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Question Number 142870 by mnjuly1970 last updated on 06/Jun/21

	Answered by qaz last updated on 06/Jun/21
	$Θ=2∫_0 ^∞ ((ln(1+u)tan^(−1) (1/u))/u)du =2{∫_0 ^1 ((ln(1+u)tan^(−1) (1/u))/u)du+∫_0 ^1 (([ln(1+u)−lnu]tan^(−1) u)/u)du} =2∫_0 ^1 (((π/2)ln(1+u)−lnutan^(−1) u)/u)du =(π^3 /(12))−2∫_0 ^1 ((lnutan^(−1) u)/u)du =(π^3 /(12))−2{(1/2)ln^2 utan^(−1) u∣_0 ^1 −(1/2)∫_0 ^1 ((ln^2 u)/(1+u^2 ))du} =(π^3 /(12))+Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 u^(2n) ln^2 udu =(π^3 /(12))+2Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^3 )) =(π^3 /(12))+(π^3 /(16)) =(7/(48))π^3 −−−−−−−−−−−− ∫_0 ^1 x^a dx=(1/(a+1)) diff a for b times ⇒ ∫_0 ^1 x^a ln^b xdx=(((−1)^b b!)/((a+1)^(b+1) ))$
	$Θ = 2 \int_{0}^{\infty} \frac{\ln (1 + u) \tan^{- 1} \frac{1}{u}}{u} du$ $= 2 {\int_{0}^{1} \frac{\ln (1 + u) \tan^{- 1} \frac{1}{u}}{u} du + \int_{0}^{1} \frac{[\ln (1 + u) - lnu] \tan^{- 1} u}{u} du}$ $= 2 \int_{0}^{1} \frac{\frac{π}{2} \ln (1 + u) - {lnutan}^{- 1} u}{u} du$ $= \frac{π^{3}}{12} - 2 \int_{0}^{1} \frac{{lnutan}^{- 1} u}{u} du$ $= \frac{π^{3}}{12} - 2 {\frac{1}{2} \ln^{2} {utan}^{- 1} u ∣_{0}^{1} - \frac{1}{2} \int_{0}^{1} \frac{\ln^{2} u}{1 + u^{2}} du}$ $= \frac{π^{3}}{12} + \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} u^{2 n} \ln^{2} udu$ $= \frac{π^{3}}{12} + 2 \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}}$ $= \frac{π^{3}}{12} + \frac{π^{3}}{16}$ $= \frac{7}{48} π^{3}$ $- - - - - - - - - - - -$ $\int_{0}^{1} x^{a} dx = \frac{1}{a + 1}$ $diff a for b times \Rightarrow \int_{0}^{1} x^{a} \ln^{b} xdx = \frac{{(- 1)}^{b} b!}{{(a + 1)}^{b + 1}}$
	Commented by Dwaipayan Shikari last updated on 06/Jun/21

	$\frac{π}{2} t a n \frac{π}{2} x = \frac{1}{1 - x} - \frac{1}{1 + x} + \frac{1}{3 - x} - \frac{1}{3 + x} + . . .$ $\frac{π^{2}}{4} {s e c}^{2} (\frac{π}{2} x) = \frac{1}{{(1 - x)}^{2}} + \frac{1}{{(1 + x)}^{2}} + . . .$ $\frac{π^{3}}{8} {s e c}^{2} (\frac{π}{2} x) t a n (\frac{π}{2} x) = \frac{1}{{(1 - x)}^{3}} - \frac{1}{{(1 + x)}^{3}} + . . .$ $O n e c a n g e t d i f f e r e n t v a l u e s b y D . B . W . R . x a t x = \frac{1}{2}$
	Commented by qaz last updated on 06/Jun/21

	$β (1) = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1} = \frac{π}{4}$ $β (2) = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}} = G$ $β (3) = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} = \frac{π^{3}}{32}$ $β (5) = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{5}} = \frac{5 π^{5}}{1536}$ $β (4) = ? β (6) = ? β (7) = ? β (8) = ? β (9) = ? β (10) = ?$
	Commented by mnjuly1970 last updated on 06/Jun/21

	$t h a n k y o u s o m u c h . . . m r q a z . .$
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