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I-0-1-ln-x-1-x-2-dx-0-1-ln-x-n-0-1-n-x-2n-dx-n-0-1-n-0-1-x-2n-ln-x-dx-n-0-

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Question Number 144813 by mnjuly1970 last updated on 29/Jun/21
I:=∫_0 ^( 1) ((ln (x))/(1 + x^( 2) )) dx := ∫_0 ^( 1) ln(x ) Σ_(n=0) ^∞ (−1)^( n) x^( 2n) dx := Σ_(n=0) ^∞ ( −1 )^( n) ∫_0 ^( 1) x^( 2n) ln( x )dx : = Σ_(n=0) ^∞ ( −1 )^( n) { [(x^( 2n+1) /(2n +1)) ln ( x )]_0 ^( 1) −(1/((2n +1 )^( 2) )) } : = Σ_(n=1) ^( ∞) ((( −1 )^( n−1) )/(( 2n +1)^( 2) )) = −G (Catalan constant )
$$ \\ $$$$\:\:\:\:\:\:\mathrm{I}:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\:\left({x}\right)}{\mathrm{1}\:+\:{x}^{\:\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\::=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{ln}\left({x}\:\right)\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\:{n}} \:{x}^{\:\mathrm{2}{n}} \:{dx} \\ $$$$\:\:\:\:\:\:\:\::=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(\:−\mathrm{1}\:\right)^{\:{n}} \:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:\mathrm{2}{n}} \:\mathrm{ln}\left(\:{x}\:\right){dx} \\ $$$$\:\:\:\:\:\:\:\::\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\:−\mathrm{1}\:\right)^{\:{n}} \left\{\:\left[\frac{{x}^{\:\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}\:+\mathrm{1}}\:\mathrm{ln}\:\left(\:{x}\:\right)\right]_{\mathrm{0}} ^{\:\mathrm{1}} −\frac{\mathrm{1}}{\left(\mathrm{2}{n}\:+\mathrm{1}\:\right)^{\:\mathrm{2}} \:}\:\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\::\:=\:\underset{{n}=\mathrm{1}} {\overset{\:\infty} {\sum}}\frac{\left(\:−\mathrm{1}\:\right)^{\:{n}−\mathrm{1}} }{\left(\:\mathrm{2}{n}\:+\mathrm{1}\right)^{\:\mathrm{2}} }\:=\:−\mathrm{G}\:\:\left(\mathrm{Catalan}\:\mathrm{constant}\:\right) \\ $$
Commented by Dwaipayan Shikari last updated on 29/Jun/21
∫_0 ^1 ((log(x))/(1+x^2 ))dx=∫_0 ^1 ((1−x^2 )/(1−x^4 ))log(x)dx =(1/(16))∫_0 ^1 ((u^(−(3/4)) −u^(−(1/4)) )/(1−u))log(u)du =(1/(16))(ψ′((1/4))−ψ′((3/4)))=−G
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{4}} }{log}\left({x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{−\frac{\mathrm{3}}{\mathrm{4}}} −{u}^{−\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{1}−{u}}{log}\left({u}\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\left(\psi'\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\psi'\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right)=−{G} \\ $$

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