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Calculate-0-1-ln-x-1-x-2-dx-G-Solution-0-1-k-0-1-k-x-2k-ln-x-dx-k-0-1-k-0-1-x-




Question Number 177604 by mnjuly1970 last updated on 07/Oct/22
          Calculate       𝛗=∫_0 ^( 1) (( ln(x ))/(1 + x^( 2) )) dx =^?  −G      ∼ Solution ∼        𝛗 = ∫_0 ^( 1) {Σ_(k=0) ^∞ (−1)^( k) x^( 2k) ln(x) }dx           = Σ_(k=0) ^∞  (−1 )^( k)  ∫_0 ^( 1) x^( 2k)  ln(x) dx          =^(i.b.p)  Σ_(k=0) ^∞ (−1 )^( k) {[(x^( 1+2k) /(1+2k)) ln(x ) ]_0 ^1 −(1/(1+2k))∫_0 ^( 1) x^( 2k) dx }          = Σ_(k=0) ^∞ (( (−1)^(1+k) )/(( 1 + 2k )^( 2) ))  = − G ( Catalan constant )           ∴          𝛗 =  − G                ■ m.n
$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{Calculate} \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{ln}\left({x}\:\right)}{\mathrm{1}\:+\:{x}^{\:\mathrm{2}} }\:\mathrm{d}{x}\:\overset{?} {=}\:−\mathrm{G} \\ $$$$\:\:\:\:\sim\:\mathrm{Solution}\:\sim \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\:{k}} {x}^{\:\mathrm{2}{k}} \mathrm{ln}\left({x}\right)\:\right\}\mathrm{d}{x} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(−\mathrm{1}\:\right)^{\:{k}} \:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:\mathrm{2}{k}} \:\mathrm{ln}\left({x}\right)\:\mathrm{d}{x} \\ $$$$\:\:\:\:\:\:\:\:\overset{{i}.{b}.{p}} {=}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\:\right)^{\:{k}} \left\{\left[\frac{{x}^{\:\mathrm{1}+\mathrm{2}{k}} }{\mathrm{1}+\mathrm{2}{k}}\:\mathrm{ln}\left({x}\:\right)\:\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}{k}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:\mathrm{2}{k}} \mathrm{d}{x}\:\right\} \\ $$$$\:\:\:\:\:\:\:\:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{\mathrm{1}+{k}} }{\left(\:\mathrm{1}\:+\:\mathrm{2}{k}\:\right)^{\:\mathrm{2}} }\:\:=\:−\:\mathrm{G}\:\left(\:\mathrm{C}{atalan}\:{constant}\:\right)\: \\ $$$$\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}\:=\:\:−\:\mathrm{G}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\:\mathrm{m}.\mathrm{n}\: \\ $$$$\:\:\:\:\: \\ $$

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