View source for Li 2(z)=-Li 2(1/z)-(1/2)(log z)^2 + i pi log(z) + pi^2/3

==Theorem==
The following formula holds:
$$\mathrm{Li}_2(z)=-\mathrm{Li}_2 \left( \dfrac{1}{z} \right) - \dfrac{\log(z)^2}{2} + i \pi \log(z) + \dfrac{\pi^2}{3},$$
where $\mathrm{Li}_2$ denotes the [[dilogarithm]], $\log$ denotes the [[logarithm]], $i$ denotes the [[imaginary number]], and $\pi$ denotes [[pi]].

==Proof==

==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=Li2(z)=zPhi(z,2,1)|next=findme}}: $\S 1.11.1 (23)$

[[Category:Theorem]]
[[Category:Unproven]]