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Question Number 113756 by Dwaipayan Shikari last updated on 15/Sep/20

∫_0 ^1 ((log(x+1))/x)dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left({x}+\mathrm{1}\right)}{{x}}{dx} \\ $$

Commented by Dwaipayan Shikari last updated on 15/Sep/20

∫_0 ^1 Σ_(n=1) ^∞ (((−1)^n x^(n−1) )/n)dx Σ_(n=1) ^∞ (−1)^n ∫_0 ^1 (x^(n−1) /n)dx Σ_(n=1) ^∞ (−1)^n (1/n^2 )=(π^2 /(12))

$$\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{{n}−\mathrm{1}} }{{n}}{dx} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}−\mathrm{1}} }{{n}}{dx} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$ \\ $$

Answered by mathdave last updated on 15/Sep/20

solution let x=−x ∫_0 ^(−1) ((ln(1−x))/x)dx=−Li_2 (−1)=(π^2 /(12))

$${solution}\: \\ $$$${let}\:{x}=−{x} \\ $$$$\int_{\mathrm{0}} ^{−\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx}=−{Li}_{\mathrm{2}} \left(−\mathrm{1}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$

Answered by mindispower last updated on 15/Sep/20

∫_0 ^1 (1/x)Σ_(k≥0) (−1)^k (x^(k+1) /(k+1))dx =Σ_(k≥0) (((−1)^k )/(k+1))∫_0 ^1 x^k dx=Σ(((−1)^k )/((k+1)^2 )) =Σ_(k≥0) ((1/((2k+1)^2 ))−(1/((2k+2)^2 ))) =Σ_(k≥0) (1/((2k+1)^2 ))−(1/4)Σ(1/((k+1)^2 )) (ζ(2)−((ζ(2))/4)−((ζ(2))/4))=((ζ(2))/2)=(π^2 /(12))

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}}\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} \frac{{x}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}{dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{k}} {dx}=\Sigma\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{2}\right)^{\mathrm{2}} }\right) \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}}\Sigma\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\left(\zeta\left(\mathrm{2}\right)−\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{4}}−\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{4}}\right)=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$ \\ $$

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