TITLE: When is this multiple integral finite?
QUESTION [10 upvotes]: Consider the following integral:
$$
I_k(\alpha)=\int_{[0,1]^k}|x_1-x_2|^{\alpha}|x_2-x_3|^{\alpha}\ldots|x_{k-1}-x_k|^{\alpha}|x_k-x_1|^{\alpha}d\mathbf{x}.
$$
where $k=2,3,4,\ldots$
The question is to find $\beta_k=\inf\{\alpha\mid I_k(\alpha)<\infty\}$.
Remark:
$I_k(\alpha)$ is a decreasing function in $\alpha$.
Obviously $\beta_k\ge-1$. It is also known that $\beta_2=-1/2$,
$\beta_3=-2/3$ and $\beta_k\le-1/2$. These are explained as follows.
The case $k=2$ is trivial.
When $k=3$ and $\alpha>-2/3$, one can use the symmetry of the integrand to derive that $I_3(\alpha)=\frac{2}{(1+\alpha)(2+3\alpha)}\mathrm{B}(1+\alpha,1+\alpha)$, where $\mathrm{B}(\cdot,\cdot)$ is the beta function.
When $k\ge 4$, I don't know any explicit formula. Using Cauchy-Schwarz to separate one factor from the circular integrand, one can derive the bound $I_k(\alpha)\le [(1+2\alpha)(1+\alpha)]^{-k/2}$ for $\alpha>-1/2$.
A weaker question which is also useful for me is to show whether $I_k(-1/2)<\infty$ for $k\ge 3$, or even $k=4$.
The answer should be $\beta_k=−(k−1)/k$.

REPLY [3 votes]: The integral $I_4(-1/2)$ is finite.
Write the integral as
$$I_4(-1/2)=\int_{[0,1]^4}\frac{dp\ dq\ dr\ ds}{\sqrt{\big|(p-q)(q-r)(r-s)(s-p)\big|}}$$
Assume wlog that $p$ is the largest, so
$$\frac{I_4(-1/2)}{4} = \int_{s<r<q<p} + \int_{r<s<q<p} + \int_{s<q<r<p} + \int_{q<s<r<p} + \int_{r<q<s<p} + \int_{q<r<s<p}$$
With Mathematica, most of this evaluates quickly to
$$\frac{I_4(-1/2)}{4} = 3\pi\  +\ \pi^2/4\ +\ \log(4)\ +\ \log(4)\ +\ \int_{r<q<s<p}\ +\ 3\pi$$
So
$$I_4(-1/2) = 24\pi +\pi^2 +8\log(4) +4 \int_{0<r<q<s<p<1}\frac{dp\ dq\ dr\ ds}{\sqrt{(p-q)(q-r)(p-s)(s-r)}}$$
Integrating with respect to $p$ and $r$ reduces the last integral to
$$\int_{0<q<s<1}2\log\bigg(\frac{\sqrt{1-q}+\sqrt{1-s}}{\sqrt{s-q}}\bigg)\log\bigg(
   \frac{\sqrt{s/q}+1}{\sqrt{s/q}-1}\bigg)dq\ ds$$
Finally, that last integral evaluates to $\pi^2/4$, so that
$$I_4(-1/2)=24\pi + 2\pi^2 + 8\log(4).$$ 
(Added by the question poster) Following the observation of fedja, the general answer should be $\beta_k=-(k-1)/k$.