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0-1-ln-x-1-x-2-x-dx-

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Question Number 140961 by iloveisrael last updated on 14/May/21
∫_0 ^( 1) ((ln (x+(√(1−x^2 ))))/x) dx =?
$$\:\underset{\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\:\frac{\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{{x}}\:{dx}\:=?\: \\ $$
Answered by qaz last updated on 14/May/21
∫_0 ^1 ((ln(x+(√(1−x^2 ))))/x)dx =∫_0 ^1 dy∫_0 ^1 (dx/(yx+(√(1−x^2 ))))+(1/2)∫_0 ^1 ((ln(1−x)+ln(1+x))/x)dx =∫_0 ^1 dy∫_0 ^1 (dx/(yx+(√(1−x^2 ))))+(1/2)(−(π^2 /6)+(π^2 /(12))) =∫_0 ^1 dy∫_0 ^(π/2) ((cos z)/(ysin z+cos z))dz−(π^2 /(24)) =∫_0 ^1 dy∫_0 ^(π/2) ((d(tan z))/((ytan z+1)(tan^2 z+1)))−(π^2 /(24)) =∫_0 ^1 dy∫_0 ^∞ (du/((yu+1)(u^2 +1)))−(π^2 /(24)) =∫_0 ^1 (1/(1+y^2 ))∫_0 ^∞ ((y^2 /(yu+1))+((−yu+1)/(u^2 +1)))du−(π^2 /(24)) =∫_0 ^1 (1/(1+y^2 ))(yln((yu+1)/( (√(u^2 +1))))∣_0 ^∞ +(π/2))dy−(π^2 /(24)) =∫_0 ^1 (1/(1+y^2 ))(ylny+(π/2))dy−(π^2 /(24)) =−(π^2 /(48))+(π^2 /8)−(π^2 /(24)) =(π^2 /(16)) −−−−−−−−−−−−−−−−−− ∫_0 ^1 ((ylny)/(1+y^2 ))dy ={(1/2)ln(1+y^2 )lny−(1/2)∫((ln(1+y^2 ))/y)dy}_0 ^1 =−(1/2)Σ_(n=1) ^∞ (((−1)^(n−1) )/n)∫_0 ^1 y^(2n−1) dy =−(1/4)Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 )=−(π^2 /(48))
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {dy}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{{yx}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)+{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {dy}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{{yx}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {dy}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{cos}\:{z}}{{y}\mathrm{sin}\:{z}+\mathrm{cos}\:{z}}{dz}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {dy}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{d}\left(\mathrm{tan}\:{z}\right)}{\left({y}\mathrm{tan}\:{z}+\mathrm{1}\right)\left(\mathrm{tan}\:^{\mathrm{2}} {z}+\mathrm{1}\right)}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {dy}\int_{\mathrm{0}} ^{\infty} \frac{{du}}{\left({yu}+\mathrm{1}\right)\left({u}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \left(\frac{{y}^{\mathrm{2}} }{{yu}+\mathrm{1}}+\frac{−{yu}+\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}\right){du}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} }\left({yln}\frac{{yu}+\mathrm{1}}{\:\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}}\mid_{\mathrm{0}} ^{\infty} +\frac{\pi}{\mathrm{2}}\right){dy}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} }\left({ylny}+\frac{\pi}{\mathrm{2}}\right){dy}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{48}}+\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$$$−−−−−−−−−−−−−−−−−− \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ylny}}{\mathrm{1}+{y}^{\mathrm{2}} }{dy} \\ $$$$=\left\{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{y}^{\mathrm{2}} \right){lny}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{ln}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}{{y}}{dy}\right\}_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{\mathrm{2}{n}−\mathrm{1}} {dy} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }=−\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$
Answered by iloveisrael last updated on 15/May/21
with integral parameterized by a>0 I(a)=∫_0 ^( 1) ((ln (ax+(√(1−x^2 ))))/x) dx I ′(a)= ∫_0 ^( 1) (1/(ax+(√(1−x^2 )))) dx I ′(a) = [ ((a ln ((([(a^2 +1)x^2 −1][(√(1−x^2 ))+ax])/( (√(1−x^2 ))−ax)))+2arcsin x)/(2(a^2 +1)))]_(x=0) ^(x=1) = ((a ln a)/(a^2 +1)) +(π/2) (1/(a^2 +1)) I(1)−I(0) = ∫_0 ^( 1) ((a ln a)/(a^2 +1)) da +(π/2) ∫_0 ^( 1) (1/(a^2 +1)) da (1) ∫_0 ^( 1) ((a ln a)/(a^2 +1)) da = Σ_(k=1) ^∞ (−1)^k ∫_0 ^( 1) a^(2k+1) ln a da = −(1/4) Σ_(k=0) ^∞ (((−1)^k )/((k+1)^2 )) = −(1/4)η(2)=−(π^2 /(48)) (2)(π/2)∫_0 ^( 1) (da/(1+a^2 )) = (π^2 /8) I(0) = (1/2)∫_0 ^( 1) ((ln (1−x^2 ))/x) dx = −(1/2) Σ_(k=1) ^∞ (1/k) ∫_0 ^( 1) x^(2k−1) dx = −(1/4)Σ_(k=1) ^∞ (1/k^2 ) = −(1/4)ζ(2)=−(π^2 /(24)) I(1)= −(π^2 /(48))+(π^2 /8)−(π^2 /(24)) = (π^2 /(16))
$${with}\:{integral}\:{parameterized}\:{by}\:{a}>\mathrm{0} \\ $$$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\:\left({ax}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{{x}}\:{dx}\: \\ $$$${I}\:'\left({a}\right)=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{1}}{{ax}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}\: \\ $$$${I}\:'\left({a}\right)\:=\:\left[\:\frac{{a}\:\mathrm{ln}\:\left(\frac{\left[\left({a}^{\mathrm{2}} +\mathrm{1}\right){x}^{\mathrm{2}} −\mathrm{1}\right]\left[\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{ax}\right]}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−{ax}}\right)+\mathrm{2arcsin}\:{x}}{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)}\right]_{{x}=\mathrm{0}} ^{{x}=\mathrm{1}} \\ $$$$=\:\frac{{a}\:\mathrm{ln}\:{a}}{{a}^{\mathrm{2}} +\mathrm{1}}\:+\frac{\pi}{\mathrm{2}}\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}\: \\ $$$${I}\left(\mathrm{1}\right)−{I}\left(\mathrm{0}\right)\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{a}\:\mathrm{ln}\:{a}}{{a}^{\mathrm{2}} +\mathrm{1}}\:{da}\:+\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}\:{da} \\ $$$$\left(\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{a}\:\mathrm{ln}\:{a}}{{a}^{\mathrm{2}} +\mathrm{1}}\:{da}\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{a}^{\mathrm{2}{k}+\mathrm{1}} \:\mathrm{ln}\:{a}\:{da} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{4}}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:−\frac{\mathrm{1}}{\mathrm{4}}\eta\left(\mathrm{2}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$$\left(\mathrm{2}\right)\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{da}}{\mathrm{1}+{a}^{\mathrm{2}} }\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$${I}\left(\mathrm{0}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}}\:{dx} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{k}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{x}^{\mathrm{2}{k}−\mathrm{1}} \:{dx}\: \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{4}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\:−\frac{\mathrm{1}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$${I}\left(\mathrm{1}\right)=\:−\frac{\pi^{\mathrm{2}} }{\mathrm{48}}+\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$$$ \\ $$

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