Question Number 166246 by mnjuly1970 last updated on 16/Feb/22
Answered by Mathspace last updated on 18/Feb/22
![I=_(by parts) [(1−(1/x))ln^2 (1−x^2 )]_0 ^1 −∫_0 ^1 (1−(1/x)).2ln(1−x^2 )×((−2x)/(1−x^2 ))dx =0−4∫_0 ^1 ((x−1)/x)×((xln(1−x^2 ))/((1−x)(1+x)))dx =4∫_0 ^1 ((ln(1−x^2 ))/(1+x))dx =4∫_0 ^1 ((ln(1−x)+ln(1+x))/(1+x))dx =4∫_0 ^1 ((ln(1−x))/(1+x)) +4∫_0 ^1 ((ln(1+x))/(1+x))dx we have ∫_0 ^1 ((ln(1+x))/(1+x))dx=_(1+x=t) ∫_1 ^2 ((ln(t))/t)dt =[(1/2)ln^2 (t)]_1 ^2 =(1/2)ln^2 (2) ∫_0 ^(1 ) ((ln(1−x))/(1+x))dx= ∫_0 ^1 ln(1−x)Σ_(n=0) ^∞ (−1)^n x^n dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^n ln(1−x)dx =Σ_(n=0) ^∞ (−1)^n U_n U_n =[(x^(n+1) /(n+1))−(1/(n+1)))ln(1−x)]_0 ^1 −∫_0 ^1 (((x^(n+1) −1)/(n+1)))((−1)/(1−x))dx =0−(1/(n+1))∫_0 ^1 ((x^(n+1) −1)/(x−1))dx =−(1/(n+1))∫_0 ^1 (1+x+x^2 +...+x^n )dx =−(1/(n+1))[x+(x^2 /2)+....+(x^(n+1) /(n+1))]_0 ^1 =−(1/(n+1)){1+(1/2)+...+(1/(n+1))} =−(H_(n+1) /(n+1)) ⇒ ∫_0 ^1 ((ln(1−x))/(1+x))dx=−Σ_(n=0) ^∞ (−1)^n (H_(n+1) /(n+1)) =Σ_(n=1) ^∞ (−1)^n (H_n /n) ⇒ ∫_0 ^1 ((ln^2 (1−x^2 ))/x^2 )dx=2ln^2 (2)+4Σ_(n=1) ^∞ (−1)^n (H_n /n) rest to find the value of this serie...](https://www.tinkutara.com/question/Q166317.png)
$${I}=_{{by}\:{parts}} \:\:\left[\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right){ln}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right).\mathrm{2}{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)×\frac{−\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{0}−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}−\mathrm{1}}{{x}}×\frac{{xln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}{dx} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}}{dx} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)+{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}\:+\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$${we}\:{have}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx}=_{\mathrm{1}+{x}={t}} \:\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{ln}\left({t}\right)}{{t}}{dt} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left({t}\right)\right]_{\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}\:} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}{dx}= \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right)\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{x}^{{n}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {ln}\left(\mathrm{1}−{x}\right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {U}_{{n}} \\ $$$$\left.{U}_{{n}} =\left[\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right){ln}\left(\mathrm{1}−{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{x}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}}\right)\frac{−\mathrm{1}}{\mathrm{1}−{x}}{dx} \\ $$$$=\mathrm{0}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}+\mathrm{1}} −\mathrm{1}}{{x}−\mathrm{1}}{dx} \\ $$$$=−\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…+{x}^{{n}} \right){dx} \\ $$$$=−\frac{\mathrm{1}}{{n}+\mathrm{1}}\left[{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+….+\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{{n}+\mathrm{1}}\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+…+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right\} \\ $$$$=−\frac{{H}_{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}{dx}=−\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \frac{{H}_{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\frac{{H}_{{n}} }{{n}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx}=\mathrm{2}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{4}\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \frac{{H}_{{n}} }{{n}} \\ $$$${rest}\:{to}\:{find}\:{the}\:{value}\:{of}\:{this}\:{serie}… \\ $$
Commented by mnjuly1970 last updated on 19/Feb/22
$${thanks}\:{alot} \\ $$