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Question Number 29973 by abdo imad last updated on 14/Feb/18

$$find\sum _{n=1}^{\mathrm{\infty }}\frac{sin\left(n\alpha \right)}{n}{x}^{n}with-1<x<1.$$

Commented byabdo imad last updated on 16/Feb/18

$$letputS\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{sin\left(n\alpha \right)}{n}{x}^{n}duetouniformconvergence$$ $$wehave{S}^{{}^{\prime }}\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}sin\left(n\alpha \right)x^{n-1}=\sum _{n=0}^{\mathrm{\infty }}sin\left((n+1)\alpha \right)x^n$$ $$=Im\left(\sum _{n=0}^{\mathrm{\infty }}{e}^{i(n+1)\alpha }{x}^n\right)=Im\left(e^{i\alpha }\sum _{n=0}^{\mathrm{\infty }}{\left(x{e}^{i\alpha }\right)}^n\right)wehave$$ $$\mid x{e}^{i\alpha }\mid <1\Rightarrow e^{i\alpha }\sum _{n=0}^{\mathrm{\infty }}{\left({xe}^{i\alpha }\right)}^n=e^{i\alpha }\frac{1}{1-{xe}^{i\alpha }}=\frac{1}{e^{-i\alpha }-x}$$ $$=\frac{1}{cos\alpha -isin\alpha -x}=\frac{1}{cos\alpha -x-isin\alpha }$$ $$=\frac{cos\alpha -x+isin\alpha }{{(cos\alpha -x)}^2+{sin}^2\alpha }\Rightarrow Im\left(\mathrm{\Sigma }(...)\right)=\frac{sin\alpha }{{(x-cos\alpha )}^2+{sin}^2\alpha }\Rightarrow$$ $$S\left(x\right)={\int }_0^x\frac{sin\alpha }{{(t-cos\alpha )}^2+{sin}^2\alpha }dt+\lambda but\lambda =S\left(0\right)=0$$ $$thech.t-cos\alpha =sin\alpha ugive$$ $$S\left(x\right)={\int }_{-cotan\alpha }^{\frac{x-cos\alpha }{sin\alpha }}\frac{sin\alpha }{{sin}^2{u}^2+{sin}^2\alpha }sin\alpha du$$ $$={\int }_{-cotan\alpha }^{\frac{x-cos\alpha }{sin\alpha }}\frac{du}{1+u^2}={\left[arctanu\right]}_{-cotan\alpha }^{\frac{x-cos\alpha }{sin\alpha }}$$ $$=artan\left(\frac{x-cos\alpha }{sin\alpha }\right)+artan\left(cotan\alpha \right).$$

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