Question Number 30798 by prof Abdo imad last updated on 25/Feb/18
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx} \\ $$
Commented by prof Abdo imad last updated on 08/Mar/18
![for ∣t∣<1 ln^′ (1−t)=((−1)/(1−t))=−Σ_(n=0) ^∞ t^n ⇒ ln(1−t)=−Σ_(n=0) ^∞ (t^(n+1) /(n+1))=−Σ_(n=1) ^∞ (t^n /n) =Σ_(n=1) ^∞ (t^(n−1) /n) let integrate by parts I= [(−(1/x)+1)ln(1−x^2 )]_(x→0) ^1 −∫_0 ^1 (1−(1/x))((−2x)/(1−x^2 ))dx =o +∫_0 ^1 2x(1−(1/x))(dx/(1−x^2 )) because lim_(x→0) (1−(1/x))ln(1−x^2 )=−lim_(x→0) ((ln(1−x^2 ))/x) =−lim_(x→0) ((f(x)−f(0))/x) =f^′ (0) with f(x)=ln(1−x^2 ) ⇒f^′ (x)=((−2x)/(1−x^2 )) ⇒f^′ (0)=0 lim_(x→1) (1−(1/x))ln(1−x^2 )=0 withch=1−x=t so I= ∫_0 ^1 ((2x(x−1)dx)/(x(1−x)(1+x)))=∫_0 ^1 ((−2dx)/(1+x)) =−2 [ln∣1+x∣]_0 ^1 =−2ln(2) . ∫_0 ^1 ((ln(1−x^2 ))/x^2 ) dx=−2ln(2) .](https://www.tinkutara.com/question/Q31434.png)
$$\:{for}\:\mid{t}\mid<\mathrm{1}\:\:{ln}^{'} \left(\mathrm{1}−{t}\right)=\frac{−\mathrm{1}}{\mathrm{1}−{t}}=−\sum_{{n}=\mathrm{0}} ^{\infty} \:{t}^{{n}} \:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{t}\right)=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{t}^{{n}} }{{n}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{t}^{{n}−\mathrm{1}} }{{n}}\:\:{let}\:{integrate}\:{by}\:{parts} \\ $$$${I}=\:\left[\left(−\frac{\mathrm{1}}{{x}}+\mathrm{1}\right){ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\right]_{{x}\rightarrow\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\frac{−\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\: \\ $$$$={o}\:\:+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{x}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\frac{{dx}}{\mathrm{1}−{x}^{\mathrm{2}} }\:{because} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right){ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)=−{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}} \\ $$$$=−{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{f}\left({x}\right)−{f}\left(\mathrm{0}\right)}{{x}}\:={f}^{'} \left(\mathrm{0}\right)\:{with} \\ $$$${f}\left({x}\right)={ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:\Rightarrow{f}^{'} \left({x}\right)=\frac{−\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\Rightarrow{f}^{'} \left(\mathrm{0}\right)=\mathrm{0} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right){ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)=\mathrm{0}\:{withch}=\mathrm{1}−{x}={t}\:{so} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{x}\left({x}−\mathrm{1}\right){dx}}{{x}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{−\mathrm{2}{dx}}{\mathrm{1}+{x}} \\ $$$$=−\mathrm{2}\:\left[{ln}\mid\mathrm{1}+{x}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} =−\mathrm{2}{ln}\left(\mathrm{2}\right)\:. \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }\:{dx}=−\mathrm{2}{ln}\left(\mathrm{2}\right)\:. \\ $$