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Question Number 144701 by mathmax by abdo last updated on 28/Jun/21

$$\mathrm{calculate}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\cos \left(\mathrm{n}\theta \right)}{{\mathrm{n}}^2}$$

Answered by Dwaipayan Shikari last updated on 28/Jun/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\sin \left(n\theta \right)}{n}=\frac{1}{2i}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\left(e^{i\theta }\right)}^n}{n}-\frac{{\left(e^{-i\theta }\right)}^n}{n}$$$$=-\frac{1}{2i}\log \left(\frac{1-e^{i\theta }}{1-e^{-i\theta }}\right)=\frac{\pi -\theta }{2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^{\theta }\frac{\sin \left(n\theta \right)}{n}=\frac{2\pi \theta -{\theta }^2}{4}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\cos \left(0\right)}{n^2}-\frac{\cos \left(n\theta \right)}{n^2}=\frac{2\pi \theta -{\theta }^2}{4}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\cos \left(n\theta \right)}{n^2}=\frac{{\pi }^2}{6}-\frac{\pi \theta }{2}+\frac{{\theta }^2}{4}$$

Answered by mathmax by abdo last updated on 28/Jun/21

$$\phi \left(\theta \right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\cos \left(\mathrm{n}\theta \right)}{{\mathrm{n}}^2}\Rightarrow {\phi }^{{}^{\prime }}\left(\theta \right)=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\sin \left(\mathrm{n}\theta \right)}{\mathrm{n}}$$$$=-\mathrm{Im}\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{in}\theta }}{\mathrm{n}}\right)=-\mathrm{Im}\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\left({\mathrm{e}}^{\mathrm{i}\theta }\right)}^{\mathrm{n}}}{\mathrm{n}}\right)=-\mathrm{Im}\left(\log (1-{\mathrm{e}}^{\mathrm{i}\theta })\right)$$$$\mathrm{we}\mathrm{have}\log (1-{\mathrm{e}}^{\mathrm{i}\theta })=\log (1-\cos \theta -\mathrm{isin}\theta )$$$$=\log ({2\sin }^2\left(\frac{\theta }{2}\right)-2\mathrm{isin}\left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right))=\log (-2\mathrm{isin}\left(\frac{\theta }{2}\right){\mathrm{e}}^{\frac{\mathrm{i}\theta }{2}})$$$$=\log 2+\log (-\mathrm{i})+\log \left(\sin \left(\frac{\theta }{2}\right)\right)+\frac{\mathrm{i}\theta }{2}$$$$=\log \left(2\sin \left(\frac{\theta }{2}\right)\right)-\frac{\mathrm{i}\pi }{2}+\frac{\mathrm{i}\theta }{2}\Rightarrow {\phi }^{{}^{\prime }}\left(\theta \right)=\frac{\pi -\theta }{2}\Rightarrow$$$$\phi \left(\theta \right)=\frac{\pi }{2}\theta -\frac{{\theta }^2}{4}+\mathrm{K}\mathrm{but}\mathrm{K}=\phi \left(0\right)=\frac{{\pi }^2}{6}\Rightarrow$$$$\phi \left(\theta \right)=-\frac{{\theta }^2}{4}+\frac{\pi \theta }{2}+\frac{{\pi }^2}{6}$$

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