n=1}^\infty\frac{1}{n}\int_{0}^{\infty} e^{-n\beta \hbar \omega}  \omega^{2} \mathop{d\omega}\\
&=&-\frac{3v}{2\pi^2 c_{s}^{3}}\sum_{n=1}^\infty\frac{1}{n}\frac{2}{(\beta \hbar)^3n^3}\\
&=&-\frac{3v}{2\pi^2 c_{s}^{3}}\sum_{n=1}^\infty\frac{1}{n}\frac{2}{(\beta \hbar)^3n^3}\\
&=&-\frac{3v}{\pi^2 c_{s}^{3}\beta^3 \hbar^3}\sum_{n=1}^\infty\frac{1}{n^4}\\
&=&-\frac{3v}{\pi^2 c_{s}^{3}\beta^3 \hbar^3}\frac{\pi^4}{90}\\
&=&-\frac{\pi^2v}{30 c_{s}^{3}\beta^3 \hbar^3}
\end{eqnarray*}
A: Using
$$\ln\left(1-e^{-\beta\hbar\omega}\right) = -\sum_{n=1}^\infty\frac{e^{-\beta\hbar n\omega}}{n}$$
the integral becomes
$$I = -\frac{3v}{2\pi^2 c_s^3}\sum_{n=1}^\infty \frac{1}{n}\int_0^\infty \omega^2 e^{-\beta\hbar n\omega}\,d\omega$$
Letting $\beta\hbar n\omega =: u, d\omega = \frac{du}{\beta\hbar n}$,  the integral becomes
\begin{align}
I &= -\frac{3v}{2\pi^2 c_s^3\beta^3\hbar^3}\sum_{n=1}^\infty \frac{1}{n^4}\int_0^\infty u^2 e^{-u}\,du\\
&= -\frac{3v}{\pi^2 c_s^3\beta^3\hbar^3}\sum_{n=1}^\infty \frac{1}{n^4}\\
&= -\frac{\pi^2 v}{30 c_s^3\beta^3\hbar^3}
\end{align}
A: Another approach is with Riemann Zeta Function, and $t=\beta\hbar$
\begin{align}
\frac{3V}{2\pi^2 c_{s}^{3}}\int_{0}^{\infty} \ln(1-e^{-t\omega})\omega^2\ d\omega
&= \frac{3V}{2\pi^2 c_{s}^{3}}\int_{0}^{\infty} \int_\infty^t \dfrac{\omega e^{-\alpha\omega}}{1-e^{-\alpha\omega}} \omega^2\ d\alpha\ d\omega \\
&= \frac{3V}{2\pi^2 c_{s}^{3}}\int_\infty^t \int_{0}^{\infty} \dfrac{\omega^3}{e^{\alpha\omega}-1}\ d\omega\ d\alpha \\
&= \frac{3V}{2\pi^2 c_{s}^{3}}\int_\infty^t \dfrac{1}{\alpha^4}\ \dfrac{\pi^4}{15}\ d\alpha \\
&= \frac{3V}{2\pi^2 c_{s}^{3}}\dfrac{\pi^4}{15} \dfrac{1}{-3t^3}\\
&= \frac{3V}{2\pi^2 c_{s}^{3}}\dfrac{\pi^4}{15} \dfrac{1}{-3\beta^3\hbar^3} 
\end{align}
