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 $\beta\_k=\inf\{\alpha\mid I\_k(\alpha)<\infty\}$.**
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**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 $\beta\_k=−(k−1)/k$.**



When is this multiple integral finite?
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 $\beta\_k=-(k-1)/k$.


