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Does-anyone-know-how-to-prove-this-V-dxdydz-1-x-4-y-4-z-4-4-1-4-4-4-where-V-is-the-unit-cube-0-1-3-Thankyou-

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Question Number 197376 by megrex last updated on 15/Sep/23
Does anyone know how to prove this? ∫∫∫_V ((dxdydz)/(1+x^4 +y^4 +z^4 )) =((Γ^4 ((1/4)))/4^4 ) where V is the unit cube [0,1]^3 Thankyou.
$${Does}\:{anyone}\:{know}\:{how}\:{to}\:{prove}\:{this}? \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\int\int_{{V}} \:\frac{{dxdydz}}{\mathrm{1}+{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} }\:=\frac{\Gamma^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{4}^{\mathrm{4}} } \\ $$$${where}\:{V}\:{is}\:{the}\:{unit}\:{cube}\:\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{3}} \\ $$$${Thankyou}. \\ $$$$ \\ $$
Answered by witcher3 last updated on 15/Sep/23
(1/(1+x^4 +y^4 +z^4 ))=∫_0 ^∞ e^(−t(1+x^4 +y^4 +z^4 )) dt I=∫∫∫_V ((dxdydz)/(1+x^4 +y^4 +z^4 ))=∫∫∫_V ∫_0 ^∞ e^(−t(1+x^4 +y^4 +z^4 )) dt =∫_0 ^∞ e^(−t) {∫_0 ^1 e^(−tx^4 ) dx∫_0 ^1 e^(−ty^4 ) dy∫_0 ^1 e^(−tz^4 ) dz}dt ∫_0 ^1 e^(−ts^4 ) ds,t∈[0,∞[,∫_0 ^∞ e^(−y) =(1/4)∫_0 ^∞ (y^((1/4)−1) /( (t)^(1/4) ))e^(−y) dy =(1/(4(t)^(1/4) )).Γ((1/4)) I=∫_0 ^∞ e^(−t) {((Γ((1/4)))/(4t^(1/4) ))}^3 =((Γ^3 ((1/4)))/4^3 )∫_0 ^∞ t^(−(3/4)) e^(−t) dt =((Γ^3 ((1/4)))/4^3 )Γ((1/4))=((Γ^4 ((1/4)))/4^3 )...may bee error of calculus I didint find 4^4
$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} +\mathrm{y}^{\mathrm{4}} +\mathrm{z}^{\mathrm{4}} }=\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{t}\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} +\mathrm{y}^{\mathrm{4}} +\mathrm{z}^{\mathrm{4}} \right)} \mathrm{dt} \\ $$$$\mathrm{I}=\int\int\int_{\mathrm{V}} \frac{\mathrm{dxdydz}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} +\mathrm{y}^{\mathrm{4}} +\mathrm{z}^{\mathrm{4}} }=\int\int\int_{\mathrm{V}} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{t}\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} +\mathrm{y}^{\mathrm{4}} +\mathrm{z}^{\mathrm{4}} \right)} \mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{t}} \left\{\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{−\mathrm{tx}^{\mathrm{4}} } \mathrm{dx}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{−\mathrm{ty}^{\mathrm{4}} } \mathrm{dy}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{−\mathrm{tz}^{\mathrm{4}} } \mathrm{dz}\right\}\mathrm{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{−\mathrm{ts}^{\mathrm{4}} } \mathrm{ds},\mathrm{t}\in\left[\mathrm{0},\infty\left[,\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{y}} \right.\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\:\sqrt[{\mathrm{4}}]{\mathrm{t}}}\mathrm{e}^{−\mathrm{y}} \mathrm{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{t}}}.\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{t}} \left\{\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{4t}^{\frac{\mathrm{1}}{\mathrm{4}}} }\right\}^{\mathrm{3}} =\frac{\Gamma^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{4}^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} \mathrm{t}^{−\frac{\mathrm{3}}{\mathrm{4}}} \mathrm{e}^{−\mathrm{t}} \mathrm{dt} \\ $$$$=\frac{\Gamma^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{4}^{\mathrm{3}} }\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\Gamma^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{4}^{\mathrm{3}} }…\mathrm{may}\:\mathrm{bee}\:\mathrm{error}\:\mathrm{of}\:\mathrm{calculus} \\ $$$$\mathrm{I}\:\mathrm{didint}\:\mathrm{find}\:\mathrm{4}^{\mathrm{4}} \\ $$$$ \\ $$
Commented by megrex last updated on 16/Sep/23
Very nicely done − thankyou for your time (I will double check my extra factor). This could easily be generalised to a d−dimensional cube with arbitrary powers of x^p −which is really nice!
$${Very}\:{nicely}\:{done}\:−\:{thankyou}\:{for}\:{your}\:{time}\: \\ $$$$\left({I}\:{will}\:{double}\:{check}\:{my}\:{extra}\:{factor}\right). \\ $$$${This}\:{could}\:{easily}\:{be}\:{generalised}\:{to}\: \\ $$$${a}\:{d}−{dimensional}\:{cube}\:{with} \\ $$$$\:{arbitrary}\:{powers}\:{of}\:{x}^{{p}} \:\: \\ $$$$\:−{which}\:{is}\:{really}\:{nice}! \\ $$$$ \\ $$$$ \\ $$
Commented by witcher3 last updated on 16/Sep/23
withe Pleasur
$$\mathrm{withe}\:\mathrm{Pleasur} \\ $$

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