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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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1 Answer

up vote 3 down vote accepted

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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A proof of that identity appears in my answer here: math.stackexchange.com/questions/117246/… –  Aryabhata Mar 16 '12 at 18:43
    
@Aryabhata: Sorry but I was not aware of it. Thanks for pointing this out. –  Jon Mar 16 '12 at 19:09
    
I was just pointing it out in case someone was interested in the proof (and the comment wasn't aimed specifically at you). This is a well known formula with an easy proof. You don't have to apologise :-) –  Aryabhata Mar 16 '12 at 19:11
    
Thank you guys, It is the last identity I didn't recognize. –  John0417 Mar 16 '12 at 19:39
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