I'm searching for a nice derivation of the formua
$\sum_{n=1}^\infty \frac{1}{n} \left( \frac{q^{2n}}{1-q^n}+\frac{\bar q^{2n}}{1-\bar q^n}\right)=-\sum_{m=2}^\infty \ln |1-q^m|^2$
given for example in http://arxiv.org/abs/arXiv:0804.1773 eq.(4.27).
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1It suffices to show only the holomorphic part: $\sum_n \frac1n \frac{q^{2n}}{1-q^n}=-\sum_m \ln \left(1-q^m\right)$. Why not simply expand the geometric series on the left, the logarithm on the right and compare the coefficients of different powers of $q$? – Start wearing purple Oct 15 '14 at 21:29
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Consider the double sum:
$$\sum_{n=1}^\infty \frac{1}{n} \frac{q^{2n}}{1-q^n} = \sum_{n=1}^\infty \frac{1}{n} \sum_{m=0}^\infty q^{n(m+2)} = \sum_{m=0}^\infty \sum_{n=1}^\infty \frac{1}{n} q^{n(m+2)} = \sum_{m=\color{red}{\mathbf{2}}}^\infty \log (1 - q^{m}) $$
This is a physics paper. They aren't worried that both sides diverge if $q^n=1$ for any $n \in \mathbb{N}$.
Did you notice earlier in the paper this was written in exponential form?
$$ Z = \mathrm{exp}\bigg[ \sum_{n \geq 1} \frac{1}{n} \frac{q^{2n}}{1-q^n} \bigg] = \prod_{m \geq 2} \frac{1}{1-q^m} = \mathrm{Tr}[q^{L_0}]$$
where $L_0$ is a generator of the Virasoro algebra.
The original definition of $Z$ was a functional determinant of the Laplacian $\det \Delta$ in Anti-De Sitter space $\mathbb{H}^3/\mathbb{Z}$ (Section 4.1 in arXiv:0804.1773v1)
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Cool stuff! The function $Z$ is almost the inverse of the Dedekind eta function. – Bruno Joyal Oct 15 '14 at 23:06