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

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

Answered by witcher3 last updated on 09/Aug/23

$${\int }_0^1\frac{{\ln }^2\left(\mathrm{t}\right)}{\sqrt{1-{\mathrm{t}}^2}}\mathrm{dt}=\mathrm{\Gamma }$$$$\ln \left(\sin \left(\mathrm{x}\right)\right)=-\ln \left(2\right)-\sum _{\mathrm{n}⩾1}\frac{\cos \left(2\mathrm{nx}\right)}{\mathrm{n}}$$$$\mathrm{I}(\mathrm{n},\mathrm{k})={\int }_0^{\frac{\pi }{2}}(\cos \left(2\mathrm{nx}\right).\cos \left(2\mathrm{kx}\right))\mathrm{dx}$$$$\mathrm{I}(\mathrm{n},\mathrm{k})=\frac{1}{2}({\int }_0^{\frac{\pi }{2}}\cos \left(2(\mathrm{n}-\mathrm{k})\mathrm{x}\right)+\cos \left(2(\mathrm{n}+\mathrm{k})\mathrm{x}\right)$$$$\mathrm{n}\ne \mathrm{k}\Rightarrow \mathrm{I}(\mathrm{n},\mathrm{k})=0$$$$\mathrm{I}(\mathrm{n},\mathrm{n})=\frac{1}{2}\left(\frac{\pi }{2}\right)=\frac{\pi }{4}$$$$\mathrm{t}=\sin \left(\mathrm{x}\right)\Rightarrow \mathrm{\Gamma }={\int }_0^{\frac{\pi }{2}}\frac{{\ln }^2\left(\sin \left(\mathrm{x}\right)\right)}{\sqrt{1-{\sin }^2\left(\mathrm{x}\right)}}\cos \left(\mathrm{x}\right)\mathrm{dx}$$$$={\int }_0^{\frac{\pi }{2}}{\ln }^2\left(\sin \left(\mathrm{x}\right)\right)\mathrm{dx}$$$$={\int }_0^{\frac{\pi }{2}}{(-\ln \left(2\right)-\sum _{\mathrm{n}⩾1}\frac{\cos \left(2\mathrm{nx}\right)}{\mathrm{n}})}^2$$$$={\int }_0^{\frac{\pi }{2}}({\ln }^2\left(2\right)+2\ln \left(2\right)\sum _{\mathrm{n}⩾1}\frac{\cos \left(2\mathrm{nx}\right)}{\mathrm{n}}+\sum _{\mathrm{n}⩾1}\sum _{\mathrm{k}⩾1}\frac{\cos \left(2\mathrm{nx}\right)\cos \left(2\mathrm{kx}\right)}{\mathrm{nk}})$$$$=\frac{\pi }{2}{\ln }^2\left(2\right)+\sum _{\mathrm{n}⩾1}\frac{2\ln \left(2\right)}{\mathrm{n}}{\int }_0^{\frac{\pi }{2}}\cos \left(2\mathrm{nx}\right)\mathrm{dx}+\sum _{\mathrm{n}⩾1}\sum _{\mathrm{k}⩾1}\frac{1}{\mathrm{nk}}{\int }_0^{\frac{\pi }{2}}\cos \left(2\mathrm{nx}\right)\cos \left(2\mathrm{kx}\right)\mathrm{dx}$$$$=\frac{\pi {\ln }^2\left(2\right)}{2}+\sum _{\mathrm{n}⩾1}\frac{1}{{\mathrm{n}}^2}.\frac{\pi }{4}=\frac{\pi }{2}({\ln }^2\left(2\right)+\frac{\zeta \left(2\right)}{2})$$$$$$$$$$$$$$

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