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Question Number 114635 by mathmax by abdo last updated on 20/Sep/20

$$\mathrm{find}{\int }_0^1{\mathrm{x}}^4\ln \left(\mathrm{x}\right)\ln (1-{\mathrm{x}}^2)\mathrm{dx}$$

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

$$\mathrm{great}\mathrm{sir}$$

Answered by maths mind last updated on 20/Sep/20

$$ln(1-x^2)=-\sum _{k⩾0}\frac{x^{2k+2}}{k+1}$$$$={\int }_0^1\sum _{k⩾0}\frac{x^{2k+6}}{(k+1)}ln\left(x\right)dx$$$$=-\sum _{k⩾0}\frac{1}{k+1}{\int }_0^1{x}^{2k+6}ln\left(x\right)dx$$$$=\sum _{k⩾0}\frac{1}{(k+1){(2k+7)}^2}$$$$\mathrm{\Sigma }(\frac{1}{25(k+1)}-\frac{2}{5{(2k+7)}^2}+\frac{a}{(2k+7)})$$$$=\left(\frac{{(2k+7)}^2-10(k+1)+25a(k+1)(2k+7)}{25(k+1){(2k+7)}^2}\right)$$$$50a=-4\Rightarrow$$$$=\frac{1}{25}\sum _{k⩾0}(\frac{1}{k+1}-\frac{10}{{(2k+7)}^2}-\frac{2}{(2k+7)})$$$$=\frac{1}{25}\sum _{k⩾0}(\frac{1}{k+1}-\frac{2}{2k+7})-2\sum _{k⩾0}\frac{1}{{(2k+7)}^2}$$$$=\frac{1}{25}\mathrm{\Sigma }\left(\frac{2k+7-2k-2}{(2k+7)(k+1)}\right)-\frac{2}{25}\sum _{k⩾0}\frac{1}{{(2k+7)}^2}$$$$=\frac{1}{5}\sum _{k⩾0}\frac{1}{2(k+\frac{7}{2})(k+1)}-\frac{2}{25}\sum _{k⩾0}\frac{1}{{(2(k+3)+1)}^2}$$$$=\frac{1}{15}\sum _{k⩾0}\frac{\frac{7}{2}-1}{(k+\frac{7}{2})(k+1)}-\frac{2}{25}\sum _{k⩾3}\frac{1}{{(2k+1)}^2}$$$$=\frac{1}{15}(\mathrm{\Psi }\left(\frac{7}{2}\right)-\mathrm{\Psi }\left(1\right))-\frac{2}{25}(\left(\sum _{k⩾0}\frac{1}{{(2k+1)}^2}\right)-1-\frac{1}{9}-\frac{1}{25})$$$$\mathrm{\Psi }\left(\frac{7}{2}\right)=\mathrm{\Psi }\left(\frac{1}{2}\right)+\frac{2}{5}+\frac{2}{3}+2=\mathrm{\Psi }\left(\frac{1}{2}\right)+\frac{46}{15}$$$$\sum _{k⩾0}{\left(\frac{1}{2k+1}\right)}^2=\frac{3}{4}\zeta \left(2\right)=\frac{{\pi }^2}{8}$$$$\mathrm{\Psi }\left(\frac{1}{2}\right)=-\gamma -2ln\left(2\right)$$$$weget$$$$\frac{1}{15}(-\gamma -2ln\left(2\right)+\frac{46}{15}-(-\gamma ))-\frac{2}{25}(\frac{{\pi }^2}{48}-\frac{259}{225})$$$$$$$$$$$$$$$$$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sur}$$

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