calculate-n-1-cos-n-n-2-

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

calculate \sum_{n = 1}^{\infty} \frac{\cos (n θ)}{n^{2}}

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

\underset{n = 1}{\sum^{\infty}} \frac{\sin (n θ)}{n} = \frac{1}{2 i} \underset{n = 1}{\sum^{\infty}} \frac{{(e^{i θ})}^{n}}{n} - \frac{{(e^{- i θ})}^{n}}{n}

= - \frac{1}{2 i} \log (\frac{1 - e^{i θ}}{1 - e^{- i θ}}) = \frac{π - θ}{2}

\underset{n = 1}{\sum^{\infty}} \int_{0}^{θ} \frac{\sin (n θ)}{n} = \frac{2 π θ - θ^{2}}{4}

\underset{n = 1}{\sum^{\infty}} \frac{\cos (0)}{n^{2}} - \frac{\cos (n θ)}{n^{2}} = \frac{2 π θ - θ^{2}}{4}

\underset{n = 1}{\sum^{\infty}} \frac{\cos (n θ)}{n^{2}} = \frac{π^{2}}{6} - \frac{π θ}{2} + \frac{θ^{2}}{4}

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

φ (θ) = \sum_{n = 1}^{\infty} \frac{\cos (n θ)}{n^{2}} \Rightarrow φ^{^{'}} (θ) = - \sum_{n = 1}^{\infty} \frac{\sin (n θ)}{n}

= - Im (\sum_{n = 1}^{\infty} \frac{e^{in θ}}{n}) = - Im (\sum_{n = 1}^{\infty} \frac{{(e^{i θ})}^{n}}{n}) = - Im (\log (1 - e^{i θ}))

we have \log (1 - e^{i θ}) = \log (1 - \cos θ - isin θ)

= \log ({2 \sin}^{2} (\frac{θ}{2}) - 2 isin (\frac{θ}{2}) \cos (\frac{θ}{2})) = \log (- 2 isin (\frac{θ}{2}) e^{\frac{i θ}{2}})

= \log 2 + \log (- i) + \log (\sin (\frac{θ}{2})) + \frac{i θ}{2}

= \log (2 \sin (\frac{θ}{2})) - \frac{i π}{2} + \frac{i θ}{2} \Rightarrow φ^{^{'}} (θ) = \frac{π - θ}{2} \Rightarrow

φ (θ) = \frac{π}{2} θ - \frac{θ^{2}}{4} + K but K = φ (0) = \frac{π^{2}}{6} \Rightarrow

φ (θ) = - \frac{θ^{2}}{4} + \frac{π θ}{2} + \frac{π^{2}}{6}