Question Number 143429 by mnjuly1970 last updated on 14/Jun/21
$$\:\: \\ $$$$\:\:\:\:\:\:\:…….\:{nice}\:…..{integral}……. \\ $$$$\:\:\:\:{Evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\xi\::=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}−{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:=? \\ $$
Answered by Dwaipayan Shikari last updated on 14/Jun/21
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}−{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({cos}\theta−{sin}\theta\right)−{log}\left({cos}\theta\right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left(\sqrt{\mathrm{2}}\:{cos}\left(\frac{\pi}{\mathrm{4}}+\theta\right)\right)−{log}\left({cos}\theta\right){d}\theta \\ $$$$=\frac{\pi}{\mathrm{4}}{log}\left(\sqrt{\mathrm{2}}\right)+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({cos}\left(\frac{\pi}{\mathrm{4}}+\theta\right)\right)\:{d}\theta−{log}\left({cos}\theta\right){d}\theta \\ $$$$=\frac{\pi}{\mathrm{8}}{log}\left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({cos}\left(\frac{\pi}{\mathrm{2}}−\theta\right)\right)−{log}\left({cos}\theta\right){d}\theta \\ $$$$=\frac{\pi}{\mathrm{8}}{log}\left(\mathrm{2}\right)−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({tan}\theta\right){d}\theta=\frac{\pi}{\mathrm{8}}{log}\left(\mathrm{2}\right)−{G} \\ $$
Commented by mnjuly1970 last updated on 14/Jun/21
$$\:\:{thank}\:{you}\:{so}\:{much}… \\ $$