find-n-1-1-n-2-by-use-of-integral-0-pi-2-ln-2cos-d-

Question Number 60976 by maxmathsup by imad last updated on 28/May/19

f i n d \sum_{n = 1}^{\infty} \frac{1}{n^{2}} b y u s e o f i n t e g r a l \int_{0}^{\frac{π}{2}} l n (2 c o s θ) d θ .

Commented by maxmathsup by imad last updated on 31/May/19

w e h a v e \int_{0}^{\frac{π}{2}} l n (2 c o s θ) d θ = \int_{0}^{\frac{π}{2}} l n (e^{i θ} + e^{- i θ}) d θ

= \int_{0}^{\frac{π}{2}} l n {e^{i θ} (1 + e^{- 2 i θ})} d θ = \int_{0}^{\frac{π}{2}} i θ d θ + \int_{0}^{\frac{π}{2}} l n (1 + e^{- 2 i θ}) d θ

= i {[\frac{θ^{2}}{2}]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} l n (1 + e^{- 2 i θ}) d θ = i \frac{π^{2}}{8} + \int_{0}^{\frac{π}{2}} l n (1 + e^{- 2 i θ}) d θ

w e h a v e f o r ∣ z ∣⩽ 1 {l n}^{^{'}} (1 + z) = \frac{1}{1 + z} = \sum_{n = 0}^{\infty} {(- 1)}^{n} z^{n} \Rightarrow

l n (1 + z) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} z^{n + 1}}{n + 1} + c (c = 0)

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} z^{n}}{n} \Rightarrow \int_{0}^{\frac{π}{2}} l n (1 + e^{- 2 i θ}) = \int_{0}^{\frac{π}{2}} (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} e^{- 2 i n θ}}{n}) d θ

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{\frac{π}{2}} e^{- 2 i n θ} d θ

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} {[- \frac{1}{2 i n} e^{- 2 i n θ}]}_{0}^{\frac{π}{2}}

= - \frac{1}{2 i} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} {{(- 1)}^{n} - 1}

= \frac{1}{i} \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = - i \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow

\int_{0}^{\frac{π}{2}} l n (2 c o s θ) d θ = i (\frac{π^{2}}{8} - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}) b u t t h i s i n t e g r a l i s

r e a l \Rightarrow \frac{π^{2}}{8} - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = 0 \Rightarrow \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π^{2}}{8}

w e h a v e \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow

\frac{3}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{8} \Rightarrow \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{4}{3} \frac{π^{2}}{8} = \frac{π^{2}}{6} .