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Question Number 60678 by maxmathsup by imad last updated on 24/May/19

calculate ∫_0 ^1 ((ln(1−x^2 ))/x) dx

$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}}\:{dx} \\ $$

Commented by Smail last updated on 24/May/19

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

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

Commented by maxmathsup by imad last updated on 26/May/19

we have ln^′ (1−u) =((−1)/(1−u)) =−Σ_(n=0) ^∞ u^n ⇒ln(1−u) =−Σ_(n=0) ^∞ (u^(n+1) /(n+1)) +c (c=0) =−Σ_(n=1) ^∞ (u^n /n) ⇒ln(1−x^2 ) =−Σ_(n=1) ^∞ (x^(2n) /n) ⇒((ln(1−x^2 ))/x) =−Σ_(n=1) ^∞ (x^(2n−1) /n) ⇒ ∫_0 ^1 ((ln(1−x^2 ))/x) dx =−Σ_(n=1) ^∞ (1/n) ∫_0 ^1 x^(2n−1) dx =−Σ_(n=1) ^∞ (1/n) (1/(2n)) =−(1/2) Σ_(n=1) ^∞ (1/n^2 ) =−(1/2)ξ(2) =−(1/2) (π^2 /6) =−(π^2 /(12)) .

$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}−{u}\right)\:=\frac{−\mathrm{1}}{\mathrm{1}−{u}}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:{u}^{{n}} \:\Rightarrow{ln}\left(\mathrm{1}−{u}\right)\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{u}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:+{c}\:\:\:\left({c}=\mathrm{0}\right) \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{u}^{{n}} }{{n}}\:\Rightarrow{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} }{{n}}\:\Rightarrow\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}−\mathrm{1}} }{{n}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}}\:{dx}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{\mathrm{2}{n}−\mathrm{1}} \:{dx}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\:\frac{\mathrm{1}}{\mathrm{2}{n}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{2}}\xi\left(\mathrm{2}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:. \\ $$

Commented by maxmathsup by imad last updated on 26/May/19

sir smail ln(1−x) =−Σ_(n=1) ^∞ (x^n /n) ...

$${sir}\:{smail}\:\:{ln}\left(\mathrm{1}−{x}\right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:... \\ $$

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