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

Question Number 169873 by sciencestudent last updated on 11/May/22

$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}=? \\ $$

Answered by ArielVyny last updated on 11/May/22

Σ_(n≥0) (−1)^n x^n =(1/(1+x)) Σ_(n≥0) (−1)^n (x^(n+1) /(n+1))=ln(1+x) Σ_(n≥0) (−1)^n ∫_0 ^1 (x^n /(n+1))=∫_0 ^1 ((ln(1+x))/x)dx Σ_(n≥0) (−1)^n (1/((n+1)^2 ))=∫_0 ^1 ((ln(1+x))/x)=(π^2 /(12))

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

Answered by Mathspace last updated on 11/May/22

(d/dx)ln(1+x)=(1/(1+x))=Σ_(n=0) ^∞ (−1)^n x^n with∣x∣<1⇒ln(1+x)=Σ_(n=0) ^∞ (((−1)^n )/(n+1))x^(n+1) +c(c=0) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^n ⇒ ∫_0 ^1 ((ln(1+x))/x)dx=∫_0 ^1 Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^(n−1) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)∫_0 ^1 x^(n−1) dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 )=−δ(2) δ(x)=Σ_(n=1) ^∞ (((−1)^n )/n^x ) =(2^(1−x) −1)ξ(x) ⇒ δ(2)=((1/2)−1)ξ(2)=−(1/2)(π^2 /6) =−(π^2 /(12)) ⇒∫_0 ^1 ((ln(1+x))/x)dx=(π^2 /(12))

$$\frac{{d}}{{dx}}{ln}\left(\mathrm{1}+{x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}}=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \\ $$$${with}\mid{x}\mid<\mathrm{1}\Rightarrow{ln}\left(\mathrm{1}+{x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} +{c}\left({c}=\mathrm{0}\right) \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{x}^{{n}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{x}^{{n}−\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} {dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }=−\delta\left(\mathrm{2}\right) \\ $$$$\delta\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} } \\ $$$$=\left(\mathrm{2}^{\mathrm{1}−{x}} −\mathrm{1}\right)\xi\left({x}\right)\:\Rightarrow \\ $$$$\delta\left(\mathrm{2}\right)=\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)\xi\left(\mathrm{2}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$

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