Question Number 156617 by mnjuly1970 last updated on 13/Oct/21
Answered by mindispower last updated on 14/Oct/21
$${e}^{−{x}} ={t} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{t}\right)\left(\mathrm{1}−{t}\right)}{\mathrm{1}+{t}}.\frac{{dt}}{{t}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\frac{\mathrm{2}{ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}}+\frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}}\right){dt} \\ $$$$=−{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−{Li}_{\mathrm{2}} \left(−\mathrm{1}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−{ln}^{\mathrm{2}} \left(\mathrm{2}\right)=\frac{\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}}.{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}} \\ $$