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Question Number 96196 by mathmax by abdo last updated on 30/May/20

$$\mathrm{let}\mathrm{g}\left(\mathrm{x}\right)=\ln \left(\mathrm{sinx}\right)\mathrm{developp}\mathrm{g}\mathrm{at}\mathrm{fourier}\mathrm{serie}$$

Answered by mathmax by abdo last updated on 31/May/20

$$\mathrm{g}\left(\mathrm{x}\right)=\ln \left(\mathrm{sinx}\right)=\ln \left(\frac{{\mathrm{e}}^{\mathrm{ix}}-{\mathrm{e}}^{-\mathrm{ix}}}{2\mathrm{i}}\right)=\ln ({\mathrm{e}}^{\mathrm{ix}}-{\mathrm{e}}^{-\mathrm{ix}})-\ln \left(2\mathrm{i}\right)$$$$=\ln \left({\mathrm{e}}^{\mathrm{ix}}(1-{\mathrm{e}}^{-2\mathrm{ix}})\right)-\ln \left(2\right)-\ln \left({\mathrm{e}}^{\frac{\mathrm{i}\pi }{2}}\right)=\mathrm{ix}+\ln (1-{\mathrm{e}}^{-2\mathrm{ix}})-\ln 2-\frac{\mathrm{i}\pi }{2}$$$$\mathrm{we}\mathrm{have}{\ln }^{{}^{\prime }}(1-\mathrm{u})=-\frac{1}{1-\mathrm{u}}=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{u}}^{\mathrm{n}}\Rightarrow \ln (1-\mathrm{u})=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}(\mathrm{c}=0)$$$$=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\mathrm{n}}}{\mathrm{n}}\Rightarrow \ln (1-{\mathrm{e}}^{-2\mathrm{ix}})=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\left({\mathrm{e}}^{-2\mathrm{ix}}\right)}^{\mathrm{n}}}{\mathrm{n}}=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-2\mathrm{inx}}}{\mathrm{n}}$$$$\Rightarrow \mathrm{g}\left(\mathrm{x}\right)=\mathrm{i}(\mathrm{x}-\frac{\pi }{2})-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-2\mathrm{inx}}}{\mathrm{n}}-\ln 2$$$$=\mathrm{i}(\mathrm{x}-\frac{\pi }{2})-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\cos \left(2\mathrm{nx}\right)}{\mathrm{n}}+\mathrm{i}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\sin \left(2\mathrm{nx}\right)}{\mathrm{n}}-\ln 2$$$$\mathrm{we}\mathrm{have}\ln \left(\mathrm{sinx}\right)\mathrm{real}\Rightarrow \ln \left(\mathrm{sinx}\right)=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\cos \left(2\mathrm{nx}\right)}{\mathrm{n}}-\ln 2\mathrm{and}$$$$\mathrm{x}-\frac{\pi }{2}+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\sin \left(2\mathrm{nx}\right)}{\mathrm{n}}=0$$

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