When is this multiple integral is finite

Question

Consider the following integral: $$ \int_0^\infty \int_0^\infty \int_0^\infty du_1 \, du_2 \, du_3 \, e^{-(u_1+u_2+u_3)} \vert u_1-u_2\vert^\alpha \vert u_2-u_3\vert^\alpha\vert u_3-u_1 \vert^\alpha. $$ The question is to find conditions on $\alpha$ to assure that the integral is finite.

Answer 1

Using $a$ for $\alpha,$ we know $a> -1$ is a necessary condition for convergence of the given integral. I think that can be improved to: $a>-2/3$ is a necessary condition.

Let $V=\{(y,z): 0< y < z < 1/2\}.$ Note that $V\times (0,1)$ is a small part of the full domain of integration. Note also that $e^{-(x+y+z)}$ is bounded below by a positive constant on $V\times (0,1).$ Since we'll be integrating over $V\times (0,1),$ we can ignore the exponential. So consider

$$\tag 1 \int_V |z-y|^a\int_0^1 |x-y|^a|z-x|^a\, dx\, dy\, dz.$$

In the inner integral, let $x=y+ (z-y)t.$ That integral becomes

$$|z-y|^{2a+1}\int_{-y/(z-y)}^{(1-y)/(z-y)} |t|^a|1-t|^a\, dx \ge |z-y|^{2a+1}\int_{0}^{1} |t|^a|1-t|^a\, dx.$$

It follows that $(1)$ is at least a constant times

$$\int_V |z-y|^{3a+1}\, dy\, dz = \int_0^{1/2}\int_0^z |z-y|^{3a+1}\, dy\, dz. $$

The $y$-integral converges iff $3a+ 1> -1$ or $a>-2/3.$ Thus $a>-2/3$ is a necessary condition for convergence of the given integral as claimed.