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Question Number 194017 by MrGHK last updated on 26/Jun/23

Answered by witcher3 last updated on 26/Jun/23

$$\mathrm{\forall }(\mathrm{x},\mathrm{y},\mathrm{z})\in {[0,1]}^3,{\left(\mathrm{xyz}\right)}^4⩽1$$$$\frac{{\mathrm{x}}^4{\mathrm{y}}^6{\mathrm{z}}^8}{1-{\mathrm{x}}^4{\mathrm{y}}^4{\mathrm{z}}^4}={\mathrm{x}}^4{\mathrm{y}}^6{\mathrm{z}}^8\mathrm{\Sigma }{\mathrm{x}}^{4\mathrm{n}}{\mathrm{y}}^{4\mathrm{n}}{\mathrm{z}}^{4\mathrm{n}}$$$$=\sum _{\mathrm{n}⩾0}{\mathrm{x}}^{4\mathrm{n}+4}{\mathrm{y}}^{4\mathrm{n}+6}{\mathrm{z}}^{4\mathrm{n}+8}$$$$\mathrm{I}=\int \int {\int }_{{[0,1]}^3}\sum _{\mathrm{n}⩾0}{\mathrm{x}}^{4\mathrm{n}+4}{\mathrm{y}}^{4\mathrm{n}+6}{\mathrm{z}}^{4\mathrm{n}+6}\mathrm{dxdydz}$$$$=\sum _{\mathrm{n}⩾0}\int \int {\int }_{{[0,1]}^3}{\mathrm{x}}^{4\mathrm{n}+4}{\mathrm{y}}^{4\mathrm{n}+6}{\mathrm{z}}^{4\mathrm{n}+8}\mathrm{dxdydz}$$$$\mathrm{cause}\sum _{\mathrm{n}⩾0}{\mathrm{a}}^{\mathrm{n}},\mathrm{cv}\mathrm{uniformaly}\mathrm{over}[-\mathrm{a},\mathrm{a}],\mathrm{\forall }\mathrm{a}\in [0,1[$$$$\mathrm{this}\mathrm{justify}\mathrm{switch}\int \mathrm{and}\mathrm{\Sigma }$$$$\mathrm{I}=\sum _{\mathrm{n}⩾0}{\int }_0^1{\mathrm{x}}^{4\mathrm{n}+4}\mathrm{dx}{\int }_0^1{\mathrm{y}}^{4\mathrm{n}+6}\mathrm{dy}{\int }_0^1{\mathrm{z}}^{4\mathrm{n}+8}\mathrm{dz}$$$$=\sum _{\mathrm{n}⩾0}\frac{1}{(4\mathrm{n}+5)(4\mathrm{n}+7)(4\mathrm{n}+9)}=\mathrm{\Sigma }{\mathrm{u}}_{\mathrm{n}}$$$${\mathrm{u}}_{\mathrm{n}}=\frac{1}{8(4\mathrm{n}+5)}-\frac{1}{4(4\mathrm{n}+7)}+\frac{1}{8(4\mathbf{n}+9)}$$$$\frac{\mathrm{n}}{8}+\frac{\mathrm{n}}{8}-\frac{\mathrm{n}}{4}=0,\mathrm{\forall }\mathrm{n}\in \mathbb{N}$$$${\mathrm{U}}_{\mathrm{n}}=-\frac{1}{32}(-\frac{1}{\mathrm{n}+\frac{5}{4}}+\frac{1}{\mathrm{n}})+\frac{1}{16}(-\frac{1}{\mathrm{n}+\frac{7}{4}}+\frac{1}{\mathrm{n}})-\frac{1}{32}(-\frac{1}{\mathrm{n}+\frac{9}{4}}+\frac{1}{\mathrm{n}})$$$$\mathrm{\Psi }(\mathrm{z}+1)=\frac{1}{\mathrm{z}}+\mathrm{\Psi }\left(\mathrm{z}\right)=-\gamma +\sum _{\mathrm{n}⩾1}\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+\mathrm{z}}$$$$\mathrm{I}=\frac{1}{315}-\frac{1}{32}\sum _{\mathrm{n}⩾1}(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+\frac{5}{4}})+(\frac{1}{\mathrm{n}}-\frac{1}{\mathbf{n}+\frac{9}{4}})+\frac{1}{16}\mathrm{\Sigma }(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+\frac{7}{4}})$$$$=\frac{1}{315}-\frac{1}{32}\mathrm{\Psi }\left(\frac{9}{4}\right)-\frac{1}{32}\mathrm{\Psi }\left(\frac{13}{4}\right)+\frac{1}{16}\mathrm{\Psi }\left(\frac{11}{4}\right)$$$$=\frac{1}{315}-\frac{1}{32}(2(\frac{4}{5}+4+\mathrm{\Psi }\left(\frac{1}{4}\right))+\frac{4}{9})+\frac{1}{16}(\frac{4}{7}+\frac{4}{3}+\mathrm{\Psi }\left(\frac{3}{4}\right)$$$$=\frac{1}{315}-\frac{1}{16}(\frac{24}{5}+\frac{4}{18})+\frac{1}{16}(\mathrm{\Psi }\left(\frac{3}{4}\right)-\mathrm{\Psi }\left(\frac{1}{4}\right))+\frac{1}{16}\left(\frac{40}{21}\right)$$$$\mathrm{\Psi }(1-\mathbf{z})-\mathrm{\Psi }\left(\mathrm{z}\right)=\pi \cot \left(\pi \mathrm{z}\right)$$$$=\frac{1}{315}-\frac{1}{4}(\frac{6}{5}+\frac{1}{18})+\frac{\pi \cot \left(\frac{\pi }{4}\right)}{16}+\frac{5}{42}$$$$=\frac{1}{315}-\frac{113}{360}+\frac{5}{42}+\frac{\pi }{16}$$$$=-\frac{23}{120}+\frac{\pi }{16}$$$$$$$$$$

Commented by witcher3 last updated on 26/Jun/23

$$\mathrm{thank}\mathrm{You}\mathrm{sir}$$

Commented by MrGHK last updated on 26/Jun/23

$$greatsolution$$

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