Calculate-Li-2-z-Li-2-1-z-

Question Number 150718 by mathdanisur last updated on 14/Aug/21

Calculate :

{Li}_{2} (z) + {Li}_{2} (1 - z) = ?

Answered by Olaf_Thorendsen last updated on 14/Aug/21

Li (z) = - \int_{0}^{z} \frac{\ln (1 - u)}{u} d u (1)

{Li}^{'} (z) = - \frac{\ln (1 - z)}{z}

{Li}^{'} (1 - z) = - \frac{\ln z}{1 - z}

{Li}^{'} (z) - {Li}^{'} (1 - z) = - [\frac{\ln (1 - z)}{z} - \frac{\ln z}{1 - z}]

{Li}^{'} (z) - {Li}^{'} (1 - z) = - \frac{d}{d z} [\ln z . \ln (1 - z)]

\Rightarrow Li (z) + Li (1 - z) = - \ln z . \ln (1 - z) + C (2)

With (1) : Li (z) = \underset{k = 1}{\sum^{\infty}} \frac{z^{k}}{k^{2}}

Li (0) = \underset{k = 1}{\sum^{\infty}} \frac{0^{k}}{k^{2}} = 0

Li (1) = \underset{k = 1}{\sum^{\infty}} \frac{1^{k}}{k^{2}} = \underset{k = 1}{\sum^{\infty}} \frac{1}{k^{2}} = \frac{π^{2}}{6}

(2) : Li (0^{+}) + Li (1 - 0^{+}) = - {\ln 0}^{+} . \ln (1 - 0^{+}) + C

0 + \frac{π^{2}}{6} = 0 + C \Rightarrow C = \frac{π^{2}}{6}

Li (z) + Li (1 - z) = - \ln z . \ln (1 - z) + \frac{π^{2}}{6}

Commented by Olaf_Thorendsen last updated on 14/Aug/21

(here Li is {Li}_{2} of course)

Commented by mathdanisur last updated on 15/Aug/21

Thank you Ser cool