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Question Number 135888 by mnjuly1970 last updated on 16/Mar/21

Answered by mindispower last updated on 19/Mar/21

recal χ_2 (x)=((li_2 (x)−li_2 (−x))/2),chi function  we have χ_2 (((1−x)/(1+x)))+χ_2 (x)=(π^2 /8)+((ln(x)ln(((1+x)/(1−x))))/2)  ...(E)  for x=(√2)−1  ⇒((1−x)/(1+x))=((2−(√2))/( (√2)))=(√2)−1  ⇔2χ_2 ((√2)−1)=(π^2 /8)−((ln^2 ((√2)−1))/2)  ln((√2)−1)=−ln(1+(√2))  ⇔2χ_2 ((√2)−1)=li_2 ((√2)−1)−li_2 (1−(√2))=(π^2 /8)−((log^2 (1+(√2)))/2)  for show (E)  just use (d/dx)χ(x)=(1/2)((ln(((1+x)/(1−x))))/x)...

$${recal}\:\chi_{\mathrm{2}} \left({x}\right)=\frac{{li}_{\mathrm{2}} \left({x}\right)−{li}_{\mathrm{2}} \left(−{x}\right)}{\mathrm{2}},{chi}\:{function} \\ $$$${we}\:{have}\:\chi_{\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)+\chi_{\mathrm{2}} \left({x}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}+\frac{{ln}\left({x}\right){ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)}{\mathrm{2}} \\ $$$$...\left({E}\right) \\ $$$${for}\:{x}=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{2}\chi_{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{{ln}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)=−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$$\Leftrightarrow\mathrm{2}\chi_{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)={li}_{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)−{li}_{\mathrm{2}} \left(\mathrm{1}−\sqrt{\mathrm{2}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{{log}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}} \\ $$$${for}\:{show}\:\left({E}\right) \\ $$$${just}\:{use}\:\frac{{d}}{{dx}}\chi\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\frac{{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)}{{x}}... \\ $$

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