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Question Number 204244 by pticantor last updated on 09/Feb/24

Answered by pticantor last updated on 09/Feb/24

$$pleasecansomeonehelpmetosolvequestion2-b$$$$forthisexercise?$$

Answered by witcher3 last updated on 10/Feb/24

$$2-\mathrm{b}$$$${\int }_0^{\pi }\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}\frac{1}{\mathrm{n}}\mathrm{f}\left(\frac{\mathrm{k}\pi }{\mathrm{n}}\right)$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\ln ({\mathrm{x}}^2-2\mathrm{xcos}\left(\frac{\mathrm{k}\pi }{\mathrm{n}}\right)+1)$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim ln}\left(\underset{\mathrm{k}=0}{\prod ^{\mathrm{n}-1}}({\mathrm{x}}^2-2\mathrm{xcos}\left(\frac{\mathrm{k}\pi }{\mathrm{n}}\right)+1)\right)=\mathrm{I}$$$${\mathrm{X}}^{2\mathrm{n}}-1=\underset{\mathrm{k}=0}{\prod ^{2\mathrm{n}-1}}(\mathrm{x}-{\mathrm{e}}^{\frac{\mathrm{ik}\pi }{\mathrm{n}}})=\underset{\mathrm{k}=0}{\prod ^{\mathrm{n}-1}}(\mathrm{x}-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{k}\pi }{\mathrm{n}}})\underset{\mathrm{k}=\mathrm{n}}{\prod ^{2\mathrm{n}-1}}(\mathrm{x}-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{k}\pi }{\mathrm{n}}})$$$$\mathrm{k}=\mathrm{n}+\mathrm{k}\mathrm{dans}\mathrm{le}2\mathrm{eme}\mathrm{produits},\mathrm{k}\mathrm{est}\mathrm{muete}$$$$=\underset{\mathrm{k}=0}{\prod ^{\mathrm{n}-1}}(\mathrm{x}-{\mathrm{e}}^{\frac{\mathrm{ik}\pi }{\mathrm{n}}})(\mathrm{x}-{\mathrm{e}}^{-\frac{\mathrm{ik}\pi }{\mathrm{n}}})=\mathrm{\Pi }({\mathrm{x}}^2-2\mathrm{xcos}\left(\frac{\mathrm{k}\pi }{\mathrm{n}}\right)+1)={\mathrm{X}}^{2\mathrm{n}}-1$$$$\Rightarrow \mathrm{I}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\ln ({\mathrm{X}}^{2\mathrm{n}}-1),\mid \mathrm{x}\mid >1$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }.\frac{1}{2}(2\mathrm{nln}(\mid \mathrm{x}\mid )+\ln (1-\frac{1}{{\mathrm{x}}^{2\mathrm{n}}}))=\ln \left({\mathrm{x}}^2\right)$$$$-1<\mathrm{x}<1;\mathrm{x}\ne 0\mathrm{I}=\int \ln \left({\mathrm{x}}^2\right)+\ln (1-\frac{2}{\mathrm{x}}\cos \left(\mathrm{t}\right)+\frac{1}{{\mathrm{x}}^2})\mathrm{dt}$$$$=\pi \ln \left({\mathrm{x}}^2\right)+\ln \left(\frac{1}{{\mathrm{x}}^2}\right)=(\pi -1)\ln \left({\mathrm{x}}^2\right)$$$$\mathrm{x}=0\mathrm{I}=0$$$$$$$$$$$$$$

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