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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 $\beta_k=−(k−1)/k$.

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    $\begingroup$ Note that for $x,y\in[0,1]$ and $\alpha,\beta>0,\alpha+\beta>1$, we have $\int_{[0,1]}|x-t|^{-\alpha}|y-t|^{-\beta}\,dt\approx |x-y|^{-(\alpha+\beta-1)}$. The rest should be clear. $\endgroup$
    – fedja
    Commented May 9, 2014 at 12:25
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    $\begingroup$ Well, haven't you noticed yourself that the integral is monotone in $\alpha$? ;) $\endgroup$
    – fedja
    Commented May 9, 2014 at 12:42
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    $\begingroup$ Even $I_4(-1/2)$ is difficult, since it includes $\int 1/\sqrt{(p-q)(p-r)(q-s)(r-s)}$ over $0<s<r<q<p<1$. I evaluated this with Mathematica at ~2.4674, but only after several attempts, and without getting a closed-form expression. $\endgroup$
    – user44143
    Commented May 9, 2014 at 14:39
  • $\begingroup$ Thanks Matt. Using symmetry and letting, e.g., $x_k$ be the largest among $x_i$'s, one can do a change of variable to reduce one dimension of the integration though. $\endgroup$
    – Uchiha
    Commented May 9, 2014 at 15:12
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    $\begingroup$ @Ray Bai. But $I_k(-1/2-\epsilon)<+\infty$ implies $I_k(-1/2)<+\infty$ and what I said is enough to establish that. The case $\alpha+\beta=1$ can also be treated if you do not mind some stupid logarithmic factors popping up everywhere (the best way to deal with them is just to estimate them by very small negative powers). $\endgroup$
    – fedja
    Commented May 9, 2014 at 16:22

1 Answer 1

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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 $\beta_k=-(k-1)/k$.

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  • $\begingroup$ Thank you for the work you have done for my question, Matt. Following fedja's suggestion I think the question could be answered. But I would accept your answer to thank you for such an effort. $\endgroup$
    – Uchiha
    Commented May 9, 2014 at 17:02
  • $\begingroup$ @RayBai, thanks for the acceptance -- the calculations amuse me. $\endgroup$
    – user44143
    Commented May 9, 2014 at 17:09

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