Question Number 110543 by mnjuly1970 last updated on 29/Aug/20
Answered by mathdave last updated on 29/Aug/20
![solution let y=x^2 and dx=(1/(2(√y)))dy then putting into I=(1/2)∫_0 ^1 ((ln(1−y)ln(1+y))/y)dy from algebraic uding this notation ab=(1/4)(a+b)^2 −(1/4)(a−b)^2 ln(1−y)ln(1+y)=(1/4)ln^2 (1−y^2 )−(1/4)ln^2 (((1−y)/(1+y))) I=(1/8)∫_0 ^1 ((ln^2 (1−y^2 ))/y)dy−(1/8)∫_0 ^1 ((ln^2 (((1−y)/(1+y))))/y)dy let I_x =∫_0 ^1 ((ln^2 (1−y^2 ))/y)dy (let x=1−y^2 and dy=−(dx/(2(√(1−x)))) I_x =(1/2)∫_0 ^1 ((ln^2 x)/(1−x))dx let I_x (a)=(1/2)∫_0 ^1 ((ln^2 x)/(1−x))dx=(1/2)Σ_(n=1) ^∞ ∫_0 ^1 x^(n−1) ln^2 xdx (∂^2 /∂a^2 )∣_(a=0) I_x (a)=(1/2)(∂^2 /∂a^2 )Σ_(n=1) ^∞ ∫_0 ^1 x^(n−1+a) dx=Σ_(n=1) ^∞ (1/n^3 ) I_x ^(′′) (0)=ζ(3)......(x) let I_(xx) =∫_0 ^1 ((ln^2 (((1−y)/(1+y ))))/y)dy let x=((1−y)/(1+y)) and dy=−(2/((1+x)^2 )) I_(xx) =2∫_0 ^1 ((ln^2 x)/(1−x^2 ))dx=2∫_0 ^1 ((ln^2 (x))/((1+x)(1−x)))dx I_(xx) =∫_0 ^1 ((ln^2 x)/(1+x))dx+∫_0 ^1 ((ln^2 x)/(1−x))dx therefore I_(xx) ^(′′) (0)=(−1)^n 2Σ_(n=1) ^∞ (1/n^3 )+2Σ_(n=1) ^∞ (1/n^3 ) I_(xx) ^(′′) (0)=2×(3/4)ζ(3)+2ζ(3)=(7/2)ζ(3).......(2) but I=(1/8)[I_x ^(′′) (0)−I_(xx) ^(′′) (0)]=(1/8)[ζ(3)−(7/2)ζ(3)] I=(1/8)[−(5/2)ζ(3)]=−(5/(16))ζ(3) ∵∫_0 ^1 ((ln(1−x^2 )ln(1+x^2 ))/x)dx=−(5/(16))ζ(3) where ζ(3)⇒apery′s constant by mathdave(29/08/2020)](Q110578.png)
$${solution} \\ $$$${let}\:{y}={x}^{\mathrm{2}} \:\:{and}\:{dx}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{y}}}{dy}\:\:{then}\:{putting}\:{into} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{y}\right)\mathrm{ln}\left(\mathrm{1}+{y}\right)}{{y}}{dy} \\ $$$${from}\:{algebraic}\:{uding}\:{this}\:{notation} \\ $$$${ab}=\frac{\mathrm{1}}{\mathrm{4}}\left({a}+{b}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\left({a}−{b}\right)^{\mathrm{2}} \\ $$$$\mathrm{ln}\left(\mathrm{1}−{y}\right)\mathrm{ln}\left(\mathrm{1}+{y}\right)=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{y}^{\mathrm{2}} \right)−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}^{\mathrm{2}} \left(\frac{\mathrm{1}−{y}}{\mathrm{1}+{y}}\right) \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{y}^{\mathrm{2}} \right)}{{y}}{dy}−\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\frac{\mathrm{1}−{y}}{\mathrm{1}+{y}}\right)}{{y}}{dy} \\ $$$${let}\:{I}_{{x}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{y}^{\mathrm{2}} \right)}{{y}}{dy}\:\:\:\:\:\left({let}\:\:{x}=\mathrm{1}−{y}^{\mathrm{2}} \:\right. \\ $$$${and}\:\:{dy}=−\frac{{dx}}{\mathrm{2}\sqrt{\mathrm{1}−{x}}} \\ $$$${I}_{{x}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} {x}}{\mathrm{1}−{x}}{dx} \\ $$$${let}\: \\ $$$${I}_{{x}} \left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} {x}}{\mathrm{1}−{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} \mathrm{ln}^{\mathrm{2}} {xdx} \\ $$$$\frac{\partial^{\mathrm{2}} }{\partial{a}^{\mathrm{2}} }\mid_{{a}=\mathrm{0}} {I}_{{x}} \left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\frac{\partial^{\mathrm{2}} }{\partial{a}^{\mathrm{2}} }\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}+{a}} {dx}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} } \\ $$$${I}_{{x}} ^{''} \left(\mathrm{0}\right)=\zeta\left(\mathrm{3}\right)......\left({x}\right)\: \\ $$$${let}\:{I}_{{xx}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\frac{\mathrm{1}−{y}}{\mathrm{1}+{y}\:}\right)}{{y}}{dy}\:\:\:\:{let}\:\:{x}=\frac{\mathrm{1}−{y}}{\mathrm{1}+{y}}\:\:{and}\:{dy}=−\frac{\mathrm{2}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} } \\ $$$${I}_{{xx}} =\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} {x}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left({x}\right)}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}\right)}{dx} \\ $$$${I}_{{xx}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} {x}}{\mathrm{1}+{x}}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} {x}}{\mathrm{1}−{x}}{dx}\:\:\:{therefore} \\ $$$${I}_{{xx}} ^{''} \left(\mathrm{0}\right)=\left(−\mathrm{1}\right)^{{n}} \mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }+\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} } \\ $$$${I}_{{xx}} ^{''} \left(\mathrm{0}\right)=\mathrm{2}×\frac{\mathrm{3}}{\mathrm{4}}\zeta\left(\mathrm{3}\right)+\mathrm{2}\zeta\left(\mathrm{3}\right)=\frac{\mathrm{7}}{\mathrm{2}}\zeta\left(\mathrm{3}\right).......\left(\mathrm{2}\right) \\ $$$${but} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{8}}\left[{I}_{{x}} ^{''} \left(\mathrm{0}\right)−{I}_{{xx}} ^{''} \left(\mathrm{0}\right)\right]=\frac{\mathrm{1}}{\mathrm{8}}\left[\zeta\left(\mathrm{3}\right)−\frac{\mathrm{7}}{\mathrm{2}}\zeta\left(\mathrm{3}\right)\right] \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{8}}\left[−\frac{\mathrm{5}}{\mathrm{2}}\zeta\left(\mathrm{3}\right)\right]=−\frac{\mathrm{5}}{\mathrm{16}}\zeta\left(\mathrm{3}\right) \\ $$$$\because\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}}{dx}=−\frac{\mathrm{5}}{\mathrm{16}}\zeta\left(\mathrm{3}\right) \\ $$$${where}\:\:\zeta\left(\mathrm{3}\right)\Rightarrow{apery}'{s}\:{constant} \\ $$$${by}\:{mathdave}\left(\mathrm{29}/\mathrm{08}/\mathrm{2020}\right) \\ $$
Commented by mnjuly1970 last updated on 29/Aug/20
$${thank}\:{you}\:{very}\:{much}...{sir}... \\ $$
Commented by mathdave last updated on 29/Aug/20
$${you}\:{are}\:{welcome} \\ $$