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I am looking for a relation between the dilog and its complex conjugate, that is can I simplify the following summation of terms $$f(z) = \text{Li}_2(z) + (\text{Li}_2(z))^*?$$

I have looked through the many identities that are known to exist among such functions on the Wolfram pages but did not find any involving the complex conjugate. If $z>1$ then $\text{Li}_2(z)$ is complex such that the combination $f(z)$ is real so it would be nice if $f(z)$ may be simplified to a dilog with an argument not appearing on the branch cut or something alike.

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  • $\begingroup$ The best we can say is $\text{Li}_2(z) + (\text{Li}_2(z))^* = 2\text{Re }\text{Li}_2(z)$, which is just a restatement. $\endgroup$ – GEdgar Sep 6 '18 at 11:18
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The real part $\Re \mathrm{Li}_2(x) = \frac{1}{2}f(x)$ for $x>1$ can be computed with the Euler reflection formula (see https://en.wikipedia.org/wiki/Polylogarithm#Dilogarithm)

$$\Re \mathrm{Li}_2(x) = \frac{\pi^2}{6} - \mathrm{Li}_2(1-x) - \ln x \ln|1-x|$$

where I have used $\Re \ln(1-x) = \ln|1-x|$ for $x > 1$

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