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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Revision as of 02:40, 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
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $27.7.3$ (note: the letter $f$ denotes the dilogarithm in this formula)
