Question Number 150718 by mathdanisur last updated on 14/Aug/21
$$\mathrm{Calculate}: \\ $$$$\mathrm{Li}_{\mathrm{2}} \left(\boldsymbol{\mathrm{z}}\right)\:+\:\mathrm{Li}_{\mathrm{2}} \left(\mathrm{1}\:-\:\boldsymbol{\mathrm{z}}\right)\:=\:? \\ $$
Answered by Olaf_Thorendsen last updated on 14/Aug/21
![Li(z) = −∫_0 ^z ((ln(1−u))/u) du (1) Li′(z) = −((ln(1−z))/z) Li′(1−z) = −((lnz)/(1−z)) Li′(z)−Li′(1−z) = −[((ln(1−z))/z)−((lnz)/(1−z))] Li′(z)−Li′(1−z) = −(d/dz)[lnz.ln(1−z)] ⇒ Li(z)+Li(1−z) = −lnz.ln(1−z)+C (2) With (1) : Li(z) = Σ_(k=1) ^∞ (z^k /k^2 ) Li(0) = Σ_(k=1) ^∞ (0^k /k^2 ) = 0 Li(1) = Σ_(k=1) ^∞ (1^k /k^2 ) = Σ_(k=1) ^∞ (1/k^2 ) = (π^2 /6) (2) : Li(0^+ )+Li(1−0^+ ) = −ln0^+ .ln(1−0^+ )+C 0+(π^2 /6) = 0+C ⇒ C = (π^2 /6) Li(z)+Li(1−z) = −lnz.ln(1−z)+(π^2 /6)](Q150725.png)
$$\mathrm{Li}\left({z}\right)\:=\:−\int_{\mathrm{0}} ^{{z}} \frac{\mathrm{ln}\left(\mathrm{1}−{u}\right)}{{u}}\:{du}\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Li}'\left({z}\right)\:=\:−\frac{\mathrm{ln}\left(\mathrm{1}−{z}\right)}{{z}} \\ $$$$\mathrm{Li}'\left(\mathrm{1}−{z}\right)\:=\:−\frac{\mathrm{ln}{z}}{\mathrm{1}−{z}} \\ $$$$\mathrm{Li}'\left({z}\right)−\mathrm{Li}'\left(\mathrm{1}−{z}\right)\:=\:−\left[\frac{\mathrm{ln}\left(\mathrm{1}−{z}\right)}{{z}}−\frac{\mathrm{ln}{z}}{\mathrm{1}−{z}}\right] \\ $$$$\mathrm{Li}'\left({z}\right)−\mathrm{Li}'\left(\mathrm{1}−{z}\right)\:=\:−\frac{{d}}{{dz}}\left[\mathrm{ln}{z}.\mathrm{ln}\left(\mathrm{1}−{z}\right)\right] \\ $$$$\Rightarrow\:\mathrm{Li}\left({z}\right)+\mathrm{Li}\left(\mathrm{1}−{z}\right)\:=\:−\mathrm{ln}{z}.\mathrm{ln}\left(\mathrm{1}−{z}\right)+\mathrm{C}\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{With}\:\left(\mathrm{1}\right)\::\:\mathrm{Li}\left({z}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{z}^{{k}} }{{k}^{\mathrm{2}} } \\ $$$$\mathrm{Li}\left(\mathrm{0}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{0}^{{k}} }{{k}^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$$\mathrm{Li}\left(\mathrm{1}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}^{{k}} }{{k}^{\mathrm{2}} }\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\left(\mathrm{2}\right)\::\:\mathrm{Li}\left(\mathrm{0}^{+} \right)+\mathrm{Li}\left(\mathrm{1}−\mathrm{0}^{+} \right)\:=\:−\mathrm{ln0}^{+} .\mathrm{ln}\left(\mathrm{1}−\mathrm{0}^{+} \right)+\mathrm{C} \\ $$$$\mathrm{0}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\:\mathrm{0}+\mathrm{C}\:\Rightarrow\:\mathrm{C}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$ \\ $$$$\mathrm{Li}\left({z}\right)+\mathrm{Li}\left(\mathrm{1}−{z}\right)\:=\:−\mathrm{ln}{z}.\mathrm{ln}\left(\mathrm{1}−{z}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$
Commented by Olaf_Thorendsen last updated on 14/Aug/21
$$\left(\mathrm{here}\:\mathrm{Li}\:\mathrm{is}\:\mathrm{Li}_{\mathrm{2}} \:\mathrm{of}\:\mathrm{course}\right) \\ $$
Commented by mathdanisur last updated on 15/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser}\:\mathrm{cool} \\ $$