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Question Number 116965 by Dwaipayan Shikari last updated on 08/Oct/20

$$\underset{k=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{k+1}\frac{sin\left(k\theta \right)}{k\theta }$$

Answered by Bird last updated on 08/Oct/20

$$letg\left(u\right)=\sum _{k=1}^{\mathrm{\infty }}{(-1)}^{k+1}\frac{sin\left(kx\right)}{kx}{u}^{k}with\mid u\mid <1$$$$\Rightarrow g^{{}^{\prime }}\left(u\right)=\frac{1}{x}\sum _{k=1}^{\mathrm{\infty }}{(-1)}^{k+1}sin\left(kx\right)u^{k-1}$$$$=\frac{1}{x}\sum _{k=0}^{\mathrm{\infty }}{(-1)}^{k}sin\left((k+1)x\right)u^k$$$$=\frac{1}{x}Im\left(\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k{e}^{i)k+1)x}{u}^k\right)$$$$but\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k{e}^{ix}{e}^{ikx}{u}^k$$$$=e^{ix}\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k{\left(u{e}^{ix}\right)}^k$$$$=e^{ix}\times \frac{1}{1+{ue}^{ix}}$$$$=\frac{cosx+isinx}{1+ucosx+iusinx}$$$$=\frac{(cosx+isinx)(1+ucosx-iusinx)}{{(1+ucosx)}^2+u^2{sin}^{2}x}$$$$=\frac{cosx(1+ucosx)-iucosxsinx+isinx(1+ucosx)+{usin}^{2}x}{1+2ucosx+u^2}$$$$\Rightarrow Im(\mathrm{\Sigma }...)=\frac{sinx}{u^2+2ucosx+1}$$$$\Rightarrow g^{{}^{\prime }}\left(u\right)=\frac{sinx}{x(u^2+2ucosx+1)}$$$$\Rightarrow g\left(u\right)=\frac{sinx}{x}\int \frac{du}{u^2+2ucosx+1}$$$$but\int \frac{du}{u^2+2ucosx+1}$$$$=\int \frac{du}{u^2+2ucosx+{cos}^{2}x+{sin}^{2}x}$$$$=\int \frac{du}{{(u+cosx)}^2+{sin}^{2}x}$$$${=}_{u+cosx=sinxz}\int \frac{sinxdz}{{sin}^{2}x(1+z^2)}$$$$=\frac{1}{sinx}arctan\left(\frac{u+cosx}{sinx}\right)+C\Rightarrow$$$$g\left(u\right)=\frac{1}{x}arctan\left(\frac{u+cosx}{sinx}\right)+C$$$$u=1\Rightarrow S\left(x\right)=\frac{1}{x}arctan\left(\frac{1+cosx}{sinx}\right)+C$$$$=\frac{1}{x}arctan\left(\frac{2{cos}^2\left(\frac{x}{2}\right)}{2cos\left(\frac{x}{2}\right)sin\left(\frac{x}{2}\right)}\right)+C$$$$=\frac{1}{x}arctsn\left(\frac{1}{tan\left(\frac{x}{2}\right)}\right)+C$$$$=\frac{1}{x}(\frac{\pi }{2}-\frac{x}{2})+C$$$$=\frac{\pi }{2x}-\frac{1}{2}+C$$$$x=\pi \Rightarrow S\left(\pi \right)=0=C\Rightarrow$$$$S\left(x\right)=\frac{\pi -x}{2x}$$

Commented by Dwaipayan Shikari last updated on 08/Oct/20

$$Thankingyou$$

Commented by Bird last updated on 08/Oct/20

$$youarewelcome$$

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