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$$\iiint'_{\mathbb{R}^3} \frac{du \; dv \; dw}{|u^3+v^3+w^3|^{\frac{2}{3}}}$$ where the $'$ integration constraint is that only $u,v,w$ for which $|(u^3+v^3)(u^3+w^3)(v^3+w^3)| \leq 1 $ are taken into account. Is it possible to determine whether this integral converges or not?

Do you think that under the transformation $x:=u^3+v^3,y:=u^3+w^3$ etc.. one could even compute a value for the integral ?

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Did this integral come from somewhere...? –  anon Aug 25 '11 at 1:39
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1 Answer

First of all, the integration region looks nasty:

Visualization of integration region as stated

As you suggest, let $x=u^3+v^3$, $y=u^3+w^3$ and $z=v^3+w^3$. Then

$$ \mathrm{d} u \,\, \mathrm{d} v \,\, \mathrm{d} w = \frac{4 \mathrm{d} x \,\, \mathrm{d} y \,\, \mathrm{d} z }{ ( (x+y-z)^2 (x+z-y)^2 (y+z-x)^2 )^{1/3}} $$ The region becomes $\vert x y z \vert < 1$, and integral becomes

$$ \int_{( x \cdot y \cdot z )^2 < 1} \frac{4 \mathrm{d} x \,\, \mathrm{d} y \,\, \mathrm{d} z }{ ( 4(x+y+z)^2 (x+y-z)^2 (x+z-y)^2 (y+z-x)^2 )^{1/3}} $$

From this form it seems like the integral would diverge at the origin, since the integrand scales as as $t^{-8/3}$ if all coordinates are rescaled by scale factor $t$.


Added: The integrand, as well as the region of integration, is symmetric with respect to each of $x\to-x$, $y \to -y$, $z \to -z$ as well as any permutations of $(x, y, z)$. Hence we can restrict the integration to $0 < x < y < z$ for the sake of determining the convergence. Consider $$ \int_0^\infty \mathrm{d} x \int_x^\infty \mathrm{d} y \int_y^\infty \mathrm{d} z \frac{\chi_{x y z < 1}}{{ ( (x+y+z)^2 (x+y-z)^2 (x+z-y)^2 (y+z-x)^2 )^{1/3}} } $$ In this region $x y z < 1$ implies $0<x<1 \land x < y < \frac{1}{\sqrt{x}} \land y < z < \frac{1}{x y}$:

$$ \int_0^1 \mathrm{d} x \int_x^{\frac{1}{\sqrt{x}}} \mathrm{d} y \int_y^{\frac{1}{x y}} \mathrm{d} z \frac{1}{{ ( (x+y+z)^2 (x+y-z)^2 (x+z-y)^2 (y+z-x)^2 )^{1/3}} } $$ The only factor in the denominator which could vanish here is $(x+y-z)^2$. This occurs in the following sub-region: $$ 0 < x < \frac{1}{\sqrt[3]{2}} \land x < y < -\frac{x}{2} + \sqrt{\frac{1}{x} + \frac{x^2}{4}} $$

Numerically, the integral does seem to diverge, but the proof escape me.

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The scaling argument at the end of your post escapes me. With this reasoning, the function |x|^(-8/3) would not be integrable around the origin in R^3... but it is. –  Did Aug 24 '11 at 22:34
    
@Didier Thank you for your comment, I thought about my last argument on the way from work, and also concluded that it is invalid. –  Sasha Aug 24 '11 at 23:26
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Man, that looks scary as heck indeed... :o –  J. M. Aug 25 '11 at 3:10
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