Nice derivation of $\sum_{n=1}^\infty \frac{1}{n} \left( \frac{q^{2n}}{1-q^n}+\frac{\bar q^{2n}}{1-\bar q^n}\right)=-\sum_{m=2}^\infty \ln |1-q^m|^2$

Question

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).

Answer 1

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 $n \in \mathbb{N}$.

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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)