$$(1)=\int_{n-1}^{\infty}\left(\int_{1}^{\infty}\cdots\int_{1}^{\infty}(v_1\cdots v_{n-1})^{-\beta-1}[v_1+\cdots+v_{n-1}\geq u] \;dv_1\ldots dv_{n-1}\right)du,$$ for any integer $n>1$ and a fixed $1<\beta<2$, where $[ \cdot ]$ denotes the Iverson bracket, i.e. $[ P ]=1$ if $P$ is true and zero otherwise. I am wondering if the integral is bounded. I have computed it explicitly for $n=2,3$, (see below) but it might be the case that it is unbounded for higher $n$.
The obvious upper bounds I have tried give infinity, such as $[ \cdot ]\leq1$. Similarly, trying to find a lower bound that ``blows up'' is difficult. I have tried bounding the Iverson bracket in (1) by $[ v_i\geq u/(n-1)\forall 1\leq i\leq n-1 ]$. Notice that the Iverson bracket in (1) can be replaced with the Heavyside function $H(v_1+\cdots+v_{n-1} - u )$. I have tried bounds using smooth approximations of the Heavyside function but these haven't led anywhere.
For $n=2$, we get $$(1)=\int_{1}^{\infty}\int_{1}^{\infty}v^{-\beta-1}[v>u]\;dvdu=\int_{1}^{\infty}\int_{1}^{\infty}v^{-\beta-1}\;dvdu=\frac{1}{\beta(\beta-1)}$$ For $n=3$, we get $$(1)=\int_{2}^{\infty}\left(\int_{1}^{\infty}\int_{u-v_2}^{\infty}(v_1v_2)^{-\beta-1}dv_1dv_2+\int_{u-1}^{\infty}\int_{1}^{\infty}(v_1v_2)^{-\beta-1} dv_1dv_2\right)du=\frac{2^{1-\beta}}{\beta(\beta-1)}+\frac{1}{\beta^2(\beta-1)}.$$
Thanks, in advance for the help.
We can switch the integral and observe that for any fixed $(v_1,\dots,v_{n-1})$ such that $v_1+\dots+v_{n-1}\geqslant u$, we have $$\int_{n-1}^{ +\infty}\left[v_1+\dots+v_{n-1}\geqslant u\right] \mathrm du =v_1+\dots+v_{n-1}-(n-1).$$ It thus suffices to compute the integrals $$I_n=\iint_{[1, +\infty)^{ n-1} }(v_1\dots v_{n-1})^{-\beta-1} v_{n-1}\mathrm dv_1\dots\mathrm dv_{n-1} \mbox{ and} $$
$$J_n=\iint_{[1, +\infty)^{ n-1} }(v_1\dots v_{n-1})^{-\beta-1} \mathrm dv_1\dots\mathrm dv_{n-1} $$ since, by symmetry, $$(1)=(n-1)I_n+(n-1)J_n.$$