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Question Number 139696 by mnjuly1970 last updated on 30/Apr/21

Answered by Dwaipayan Shikari last updated on 30/Apr/21

$$\vartheta \left(\theta \right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}sin\left(n\theta \right)}{n}$$$$\vartheta \left(\theta \right)=\frac{1}{2i}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}{e}^{in\theta }}{n}-\frac{1}{2i}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}{e}^{-in\theta }}{n}$$$$=\frac{1}{2i}log(1+e^{i\theta })-\frac{1}{2i}log(1+e^{-i\theta })=\frac{1}{2i}log\left(\frac{1+e^{i\theta }}{1+e^{-i\theta }}\right)=\frac{1}{2i}log\left(e^{i\theta }\right)$$$$=\frac{\theta }{2}$$$${\int }_0^{\theta }\vartheta \left(\theta \right)d\theta ={\int }_0^{\theta }\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}sin\left(n\theta \right)}{n}d\theta$$$$\Rightarrow \frac{{\theta }^2}{4}=-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}cos\left(n\theta \right)}{n^2}+\frac{{\pi }^2}{12}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}cos\left(n\theta \right)}{n^2}=\frac{{\pi }^2}{12}-\frac{{\theta }^2}{4}$$

Commented by mnjuly1970 last updated on 30/Apr/21

$$thanksalotmrpayan...$$

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