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Question Number 123977 by mnjuly1970 last updated on 29/Nov/20

	Answered by mathmax by abdo last updated on 29/Nov/20
	$let A =∫_0 ^(π/2) ln(cosx)dx we have A =−(π/2)ln(2) also A =∫_0 ^(π/2) ln(((e^(ix) +e^(−ix) )/2))dx =∫_0 ^(π/2) ln(e^(ix) +e^(−ix) )dx−(π/2)ln(2) ⇒ ∫_0 ^(π/2) ln(e^(ix) +e^(−ix) )dx=0 ⇒∫_0 ^(π/2) ln(e^(ix) (1+e^(−2ix) ))=0 ⇒ [i(x^2 /2)]_0 ^(π/2) +∫_0 ^(π/2) ln(1+e^(−2ix) )dx we have ln^′ (1+u)=(1/(1+u))=Σ_(n=0) ^∞ (−1)^n u^n ⇒ln(1+u)=Σ_(n=0) ^∞ (((−1)^n u^(n+1) )/(n+1)) =Σ_(n=1) ^∞ (((−1)^(n−1) u^n )/n) ⇒ ∫_0 ^(π/2) ln(1+e^(−2ix) )=Σ_(n=1) ^∞ (((−1)^(n−1) )/n)∫_0 ^(π/2) e^(−2inx) dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)×(1/(−2in))[e^(−2inx) ]_0 ^(π/2) =(i/2) Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ){(−1)^n −1} =−iΣ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ i(π^2 /8)−iΣ_(n=0) ^∞ (1/((2n+1)^2 ))=0 ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 ))=(π^2 /8) we have Σ_(n=1) ^∞ (1/n^2 )=Σ_(n=1) ^∞ (1/(4n^2 ))+Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ (3/4)Σ_(n=1) ^∞ (1/n^2 )=(π^2 /8) ⇒Σ_(n=1) ^∞ (1/n^2 )=(4/3)×(π^2 /8) =(π^2 /6)$
	$let A = \int_{0}^{\frac{π}{2}} \ln (cosx) dx we have A = - \frac{π}{2} \ln (2) also$ $A = \int_{0}^{\frac{π}{2}} \ln (\frac{e^{ix} + e^{- ix}}{2}) dx = \int_{0}^{\frac{π}{2}} \ln (e^{ix} + e^{- ix}) dx - \frac{π}{2} \ln (2) \Rightarrow$ $\int_{0}^{\frac{π}{2}} \ln (e^{ix} + e^{- ix}) dx = 0 \Rightarrow \int_{0}^{\frac{π}{2}} \ln (e^{ix} (1 + e^{- 2 ix})) = 0 \Rightarrow$ ${[i \frac{x^{2}}{2}]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} \ln (1 + e^{- 2 ix}) dx we have \ln^{^{'}} (1 + u) = \frac{1}{1 + u} = \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{n}$ $\Rightarrow \ln (1 + u) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} u^{n + 1}}{n + 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} u^{n}}{n} \Rightarrow$ $\int_{0}^{\frac{π}{2}} \ln (1 + e^{- 2 ix}) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{\frac{π}{2}} e^{- 2 inx} dx$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \times \frac{1}{- 2 in} {[e^{- 2 inx}]}_{0}^{\frac{π}{2}}$ $= \frac{i}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} {{(- 1)}^{n} - 1} = - i \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow$ $i \frac{π^{2}}{8} - i \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = 0 \Rightarrow \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π^{2}}{8}$ $we have \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \sum_{n = 1}^{\infty} \frac{1}{{4 n}^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow$ $\frac{3}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{8} \Rightarrow \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{4}{3} \times \frac{π^{2}}{8} = \frac{π^{2}}{6}$
	Commented by mnjuly1970 last updated on 29/Nov/20

	$t h a n k y o u s o m u c h s i r m a x$ $v e r y n i c e a s a l w a y s . . .$
	Commented by mathmax by abdo last updated on 30/Nov/20

	$you are welcome$
	Answered by Dwaipayan Shikari last updated on 30/Nov/20

	$\int_{0}^{\frac{π}{2}} l o g (2 c o s x) d x$ $= \frac{π}{2} l o g (2) - \frac{π}{2} l o g (2) = 0$ $\sum^{\infty} \frac{1}{n^{2}} = \frac{1}{1^{2}} + \frac{1}{2^{2}} + . . . .$ $\frac{s i n x}{x} = (1 - \frac{x^{2}}{π^{2}}) (1 - \frac{x^{2}}{4 π^{2}}) (1 - \frac{x^{2}}{9 π^{2}}) . . .$ $1 - \frac{x^{2}}{3!} + . . . = 1 - \frac{x^{2}}{π^{2}} - \frac{x^{2}}{4 π^{2}} + \frac{x^{4}}{4 π^{4}} - \frac{x^{2}}{9 π^{2}} + . . .$ $\frac{x^{2}}{3!} = \frac{x^{2}}{π^{2}} + \frac{x^{2}}{4 π^{2}} + \frac{x^{2}}{9 π^{2}} + . . .$ $1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . = \frac{π^{2}}{6}$
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