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So this question is about this dilogarithm function. Assume the argument $z$ is real then I want to show the formula $\operatorname{Li}_2(e^{-z})=\frac{\pi^2}{6} + z\log z -z+O(z^2) $

as $z$ approaches $0$ from positive value

How can I show this?

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An important reference is this. Using eq.(3.3) in the paper

${\rm Li}_2(1-z)=-{\rm Li}_2(z)+\frac{\pi^2}{6}+\ln(1-z)\ln z$

from the definition

${\rm Li}_2(z)=\sum_{k=1}^\infty\frac{z^k}{k^2}$

you will get immediately eq.(3.7) in the paper, the expansion for $z=1$,

${\rm Li}_2(z)=\frac{\pi^2}{6}-\sum_{k=1}^\infty\frac{(1-z)^k}{k^2}+\ln(1-z)\sum_{k=1}^\infty\frac{z^k}{k}.$

Remembering that

$\ln(1+z)=\sum_{n=1}^\infty(-1)^{n+1}\frac{z^n}{n}$

you will recognize that

$\ln(1-z)\sum_{k=1}^\infty\frac{z^k}{k}\approx z\ln z$

while you have to retain just the first term from the other series. Putting all together you get your approximation.

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    Thank you guys, It is the last identity I didn't recognize.2012-03-16