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Question Number 143429 by mnjuly1970 last updated on 14/Jun/21

           ....... nice .....integral.......      Evaluate ::          ξ := ∫_0 ^( 1)  ((ln(1−t))/(1+t^2 )) dt =?

$$\:\: \\ $$$$\:\:\:\:\:\:\:.......\:{nice}\:.....{integral}....... \\ $$$$\:\:\:\:{Evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\xi\::=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}−{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:=? \\ $$

Answered by Dwaipayan Shikari last updated on 14/Jun/21

∫_0 ^1 ((log(1−t))/(1+t^2 ))dt  =∫_0 ^(π/4) log(cosθ−sinθ)−log(cosθ)dθ  =∫_0 ^(π/4) log((√2) cos((π/4)+θ))−log(cosθ)dθ  =(π/4)log((√2))+∫_0 ^(π/4) log(cos((π/4)+θ)) dθ−log(cosθ)dθ  =(π/8)log(2)+∫_0 ^(π/4) log(cos((π/2)−θ))−log(cosθ)dθ  =(π/8)log(2)−∫_0 ^(π/4) log(tanθ)dθ=(π/8)log(2)−G

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}−{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({cos}\theta−{sin}\theta\right)−{log}\left({cos}\theta\right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left(\sqrt{\mathrm{2}}\:{cos}\left(\frac{\pi}{\mathrm{4}}+\theta\right)\right)−{log}\left({cos}\theta\right){d}\theta \\ $$$$=\frac{\pi}{\mathrm{4}}{log}\left(\sqrt{\mathrm{2}}\right)+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({cos}\left(\frac{\pi}{\mathrm{4}}+\theta\right)\right)\:{d}\theta−{log}\left({cos}\theta\right){d}\theta \\ $$$$=\frac{\pi}{\mathrm{8}}{log}\left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({cos}\left(\frac{\pi}{\mathrm{2}}−\theta\right)\right)−{log}\left({cos}\theta\right){d}\theta \\ $$$$=\frac{\pi}{\mathrm{8}}{log}\left(\mathrm{2}\right)−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({tan}\theta\right){d}\theta=\frac{\pi}{\mathrm{8}}{log}\left(\mathrm{2}\right)−{G} \\ $$

Commented by mnjuly1970 last updated on 14/Jun/21

  thank you so much...

$$\:\:{thank}\:{you}\:{so}\:{much}... \\ $$

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