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Question Number 128841 by mnjuly1970 last updated on 10/Jan/21
$...nice calculus ... Evaluation of :: Φ= ∫_0 ^( 1) ln(x).arctan(x)dx solution:: note 1:: Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n−1))=arctan(1)=(π/4) note 2 :: Σ_(n=1) ^∞ (((−1)^(n−1) )/n) =ln(1+1)=ln(2) note 3:: Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =_(eta function) ^(Drichlet) η(2)=(π^2 /(12)) ......start....... Φ=∫_0 ^( 1) {ln(x)Σ_(n=1 ) ^∞ (((−1)^(n−1) )/(2n−1)))x^(2n−1) }dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/((2n−1)))∫_0 ^( 1) ln(x)x^(2n−1) dx =Σ_(n=1 ) ^∞ (((−1)^(n−1) )/((2n−1))){[(x^(2n) /(2n))ln(x)]_0 ^1 −(1/(2n))∫_0 ^( 1) x^(2n−1) dx} =(1/4)Σ_(n=1) ^∞ (((−1)^n )/((2n−1)n^2 ))=(1/4)Σ_(n=1) ^∞ (−1)^n [((2n−(2n−1))/((2n−1)n^2 ))] =(1/2)Σ_(n=1) ^∞ (((−1)^n )/((2n−1)n)) −(1/4)Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(1/2)Σ_(n=1) ^∞ ((2(−1)^n )/(2n−1))−(1/2)Σ_(n=1) ^∞ (((−1)^n )/n)+(1/4)η(2) =((−π)/4)+(1/2)ln(2)+(π^2 /(48)) ... ...Φ=(π^2 /(48)) −(π/4) +(1/2)ln(2) ... ...m.n.july.1970...$
$. . . n i c e c a l c u l u s . . .$ $E v a l u a t i o n o f :: Φ = \int_{0}^{1} l n (x) . a r c t a n (x) d x$ $s o l u t i o n ::$ $n o t e 1 :: \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{2 n - 1} = a r c t a n (1) = \frac{π}{4}$ $n o t e 2 :: \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n} = l n (1 + 1) = l n (2)$ $n o t e 3 :: \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n^{2}} \underset{e t a f u n c t i o n}{\overset{D r i c h l e t}{=}} η (2) = \frac{π^{2}}{12}$ $. . . . . . s t a r t . . . . . . .$ $Φ = \int_{0}^{1} {l n (x) \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{2 n - 1)} x^{2 n - 1}} d x$ $= \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{(2 n - 1)} \int_{0}^{1} l n (x) x^{2 n - 1} d x$ $= \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{(2 n - 1)} {{[\frac{x^{2 n}}{2 n} l n (x)]}_{0}^{1} - \frac{1}{2 n} \int_{0}^{1} x^{2 n - 1} d x}$ $= \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n - 1) n^{2}} = \frac{1}{4} \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} [\frac{2 n - (2 n - 1)}{(2 n - 1) n^{2}}]$ $= \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n - 1) n} - \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}}$ $= \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{2 {(- 1)}^{n}}{2 n - 1} - \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n} + \frac{1}{4} η (2)$ $= \frac{- π}{4} + \frac{1}{2} l n (2) + \frac{π^{2}}{48} . . .$ $. . . Φ = \frac{π^{2}}{48} - \frac{π}{4} + \frac{1}{2} l n (2) . . .$ $. . . m . n . j u l y . 1970 . . .$
Commented by Dwaipayan Shikari last updated on 10/Jan/21

$\int_{0}^{1} l o g (x) \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1} x^{2 n + 1} d x = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1} \int_{0}^{1} x^{2 n + 1} l o g (x) d x l o g (x) = t$ $= \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1} \int_{0}^{\infty} e^{- (2 n + 2) t} t d t$ $= \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n + 1) {(2 n + 2)}^{2}} = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n + 2) (2 n + 1)} - \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 2)}^{2}}$ $= \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n + 2)} - \frac{{(- 1)}^{n}}{2 n + 1} - \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 2)}^{2}}$ $= l o g (\sqrt{2}) - \frac{π}{4} - \frac{π^{2}}{48}$
Commented by mnjuly1970 last updated on 11/Jan/21

$t h a n k y o u m r d w a i p a y a n . . .$ $g r a t e f u l .$
	Answered by mindispower last updated on 10/Jan/21

	$1) b y p a r t = {[l n (x) . (x a r c t a n (x) - \frac{l n (1 + x^{2})}{2})]}_{0}^{1}$ $- \int_{0}^{1} a r c t a n (x) d x + \int_{0}^{1} \frac{l n (1 + x^{2})}{2 x} d x$ $\int_{0}^{1} a r c t a n (x) = \frac{π}{4} - \frac{l n (2)}{2}$ $\int_{0}^{1} \frac{l n (1 + x^{2})}{2 x} d x = \int_{0}^{1} \frac{l n (1 + x^{2})}{4 x^{2}} . d (x^{2})$ $= \frac{1}{4} \int_{0}^{1} \frac{l n (1 + t)}{t} d t = \frac{1}{4} \int_{0}^{1} \sum_{n ⩾ 1} \frac{{(- 1)}^{n - 1} t^{n - 1}}{n}$ $= \frac{1}{4} \sum_{n ⩾ 1} \frac{{(- 1)}^{n - 1}}{n^{2}} = - \frac{1}{16} ζ (2) + \frac{1}{4} (\frac{3}{4}) ζ (2)$ $= \frac{ζ (2)}{8} = \frac{π^{2}}{48} . . . w e g e t \frac{π^{2}}{48} - \frac{π}{4} + \frac{l n (2)}{2}$ $2) l n (x) = \partial_{y} x^{y} ∣_{y = 0}$ $f (y) = \int_{0}^{1} x^{y} a r c t a n (x) d x$ $b y p a r t = \frac{π}{4 (y + 1)} - \frac{1}{y + 1} \int_{0}^{1} \frac{x^{y + 1}}{1 + x^{2}}$ $= \frac{π}{4 (y + 1)} - \frac{1}{y + 1} \int_{0}^{1} \frac{x^{y + 1} - x^{y + 3}}{1 - x^{4}} d x$ $x^{4} = t$ $= \frac{π}{4 (y + 1)} - \frac{1}{y + 1} \int_{0}^{1} \frac{t^{\frac{y + 1}{4}} - t^{\frac{y + 3}{4}}}{4 (1 - t)} . t^{- \frac{3}{4}} d t$ $= \frac{1}{y + 1} (\frac{π}{4} - \int_{0}^{1} \frac{t^{\frac{y - 2}{4}} - t^{\frac{y - 1}{4}}}{1 - t} d t)$ $Ψ (s + 1) = - γ + \int_{0}^{1} \frac{1 - x^{s}}{1 - x} d x$ $w e g e t \frac{1}{y + 1} (\frac{π}{4} - Ψ (\frac{y + 3}{4}) + Ψ (\frac{y + 2}{4})) = f (y)$ $f^{'} (y) = - \frac{1}{{(y + 1)}^{2}} (\frac{π}{4} + Ψ (\frac{y + 2}{4}) - Ψ (\frac{y + 3}{4})) + (\frac{1}{4} (Ψ^{'} (\frac{y + 2}{4}) - Ψ^{'} (\frac{y + 3}{4})) . \frac{1}{y + 1}$
	Commented by mnjuly1970 last updated on 11/Jan/21

	$t h a n k s a l o t s i r p o w e r$ $p o w e r f u l a n d m i g h t y a s a l w a y s$ $. p e a c e b e u p o n y o u . . .$
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