Calculer-1-0-ln-2-t-1-t-2-dt-

Question Number 195753 by Erico last updated on 09/Aug/23

$$\mathrm{Calculer}\:\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} {t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$

Answered by witcher3 last updated on 09/Aug/23

∫_0 ^1 ((ln^2 (t))/( (√(1−t^2 ))))dt=Γ ln(sin(x))=−ln(2)−Σ_(n≥1) ((cos(2nx))/n) I(n,k)=∫_0 ^(π/2) (cos(2nx).cos(2kx))dx I(n,k)=(1/2)(∫_0 ^(π/2) cos(2(n−k)x)+cos(2(n+k)x) n≠k⇒I(n,k)=0 I(n,n)=(1/2)((π/2))=(π/4) t=sin(x)⇒Γ=∫_0 ^(π/2) ((ln^2 (sin(x)))/( (√(1−sin^2 (x)))))cos(x)dx =∫_0 ^(π/2) ln^2 (sin(x))dx =∫_0 ^(π/2) (−ln(2)−Σ_(n≥1) ((cos(2nx))/n))^2 =∫_0 ^(π/2) (ln^2 (2)+2ln(2)Σ_(n≥1) ((cos(2nx))/n)+Σ_(n≥1) Σ_(k≥1) ((cos(2nx)cos(2kx))/(nk))) =(π/2)ln^2 (2)+Σ_(n≥1) ((2ln(2))/n)∫_0 ^(π/2) cos(2nx)dx+Σ_(n≥1) Σ_(k≥1) (1/(nk))∫_0 ^(π/2) cos(2nx)cos(2kx)dx =((πln^2 (2))/2)+Σ_(n≥1) (1/n^2 ).(π/4)=(π/2)(ln^2 (2)+((ζ(2))/2))

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{t}\right)}{\:\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}\mathrm{dt}=\Gamma \\ $$$$\mathrm{ln}\left(\mathrm{sin}\left(\mathrm{x}\right)\right)=−\mathrm{ln}\left(\mathrm{2}\right)−\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{cos}\left(\mathrm{2nx}\right)}{\mathrm{n}} \\ $$$$\mathrm{I}\left(\mathrm{n},\mathrm{k}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{cos}\left(\mathrm{2nx}\right).\mathrm{cos}\left(\mathrm{2kx}\right)\right)\mathrm{dx} \\ $$$$\mathrm{I}\left(\mathrm{n},\mathrm{k}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\left(\mathrm{2}\left(\mathrm{n}−\mathrm{k}\right)\mathrm{x}\right)+\mathrm{cos}\left(\mathrm{2}\left(\mathrm{n}+\mathrm{k}\right)\mathrm{x}\right)\right. \\ $$$$\mathrm{n}\neq\mathrm{k}\Rightarrow\mathrm{I}\left(\mathrm{n},\mathrm{k}\right)=\mathrm{0} \\ $$$$\mathrm{I}\left(\mathrm{n},\mathrm{n}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{t}=\mathrm{sin}\left(\mathrm{x}\right)\Rightarrow\Gamma=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{sin}\left(\mathrm{x}\right)\right)}{\:\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}}\mathrm{cos}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{sin}\left(\mathrm{x}\right)\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(−\mathrm{ln}\left(\mathrm{2}\right)−\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{cos}\left(\mathrm{2nx}\right)}{\mathrm{n}}\right)^{\mathrm{2}} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{2ln}\left(\mathrm{2}\right)\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{cos}\left(\mathrm{2nx}\right)}{\mathrm{n}}+\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{cos}\left(\mathrm{2nx}\right)\mathrm{cos}\left(\mathrm{2kx}\right)}{\mathrm{nk}}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{2ln}\left(\mathrm{2}\right)}{\mathrm{n}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\left(\mathrm{2nx}\right)\mathrm{dx}+\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{nk}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\left(\mathrm{2nx}\right)\mathrm{cos}\left(\mathrm{2kx}\right)\mathrm{dx} \\ $$$$=\frac{\pi\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}+\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }.\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{2}}\left(\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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