Difference between revisions of "Li 2(z)+Li 2(1-z)=pi^2/6-log(z)log(1-z)"

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(Created page with "==Theorem== The following formula holds: $$\mathrm{Li}_2(z)+\mathrm{Li}_2(1-z)=\dfrac{\pi^2}{6} - \log(z)\log(1-z),$$ where $\mathrm{Li}_2$ denotes the dilogarithm, $\pi$...")
 
 
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The following formula holds:
 
The following formula holds:
 
$$\mathrm{Li}_2(z)+\mathrm{Li}_2(1-z)=\dfrac{\pi^2}{6} - \log(z)\log(1-z),$$
 
$$\mathrm{Li}_2(z)+\mathrm{Li}_2(1-z)=\dfrac{\pi^2}{6} - \log(z)\log(1-z),$$
where $\mathrm{Li}_2$ denotes the [[dilogarithm]], $\pi$ denotes [[pi]], and $log$ denotes the [[logarithm]].
+
where $\mathrm{Li}_2$ denotes the [[dilogarithm]], $\pi$ denotes [[pi]], and $\log$ denotes the [[logarithm]].
  
 
==Proof==
 
==Proof==

Latest revision as of 02:41, 21 December 2017

Theorem

The following formula holds: $$\mathrm{Li}_2(z)+\mathrm{Li}_2(1-z)=\dfrac{\pi^2}{6} - \log(z)\log(1-z),$$ where $\mathrm{Li}_2$ denotes the dilogarithm, $\pi$ denotes pi, and $\log$ denotes the logarithm.

Proof

References