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Question Number 84680 by M±th+et£s last updated on 15/Mar/20

show that ∫_0 ^1 ∫_0 ^1 ∫_0 ^1 ((log(xyz))/((1+x^2 )(1+y^2 )(1+z^2 ))) dx dy dz=((−3π^2 G)/(16))

$${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left({xyz}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}\:{dx}\:{dy}\:{dz}=\frac{−\mathrm{3}\pi^{\mathrm{2}} {G}}{\mathrm{16}} \\ $$

Answered by mind is power last updated on 15/Mar/20

=∫_0 ^1 ∫_0 ^1 ∫_0 ^1 ((ln(x)+ln(y)+ln(z))/((1+x^2 )(1+y^2 )(1+z^2 )))dxdydz by Symetrie =3∫_0 ^1 ∫_0 ^1 ∫_0 ^1 ((ln(x))/((1+x^2 )(1+y^2 )(1+z^2 )))dxdydz =3∫_0 ^1 ∫_0 ^1 ((dydz)/((1+y^2 )(1+z^2 )))∫_0 ^1 ((ln(x))/((1+x^2 )))dx −G=∫_0 ^1 ((ln(x))/(1+x^2 ))⇒ =−3G∫_0 ^1 ∫_0 ^1 ((dydz)/((1+y^2 )(1+z^2 )))=−3G∫_0 ^1 (dy/((1+y^2 )))∫_0 ^1 (dz/(1+z^2 )) =−3G.((π/4))^2 =((−3π^2 G)/(16))

$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)+{ln}\left({y}\right)+{ln}\left({z}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}{dxdydz} \\ $$$${by}\:{Symetrie} \\ $$$$=\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}{dxdydz} \\ $$$$=\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dydz}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$−{G}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\Rightarrow \\ $$$$=−\mathrm{3}{G}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dydz}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}=−\mathrm{3}{G}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dy}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dz}}{\mathrm{1}+{z}^{\mathrm{2}} } \\ $$$$=−\mathrm{3}{G}.\left(\frac{\pi}{\mathrm{4}}\right)^{\mathrm{2}} =\frac{−\mathrm{3}\pi^{\mathrm{2}} {G}}{\mathrm{16}} \\ $$

Commented by M±th+et£s last updated on 15/Mar/20

thank you so much sir

$${thank}\:{you}\:{so}\:{much}\:{sir}\: \\ $$

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