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Question Number 136921 by mnjuly1970 last updated on 27/Mar/21
$evaluation of :: 𝛗=∫_0 ^( 1) ((xln(1+x))/(1+x^2 ))dx solution: 𝛗=^(I.B.P ) [(1/2)ln(1+x^2 )ln(1+x)]_0 ^1 −(1/2){∫_0 ^( 1) ((ln(1+x^2 ))/(1+x))dx=𝚽} 𝛗=(1/2)ln^2 (2)−(1/2) 𝚽 ........✓ 𝚽=∫_0 ^( 1) ((ln(1+x^2 ))/(1+x))dx =??? h(a)=∫_0 ^( 1) ((ln(1+ax^2 ))/(1+x))dx h′(a)=∫_0 ^( 1) (∂/∂_a )(((ln(1+ax^2 ))/(1+x)))dx h′(a)=∫_0 ^( 1) (x^2 /((1+x)(1+ax^2 )))dx h′(a)=(1/(1+a))∫_0 ^( 1) (1/(1+x))dx+(1/(1+a))∫_0 ^( 1) (x/(1+ax^2 ))dx−(1/(1+a))∫_0 ^( 1) (1/(1+ax^2 ))dx =(1/(1+a))ln(2)+(1/2).(1/(a(1+a)))ln(1+a)−((√a)/(a(1+a)))[(tan^(−1) (x(√a) )]_0 ^1 =((ln(2))/(1+a))+(1/2).(1/(a(1+a))) ln(1+a)−((√a)/(a(1+a)))tan^(−1) ((√a) ) ∫_0 ^( 1) h′(a)da=ln^2 (2)+(1/2)∫_0 ^( 1) ((ln(1+a))/a)da−(1/4)ln^2 (2)−((π^2 /(16))) =ln^2 (2)+(π^2 /(24))−(1/4)ln^2 (2)−(π^2 /(16)) =(3/4)ln^2 (2)−(π^2 /(48)) ∴ h(1)=𝚽=(3/4)ln^2 (2)−(π^2 /(48)) ..✓ 𝛗=(1/2)ln^2 (2)−(3/8)ln^2 (2)+(π^2 /(96)) 𝛗=(1/8)ln^2 (2)+(π^2 /(96))$
$e v a l u a t i o n o f :: ϕ = \int_{0}^{1} \frac{x l n (1 + x)}{1 + x^{2}} d x$ $s o l u t i o n :$ $ϕ \overset{I . B . P}{=} {[\frac{1}{2} l n (1 + x^{2}) l n (1 + x)]}_{0}^{1} - \frac{1}{2} {\int_{0}^{1} \frac{l n (1 + x^{2})}{1 + x} d x = Φ}$ $ϕ = \frac{1}{2} {l n}^{2} (2) - \frac{1}{2} Φ . . . . . . . . ✓$ $Φ = \int_{0}^{1} \frac{l n (1 + x^{2})}{1 + x} d x = ? ? ?$ $h (a) = \int_{0}^{1} \frac{l n (1 + {a x}^{2})}{1 + x} d x$ $h^{'} (a) = \int_{0}^{1} \frac{\partial}{\partial_{a}} (\frac{l n (1 + {a x}^{2})}{1 + x}) d x$ $h^{'} (a) = \int_{0}^{1} \frac{x^{2}}{(1 + x) (1 + {a x}^{2})} d x$ $h^{'} (a) = \frac{1}{1 + a} \int_{0}^{1} \frac{1}{1 + x} d x + \frac{1}{1 + a} \int_{0}^{1} \frac{x}{1 + {a x}^{2}} d x - \frac{1}{1 + a} \int_{0}^{1} \frac{1}{1 + {a x}^{2}} d x$ $= \frac{1}{1 + a} l n (2) + \frac{1}{2} . \frac{1}{a (1 + a)} l n (1 + a) - \frac{\sqrt{a}}{a (1 + a)} [{({t a n}^{- 1} (x \sqrt{a})]}_{0}^{1}$ $= \frac{l n (2)}{1 + a} + \frac{1}{2} . \frac{1}{a (1 + a)} l n (1 + a) - \frac{\sqrt{a}}{a (1 + a)} {t a n}^{- 1} (\sqrt{a})$ $\int_{0}^{1} h^{'} (a) d a = {l n}^{2} (2) + \frac{1}{2} \int_{0}^{1} \frac{l n (1 + a)}{a} d a - \frac{1}{4} {l n}^{2} (2) - (\frac{π^{2}}{16})$ $= {l n}^{2} (2) + \frac{π^{2}}{24} - \frac{1}{4} {l n}^{2} (2) - \frac{π^{2}}{16}$ $= \frac{3}{4} {l n}^{2} (2) - \frac{π^{2}}{48}$ $∴ h (1) = Φ = \frac{3}{4} {l n}^{2} (2) - \frac{π^{2}}{48} . . ✓$ $ϕ = \frac{1}{2} {l n}^{2} (2) - \frac{3}{8} {l n}^{2} (2) + \frac{π^{2}}{96}$ $ϕ = \frac{1}{8} {l n}^{2} (2) + \frac{π^{2}}{96}$
	Answered by mathmax by abdo last updated on 27/Mar/21
	$Φ=∫_0 ^1 ((log(1+x^2 ))/(1+x))dx =ϕ(1) with ϕ(t)=∫_0 ^1 ((log(1+tx^2 ))/(1+x)) (t>0) we have ϕ^′ (t)=∫_0 ^1 (x^2 /((x+1)(tx^2 +1)))dx =(1/t)∫_0 ^1 ((tx^2 +1−1)/((x+1)(tx^2 +1)))dx =(1/t)∫_0 ^1 (dx/(x+1))−(1/t)∫_0 ^1 (dx/((x+1)(tx^2 +1))) but ∫_0 ^1 (dx/(x+1))=ln2 let decompose F(x)=(1/((x+1)(tx^2 +1)))=(a/(x+1)) +((bx+c)/(tx^2 +1)) a=(1/(t+1)) , lim_(x→∞) xF(x)=0 =a+(b/t) ⇒0=at+b ⇒b=−(t/(t+1)) F(0)=1=a+c ⇒c=1−(1/(t+1))=(t/(t+1)) ⇒ F(x)=(1/((t+1)(x+1)))+((−(t/(t+1))x+(t/(t+1)))/(tx^2 +1)) =(1/(t+1)){(1/(x+1))−((tx−t)/(tx^2 +1))} ⇒∫_0 ^1 F(x)dx=(1/(t+1)){ln2−∫_0 ^1 ((tx−t)/(tx^2 +1))dx} and ∫_0 ^1 ((tx−t)/(tx^2 +1))dx =(1/2)∫_0 ^1 ((2tx)/(tx^2 +1))dx−∫_0 ^1 (t/(tx^2 +1))dx((√t)x=y) =(1/2)[ln(tx^2 +1)]_0 ^1 −∫_0 ^(√t) (t/(y^2 +1))(dy/( (√t))) =(1/2)ln(t+1)−(√t)arctan((√t)) ⇒ ϕ^′ (t)=((log2)/t)−(1/(t(t+1))){ln2−(1/2)log(t+1)+(√t)arctan((√t))} =((log2)/t)−((log2)/(t(t+1)))−((log(t+1))/(t(t+1)))−(((√t)arctan((√t)))/(t(t+1))) =((log2)/t)(1−(1/(t+1)))−...=((log2)/t).(t/(t+1))−((log(t+1))/(t(t+1)))−(((√t)arctan((√t)))/(t(t+1))) =((log2)/(t+1))−((1/t)−(1/(t+1)))log(t+1)−(((√t)arctan((√t)))/(t(t+1))) ⇒∫_0 ^1 ϕ^′ (t)dt =log^2 (2)−∫_0 ^1 ((log(t+1))/t)dt+∫_0 ^1 ((log(t+1))/(t+1))dt −∫_0 ^1 (((√t)arctan((√t)))/(t(t+1)))dt ∫_0 ^1 ((log(t+1))/t)dt =[logt.log(t+1)]_0 ^1 −∫_0 ^1 ((logt)/(t+1))dt =−∫_0 ^1 log(t)Σ_(n=0) ^∞ (−1)^n t^n dt=−Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 t^n logt dt U_n =∫_0 ^1 t^n logt dt =[(t^(n+1) /(n+1))logt]_0 ^1 −(1/(n+1))∫_0 ^1 t^n dt =−(1/((n+1)^2 )) ⇒∫_0 ^1 ((log(t+1))/t)dt =Σ_(n=0) ^∞ (((−1)^n )/((n+1)^2 )) =−Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =−(2^(1−2) −1)ξ(2)=−(−(1/2)).(π^2 /6)=(π^2 /(12)) ∫_0 ^1 ((log(t+1))/(t+1))dt =[log^2 (t+1)]_0 ^1 −∫_0 ^1 ((log(t+1))/(t+1))dt ⇒ ∫_0 ^1 ((log(t+1))/(t+1))dt =((ln^2 (2))/2) J=∫_0 ^1 (((√t) arctan((√t)))/(t(t+1)))dt =_((√t)=y) ∫_0 ^1 ((yarctany)/(y^2 (y^2 +1)))(2y)dy =2∫_0 ^1 ((arctan(y))/(y^2 +1))dy =2{ [arctan^2 y]_0 ^1 −∫_0 ^1 ((arctany)/(y^2 +1))} =2×(π^2 /(16))−2∫(...)dy ⇒4∫_0 ^1 ((arctany)/(y^2 +1))dy =(π^2 /8) ⇒ ∫_0 ^1 ((arctany)/(y^2 +1))dy =(π^2 /(32)) ⇒ Φ =ϕ(1)=log^2 (2)−(π^2 /(12)) +((ln^2 (2))/2)−(π^2 /(16))=(3/2)ln^2 (2)−((4π^2 )/(48))−((3π^2 )/(48)) =(3/2)ln^2 (2)−((7π^2 )/(48)) ??$
	$Φ = \int_{0}^{1} \frac{\log (1 + x^{2})}{1 + x} dx = φ (1) with φ (t) = \int_{0}^{1} \frac{\log (1 + {tx}^{2})}{1 + x} (t > 0)$ $we have φ^{^{'}} (t) = \int_{0}^{1} \frac{x^{2}}{(x + 1) ({tx}^{2} + 1)} dx = \frac{1}{t} \int_{0}^{1} \frac{{tx}^{2} + 1 - 1}{(x + 1) ({tx}^{2} + 1)} dx$ $= \frac{1}{t} \int_{0}^{1} \frac{dx}{x + 1} - \frac{1}{t} \int_{0}^{1} \frac{dx}{(x + 1) ({tx}^{2} + 1)} but \int_{0}^{1} \frac{dx}{x + 1} = \ln 2$ $let decompose F (x) = \frac{1}{(x + 1) ({tx}^{2} + 1)} = \frac{a}{x + 1} + \frac{bx + c}{{tx}^{2} + 1}$ $a = \frac{1}{t + 1}, \lim_{x \to \infty} xF (x) = 0 = a + \frac{b}{t} \Rightarrow 0 = at + b \Rightarrow b = - \frac{t}{t + 1}$ $F (0) = 1 = a + c \Rightarrow c = 1 - \frac{1}{t + 1} = \frac{t}{t + 1} \Rightarrow$ $F (x) = \frac{1}{(t + 1) (x + 1)} + \frac{- \frac{t}{t + 1} x + \frac{t}{t + 1}}{{tx}^{2} + 1}$ $= \frac{1}{t + 1} {\frac{1}{x + 1} - \frac{tx - t}{{tx}^{2} + 1}} \Rightarrow \int_{0}^{1} F (x) dx = \frac{1}{t + 1} {\ln 2 - \int_{0}^{1} \frac{tx - t}{{tx}^{2} + 1} dx} and$ $\int_{0}^{1} \frac{tx - t}{{tx}^{2} + 1} dx = \frac{1}{2} \int_{0}^{1} \frac{2 tx}{{tx}^{2} + 1} dx - \int_{0}^{1} \frac{t}{{tx}^{2} + 1} dx (\sqrt{t} x = y)$ $= \frac{1}{2} {[\ln ({tx}^{2} + 1)]}_{0}^{1} - \int_{0}^{\sqrt{t}} \frac{t}{y^{2} + 1} \frac{dy}{\sqrt{t}}$ $= \frac{1}{2} \ln (t + 1) - \sqrt{t} \arctan (\sqrt{t}) \Rightarrow$ $φ^{^{'}} (t) = \frac{\log 2}{t} - \frac{1}{t (t + 1)} {\ln 2 - \frac{1}{2} \log (t + 1) + \sqrt{t} \arctan (\sqrt{t})}$ $= \frac{\log 2}{t} - \frac{\log 2}{t (t + 1)} - \frac{\log (t + 1)}{t (t + 1)} - \frac{\sqrt{t} \arctan (\sqrt{t})}{t (t + 1)}$ $= \frac{\log 2}{t} (1 - \frac{1}{t + 1}) - . . . = \frac{\log 2}{t} . \frac{t}{t + 1} - \frac{\log (t + 1)}{t (t + 1)} - \frac{\sqrt{t} \arctan (\sqrt{t})}{t (t + 1)}$ $= \frac{\log 2}{t + 1} - (\frac{1}{t} - \frac{1}{t + 1}) \log (t + 1) - \frac{\sqrt{t} \arctan (\sqrt{t})}{t (t + 1)}$ $\Rightarrow \int_{0}^{1} φ^{^{'}} (t) dt = \log^{2} (2) - \int_{0}^{1} \frac{\log (t + 1)}{t} dt + \int_{0}^{1} \frac{\log (t + 1)}{t + 1} dt$ $- \int_{0}^{1} \frac{\sqrt{t} \arctan (\sqrt{t})}{t (t + 1)} dt$ $\int_{0}^{1} \frac{\log (t + 1)}{t} dt = {[logt . \log (t + 1)]}_{0}^{1} - \int_{0}^{1} \frac{logt}{t + 1} dt$ $= - \int_{0}^{1} \log (t) \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{n} dt = - \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} t^{n} logt dt$ $U_{n} = \int_{0}^{1} t^{n} logt dt = {[\frac{t^{n + 1}}{n + 1} logt]}_{0}^{1} - \frac{1}{n + 1} \int_{0}^{1} t^{n} dt = - \frac{1}{{(n + 1)}^{2}}$ $\Rightarrow \int_{0}^{1} \frac{\log (t + 1)}{t} dt = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}} = - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}}$ $= - (2^{1 - 2} - 1) ξ (2) = - (- \frac{1}{2}) . \frac{π^{2}}{6} = \frac{π^{2}}{12}$ $\int_{0}^{1} \frac{\log (t + 1)}{t + 1} dt = {[\log^{2} (t + 1)]}_{0}^{1} - \int_{0}^{1} \frac{\log (t + 1)}{t + 1} dt \Rightarrow$ $\int_{0}^{1} \frac{\log (t + 1)}{t + 1} dt = \frac{\ln^{2} (2)}{2}$ $J = \int_{0}^{1} \frac{\sqrt{t} \arctan (\sqrt{t})}{t (t + 1)} dt =_{\sqrt{t} = y} \int_{0}^{1} \frac{yarctany}{y^{2} (y^{2} + 1)} (2 y) dy$ $= 2 \int_{0}^{1} \frac{\arctan (y)}{y^{2} + 1} dy = 2 {{[\arctan^{2} y]}_{0}^{1} - \int_{0}^{1} \frac{arctany}{y^{2} + 1}}$ $= 2 \times \frac{π^{2}}{16} - 2 \int (. . .) dy \Rightarrow 4 \int_{0}^{1} \frac{arctany}{y^{2} + 1} dy = \frac{π^{2}}{8} \Rightarrow$ $\int_{0}^{1} \frac{arctany}{y^{2} + 1} dy = \frac{π^{2}}{32} \Rightarrow$ $Φ = φ (1) = \log^{2} (2) - \frac{π^{2}}{12} + \frac{\ln^{2} (2)}{2} - \frac{π^{2}}{16} = \frac{3}{2} \ln^{2} (2) - \frac{4 π^{2}}{48} - \frac{3 π^{2}}{48}$ $= \frac{3}{2} \ln^{2} (2) - \frac{7 π^{2}}{48} ? ?$
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