find Σ_(n=1) ^∞ (1/n^2 ) by use of integral ∫


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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
$we have ∫_0 ^(π/2) ln(2cosθ)dθ =∫_0 ^(π/2) ln(e^(iθ) +e^(−iθ) )dθ =∫_0 ^(π/2) ln{e^(iθ) (1+e^(−2iθ) )}dθ =∫_0 ^(π/2) iθ dθ +∫_0 ^(π/2) ln(1+e^(−2iθ) )dθ =i[(θ^2 /2)]_0 ^(π/2) +∫_0 ^(π/2) ln(1+e^(−2iθ) )dθ =i(π^2 /8) +∫_0 ^(π/2) ln(1+e^(−2iθ) )dθ we have for ∣z∣≤1 ln^′ (1+z) =(1/(1+z)) =Σ_(n=0) ^∞ (−1)^n z^n ⇒ ln(1+z) =Σ_(n=0) ^∞ (((−1)^n z^(n+1) )/(n+1)) +c (c=0) =Σ_(n=1) ^∞ (((−1)^(n−1) z^n )/n) ⇒∫_0 ^(π/2) ln(1+e^(−2iθ) ) =∫_0 ^(π/2) (Σ_(n=1) ^∞ (((−1)^(n−1) e^(−2inθ) )/n))dθ =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) ∫_0 ^(π/2) e^(−2inθ) dθ =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) [−(1/(2in)) e^(−2inθ) ]_0 ^(π/2) =−(1/(2i)) Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ){ (−1)^n −1} =(1/i) Σ_(n=0) ^∞ (1/((2n+1)^2 )) =−i Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ ∫_0 ^(π/2) ln(2cosθ)dθ =i((π^2 /8) − Σ_(n=0) ^∞ (1/((2n+1)^2 ))) but this integral is real ⇒(π^2 /8) −Σ_(n=0) ^∞ (1/((2n+1)^2 )) =0 ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(π^2 /8) we have Σ_(n=1) ^∞ (1/n^2 ) =(1/4) Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ (3/4) Σ_(n=1) ^∞ (1/n^2 ) =(π^2 /8) ⇒ Σ_(n=1) ^∞ (1/n^2 ) =(4/3)(π^2 /8) =(π^2 /6) .$
$w e h a v e \int_{0}^{\frac{π}{2}} l n (2 c o s θ) d θ = \int_{0}^{\frac{π}{2}} l n (e^{i θ} + e^{- i θ}) d θ$ $= \int_{0}^{\frac{π}{2}} l n {e^{i θ} (1 + e^{- 2 i θ})} d θ = \int_{0}^{\frac{π}{2}} i θ d θ + \int_{0}^{\frac{π}{2}} l n (1 + e^{- 2 i θ}) d θ$ $= i {[\frac{θ^{2}}{2}]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} l n (1 + e^{- 2 i θ}) d θ = i \frac{π^{2}}{8} + \int_{0}^{\frac{π}{2}} l n (1 + e^{- 2 i θ}) d θ$ $w e h a v e f o r ∣ z ∣⩽ 1 {l n}^{^{'}} (1 + z) = \frac{1}{1 + z} = \sum_{n = 0}^{\infty} {(- 1)}^{n} z^{n} \Rightarrow$ $l n (1 + z) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} z^{n + 1}}{n + 1} + c (c = 0)$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} z^{n}}{n} \Rightarrow \int_{0}^{\frac{π}{2}} l n (1 + e^{- 2 i θ}) = \int_{0}^{\frac{π}{2}} (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} e^{- 2 i n θ}}{n}) d θ$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{\frac{π}{2}} e^{- 2 i n θ} d θ$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} {[- \frac{1}{2 i n} e^{- 2 i n θ}]}_{0}^{\frac{π}{2}}$ $= - \frac{1}{2 i} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} {{(- 1)}^{n} - 1}$ $= \frac{1}{i} \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = - i \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow$ $\int_{0}^{\frac{π}{2}} l n (2 c o s θ) d θ = i (\frac{π^{2}}{8} - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}) b u t t h i s i n t e g r a l i s$ $r e a l \Rightarrow \frac{π^{2}}{8} - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = 0 \Rightarrow \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π^{2}}{8}$ $w e h a v e \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow$ $\frac{3}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{8} \Rightarrow \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{4}{3} \frac{π^{2}}{8} = \frac{π^{2}}{6} .$
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