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Question Number 114635 by mathmax by abdo last updated on 20/Sep/20
find ∫_0 ^1 x^4 ln(x)ln(1−x^2 )dx
$$\mathrm{find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{4}} \mathrm{ln}\left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}\:\: \\ $$
Commented by mathdave last updated on 20/Sep/20
Commented by mathdave last updated on 20/Sep/20
Commented by Tawa11 last updated on 06/Sep/21
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$
Answered by maths mind last updated on 20/Sep/20
ln(1−x^2 )=−Σ_(k≥0) (x^(2k+2) /(k+1)) =∫_0 ^1 Σ_(k≥0) (x^(2k+6) /((k+1)))ln(x)dx =−Σ_(k≥0) (1/(k+1))∫_0 ^1 x^(2k+6) ln(x)dx =Σ_(k≥0) (1/((k+1)(2k+7)^2 )) Σ((1/(25(k+1)))−(2/(5(2k+7)^2 ))+(a/((2k+7)))) =((((2k+7)^2 −10(k+1)+25a(k+1)(2k+7))/(25(k+1)(2k+7)^2 ))) 50a=−4⇒ =(1/(25))Σ_(k≥0) ((1/(k+1))−((10)/((2k+7)^2 ))−(2/((2k+7)))) =(1/(25))Σ_(k≥0) ((1/(k+1))−(2/(2k+7)))−2Σ_(k≥0) (1/((2k+7)^2 )) =(1/(25))Σ(((2k+7−2k−2)/((2k+7)(k+1))))−(2/(25))Σ_(k≥0) (1/((2k+7)^2 )) =(1/5)Σ_(k≥0) (1/(2(k+(7/2))(k+1)))−(2/(25))Σ_(k≥0) (1/((2(k+3)+1)^2 )) =(1/(15))Σ_(k≥0) (((7/2)−1)/((k+(7/2))(k+1)))−(2/(25))Σ_(k≥3) (1/((2k+1)^2 )) =(1/(15))(Ψ((7/2))−Ψ(1))−(2/(25))((Σ_(k≥0) (1/((2k+1)^2 )))−1−(1/9)−(1/(25))) Ψ((7/2))=Ψ((1/2))+(2/5)+(2/3)+2=Ψ((1/2))+((46)/(15)) Σ_(k≥0) ((1/(2k+1)))^2 =(3/4)ζ(2)=(π^2 /8) Ψ((1/2))=−γ−2ln(2) we get (1/(15))(−γ−2ln(2)+((46)/(15))−(−γ))−(2/(25))((π^2 /(48))−((259)/(225)))
$${ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)=−\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{x}^{\mathrm{2}{k}+\mathrm{2}} }{{k}+\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{x}^{\mathrm{2}{k}+\mathrm{6}} }{\left({k}+\mathrm{1}\right)}{ln}\left({x}\right){dx} \\ $$$$=−\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{k}+\mathrm{6}} {ln}\left({x}\right){dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{7}\right)^{\mathrm{2}} } \\ $$$$\Sigma\left(\frac{\mathrm{1}}{\mathrm{25}\left({k}+\mathrm{1}\right)}−\frac{\mathrm{2}}{\mathrm{5}\left(\mathrm{2}{k}+\mathrm{7}\right)^{\mathrm{2}} }+\frac{{a}}{\left(\mathrm{2}{k}+\mathrm{7}\right)}\right) \\ $$$$=\left(\frac{\left(\mathrm{2}{k}+\mathrm{7}\right)^{\mathrm{2}} −\mathrm{10}\left({k}+\mathrm{1}\right)+\mathrm{25}{a}\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{7}\right)}{\mathrm{25}\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{7}\right)^{\mathrm{2}} }\right) \\ $$$$\mathrm{50}{a}=−\mathrm{4}\Rightarrow \\ $$$$=\frac{\mathrm{1}}{\mathrm{25}}\underset{{k}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{{k}+\mathrm{1}}−\frac{\mathrm{10}}{\left(\mathrm{2}{k}+\mathrm{7}\right)^{\mathrm{2}} }−\frac{\mathrm{2}}{\left(\mathrm{2}{k}+\mathrm{7}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{25}}\underset{{k}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{{k}+\mathrm{1}}−\frac{\mathrm{2}}{\mathrm{2}{k}+\mathrm{7}}\right)−\mathrm{2}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{7}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{25}}\Sigma\left(\frac{\mathrm{2}{k}+\mathrm{7}−\mathrm{2}{k}−\mathrm{2}}{\left(\mathrm{2}{k}+\mathrm{7}\right)\left({k}+\mathrm{1}\right)}\right)−\frac{\mathrm{2}}{\mathrm{25}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{7}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{2}\left({k}+\frac{\mathrm{7}}{\mathrm{2}}\right)\left({k}+\mathrm{1}\right)}−\frac{\mathrm{2}}{\mathrm{25}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}\left({k}+\mathrm{3}\right)+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{15}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\frac{\mathrm{7}}{\mathrm{2}}−\mathrm{1}}{\left({k}+\frac{\mathrm{7}}{\mathrm{2}}\right)\left({k}+\mathrm{1}\right)}−\frac{\mathrm{2}}{\mathrm{25}}\underset{{k}\geqslant\mathrm{3}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{15}}\left(\Psi\left(\frac{\mathrm{7}}{\mathrm{2}}\right)−\Psi\left(\mathrm{1}\right)\right)−\frac{\mathrm{2}}{\mathrm{25}}\left(\left(\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{25}}\right) \\ $$$$\Psi\left(\frac{\mathrm{7}}{\mathrm{2}}\right)=\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{3}}+\mathrm{2}=\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{46}}{\mathrm{15}} \\ $$$$\underset{{k}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\right)^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\gamma−\mathrm{2}{ln}\left(\mathrm{2}\right) \\ $$$${we}\:{get} \\ $$$$\frac{\mathrm{1}}{\mathrm{15}}\left(−\gamma−\mathrm{2}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{46}}{\mathrm{15}}−\left(−\gamma\right)\right)−\frac{\mathrm{2}}{\mathrm{25}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{48}}−\frac{\mathrm{259}}{\mathrm{225}}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa11 last updated on 06/Sep/21
great sur
$$\mathrm{great}\:\mathrm{sur} \\ $$

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