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-Schwartz 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 $\beta_k=−(k−1)/k$.**