Question Number 194017 by MrGHK last updated on 26/Jun/23
Answered by witcher3 last updated on 26/Jun/23
![∀(x,y,z)∈[0,1]^3 ,(xyz)^4 ≤1 ((x^4 y^6 z^8 )/(1−x^4 y^4 z^4 ))=x^4 y^6 z^8 Σx^(4n) y^(4n) z^(4n) =Σ_(n≥0) x^(4n+4) y^(4n+6) z^(4n+8) I=∫∫∫_([0,1]^3 ) Σ_(n≥0) x^(4n+4) y^(4n+6) z^(4n+6) dxdydz =Σ_(n≥0) ∫∫∫_([0,1]^3 ) x^(4n+4) y^(4n+6) z^(4n+8) dxdydz cause Σ_(n≥0) a^n ,cv uniformaly over [−a,a],∀a∈[0,1[ this justify switch ∫ and Σ I=Σ_(n≥0) ∫_0 ^1 x^(4n+4) dx∫_0 ^1 y^(4n+6) dy∫_0 ^1 z^(4n+8) dz =Σ_(n≥0) (1/((4n+5)(4n+7)(4n+9)))=Σu_n u_n =(1/(8(4n+5)))−(1/(4(4n+7)))+(1/(8(4n+9))) (n/8)+(n/8)−(n/4)=0,∀n∈N U_n =−(1/(32))(−(1/(n+(5/4)))+(1/n))+(1/(16))(−(1/(n+(7/4)))+(1/n))−(1/(32))(−(1/(n+(9/4)))+(1/n)) Ψ(z+1)=(1/z)+Ψ(z)=−γ+Σ_(n≥1) (1/n)−(1/(n+z)) I=(1/(315))−(1/(32))Σ_(n≥1) ((1/n)−(1/(n+(5/4))))+((1/n)−(1/(n+(9/4))))+(1/(16))Σ((1/n)−(1/(n+(7/4)))) =(1/(315))−(1/(32))Ψ((9/4))−(1/(32))Ψ(((13)/4))+(1/(16))Ψ(((11)/4)) =(1/(315))−(1/(32))(2((4/5)+4+Ψ((1/4)))+(4/9))+(1/(16))((4/7)+(4/3)+Ψ((3/4)) =(1/(315))−(1/(16))(((24)/5)+(4/(18)))+(1/(16))(Ψ((3/4))−Ψ((1/4)))+(1/(16))(((40)/(21))) Ψ(1−z)−Ψ(z)=πcot(πz) =(1/(315))−(1/4)((6/5)+(1/(18)))+((πcot((π/4)))/(16))+(5/(42)) =(1/(315))−((113)/(360))+(5/(42))+(π/(16)) =−((23)/(120))+(π/(16))](https://www.tinkutara.com/question/Q194025.png)
$$\forall\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\in\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{3}} ,\left(\mathrm{xyz}\right)^{\mathrm{4}} \leqslant\mathrm{1} \\ $$$$\frac{\mathrm{x}^{\mathrm{4}} \mathrm{y}^{\mathrm{6}} \mathrm{z}^{\mathrm{8}} }{\mathrm{1}−\mathrm{x}^{\mathrm{4}} \mathrm{y}^{\mathrm{4}} \mathrm{z}^{\mathrm{4}} }=\mathrm{x}^{\mathrm{4}} \mathrm{y}^{\mathrm{6}} \mathrm{z}^{\mathrm{8}} \Sigma\mathrm{x}^{\mathrm{4n}} \mathrm{y}^{\mathrm{4n}} \mathrm{z}^{\mathrm{4n}} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\mathrm{x}^{\mathrm{4n}+\mathrm{4}} \mathrm{y}^{\mathrm{4n}+\mathrm{6}} \mathrm{z}^{\mathrm{4n}+\mathrm{8}} \\ $$$$\mathrm{I}=\int\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{3}} } \underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\mathrm{x}^{\mathrm{4n}+\mathrm{4}} \mathrm{y}^{\mathrm{4n}+\mathrm{6}} \mathrm{z}^{\mathrm{4n}+\mathrm{6}} \mathrm{dxdydz} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\int\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{3}} } \mathrm{x}^{\mathrm{4n}+\mathrm{4}} \mathrm{y}^{\mathrm{4n}+\mathrm{6}} \mathrm{z}^{\mathrm{4n}+\mathrm{8}} \mathrm{dxdydz} \\ $$$$\mathrm{cause}\:\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\mathrm{a}^{\mathrm{n}} ,\mathrm{cv}\:\mathrm{uniformaly}\:\mathrm{over}\:\left[−\mathrm{a},\mathrm{a}\right],\forall\mathrm{a}\in\left[\mathrm{0},\mathrm{1}\left[\right.\right. \\ $$$$\mathrm{this}\:\mathrm{justify}\:\mathrm{switch}\:\int\:\mathrm{and}\:\Sigma \\ $$$$\mathrm{I}=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{4n}+\mathrm{4}} \mathrm{dx}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{y}^{\mathrm{4n}+\mathrm{6}} \mathrm{dy}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{z}^{\mathrm{4n}+\mathrm{8}} \mathrm{dz} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{4n}+\mathrm{5}\right)\left(\mathrm{4n}+\mathrm{7}\right)\left(\mathrm{4n}+\mathrm{9}\right)}=\Sigma\mathrm{u}_{\mathrm{n}} \\ $$$$\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{4n}+\mathrm{5}\right)}−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{4n}+\mathrm{7}\right)}+\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{4}\boldsymbol{\mathrm{n}}+\mathrm{9}\right)} \\ $$$$\frac{\mathrm{n}}{\mathrm{8}}+\frac{\mathrm{n}}{\mathrm{8}}−\frac{\mathrm{n}}{\mathrm{4}}=\mathrm{0},\forall\mathrm{n}\in\mathbb{N} \\ $$$$\mathrm{U}_{\mathrm{n}} =−\frac{\mathrm{1}}{\mathrm{32}}\left(−\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{5}}{\mathrm{4}}}+\frac{\mathrm{1}}{\mathrm{n}}\right)+\frac{\mathrm{1}}{\mathrm{16}}\left(−\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{7}}{\mathrm{4}}}+\frac{\mathrm{1}}{\mathrm{n}}\right)−\frac{\mathrm{1}}{\mathrm{32}}\left(−\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{9}}{\mathrm{4}}}+\frac{\mathrm{1}}{\mathrm{n}}\right) \\ $$$$\Psi\left(\mathrm{z}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{z}}+\Psi\left(\mathrm{z}\right)=−\gamma+\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{z}} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{315}}−\frac{\mathrm{1}}{\mathrm{32}}\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{5}}{\mathrm{4}}}\right)+\left(\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}+\frac{\mathrm{9}}{\mathrm{4}}}\right)+\frac{\mathrm{1}}{\mathrm{16}}\Sigma\left(\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{7}}{\mathrm{4}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{315}}−\frac{\mathrm{1}}{\mathrm{32}}\Psi\left(\frac{\mathrm{9}}{\mathrm{4}}\right)−\frac{\mathrm{1}}{\mathrm{32}}\Psi\left(\frac{\mathrm{13}}{\mathrm{4}}\right)+\frac{\mathrm{1}}{\mathrm{16}}\Psi\left(\frac{\mathrm{11}}{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{315}}−\frac{\mathrm{1}}{\mathrm{32}}\left(\mathrm{2}\left(\frac{\mathrm{4}}{\mathrm{5}}+\mathrm{4}+\Psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right)+\frac{\mathrm{4}}{\mathrm{9}}\right)+\frac{\mathrm{1}}{\mathrm{16}}\left(\frac{\mathrm{4}}{\mathrm{7}}+\frac{\mathrm{4}}{\mathrm{3}}+\Psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{315}}−\frac{\mathrm{1}}{\mathrm{16}}\left(\frac{\mathrm{24}}{\mathrm{5}}+\frac{\mathrm{4}}{\mathrm{18}}\right)+\frac{\mathrm{1}}{\mathrm{16}}\left(\Psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right)+\frac{\mathrm{1}}{\mathrm{16}}\left(\frac{\mathrm{40}}{\mathrm{21}}\right) \\ $$$$\Psi\left(\mathrm{1}−\boldsymbol{\mathrm{z}}\right)−\Psi\left(\mathrm{z}\right)=\pi\mathrm{cot}\left(\pi\mathrm{z}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{315}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{6}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{18}}\right)+\frac{\pi\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}\right)}{\mathrm{16}}+\frac{\mathrm{5}}{\mathrm{42}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{315}}−\frac{\mathrm{113}}{\mathrm{360}}+\frac{\mathrm{5}}{\mathrm{42}}+\frac{\pi}{\mathrm{16}} \\ $$$$=−\frac{\mathrm{23}}{\mathrm{120}}+\frac{\pi}{\mathrm{16}} \\ $$$$ \\ $$$$ \\ $$
Commented by witcher3 last updated on 26/Jun/23
$$\mathrm{thank}\:\mathrm{You}\:\mathrm{sir} \\ $$
Commented by MrGHK last updated on 26/Jun/23
$${great}\:{solution} \\ $$