Question Number 60658 by Mr X pcx last updated on 23/May/19
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx} \\ $$
Commented by maxmathsup by imad last updated on 29/May/19
![we have ln(1−x) =−Σ_(n=1) ^∞ (x^n /n) and ln(1−x^2 ) =−Σ_(n=1) ^∞ (x^(2n) /n) if ∣x∣<1 ln(1−x)ln(1−x^2 ) =(Σ_(n=1) ^∞ (x^n /n))(Σ_(n=1) ^∞ (x^(2n) /n)) let a_n =(x^n /n) and b_n =(x^(2n) /n) ⇒ ln(1−x)ln(1−x^2 ) =(Σa_n )(Σb_n ) =Σ_(n=1) ^∞ c_n with c_n =Σ_(i+j=n) a_i b_j =Σ_(i=1) ^(n−1) ai b_(n−i) =Σ_(i=1) ^(n−1) (x^i /i) (x^(2n−2i) /(n−i)) =Σ_(i=1) ^n (x^(2n−i) /(i(n−i))) ⇒ ∫_0 ^1 ln(x)ln(1−x)(1−x^2 )dx =∫_0 ^1 (Σ_(n=1) ^∞ Σ_(i=1) ^(n−1) (x^(2n−i) /(i(n−i))))ln(x)dx =Σ_(n=1) ^∞ (Σ_(i=1) ^(n−1) (1/(i(n−i))) ∫_0 ^1 x^(2n−i) ln(x)dx) ∫_0 ^1 x^(2n−i) ln(x)dx =_(by parts) [(1/(2n−i +1)) x^(2n−i+1) ln(x)]_0 ^1 −∫_0 ^1 (1/((2n−i+1))) x^(2n−i) dx =−(1/((2n−i +1)^2 )) ⇒ I =−Σ_(n=1) ^∞ ( Σ_(i=1) ^(n−1) (1/(i(n−i)(2n−i+1)^2 ))) ...be continued....](Q61152.png)
$${we}\:{have}\:{ln}\left(\mathrm{1}−{x}\right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:\:\:{and}\:\:\:\:{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} }{{n}}\:\:\:{if}\:\mid{x}\mid<\mathrm{1} \\ $$$${ln}\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:=\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\right)\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} }{{n}}\right)\:{let}\:{a}_{{n}} =\frac{{x}^{{n}} }{{n}}\:\:{and}\:{b}_{{n}} =\frac{{x}^{\mathrm{2}{n}} }{{n}}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:=\left(\Sigma{a}_{{n}} \right)\left(\Sigma{b}_{{n}} \right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{c}_{{n}} \:\:\:{with} \\ $$$${c}_{{n}} =\sum_{{i}+{j}={n}} \:\:\:{a}_{{i}} {b}_{{j}} =\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{ai}\:{b}_{{n}−{i}} \:\:=\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{{x}^{{i}} }{{i}}\:\frac{{x}^{\mathrm{2}{n}−\mathrm{2}{i}} }{{n}−{i}}\:=\sum_{{i}=\mathrm{1}} ^{{n}} \:\:\frac{{x}^{\mathrm{2}{n}−{i}} }{{i}\left({n}−{i}\right)} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{1}} ^{\infty} \sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}{n}−{i}} }{{i}\left({n}−{i}\right)}\right){ln}\left({x}\right){dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\left(\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} \frac{\mathrm{1}}{{i}\left({n}−{i}\right)}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{\mathrm{2}{n}−{i}} {ln}\left({x}\right){dx}\right)\:\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{\mathrm{2}{n}−{i}} \:{ln}\left({x}\right){dx}\:=_{{by}\:{parts}} \:\:\:\:\left[\frac{\mathrm{1}}{\mathrm{2}{n}−{i}\:+\mathrm{1}}\:{x}^{\mathrm{2}{n}−{i}+\mathrm{1}} {ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{2}{n}−{i}+\mathrm{1}\right)}\:{x}^{\mathrm{2}{n}−{i}} \:{dx} \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}−{i}\:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${I}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \left(\:\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{{i}\left({n}−{i}\right)\left(\mathrm{2}{n}−{i}+\mathrm{1}\right)^{\mathrm{2}} }\right)\:\:\:\:...{be}\:{continued}.... \\ $$