I asked this question on math.stackexchange:

Does this integral converge when $\frac{1}{p}+\frac{1}{q}\ge1$?

No answers or very useful comments there. May be it is more appropraite for mathoverflow.

Fix a small $\delta>0$ and let $p,q>1$. Consider the integral

$$I(p,q):=\int\limits_{1-\delta}^{1+\delta} \int\limits_{y/2}^{2y}\frac{1}{|y-x|^{\frac{1}{p}}|1-x|^{\frac{1}{q}}} \,\mathrm{d}x\,\mathrm{d}y. $$

I am trying to show that $I(p,q)$ diverges if $\frac{1}{p}+\frac{1}{q}\geq 1$. I am not sure this is even the case ? Any hints on how to handle this?

Remark: This seems to be related to the failure of the Hardy-Littlewood-Sobolev inequality (HLS) at the endpoint $p=1$. HLS reads:

If $1<p,q<\infty$, $f\in L^p$ and

$$Tf(x):=\int_{\mathbb{R}^n} \frac{f(y)}{|x-y|^{\gamma}}dy$$

Then $$\|Tf\|_q\leq \|f\|_p$$ if and only if $$\frac{1}{p}-\frac{1}{q}=1-\frac{\gamma}{n}.$$

Many thanks.

  • 1
    $\begingroup$ The numeric calculation with Mathematica 12.2 NIntegrate[ 1/RealAbs[y - x]^(1/2)/RealAbs[1 - x]^(2/3), {y, 3/4, 5/4}, {x, y/2, 2 y}, Exclusions -> {y == x}, AccuracyGoal -> 4, PrecisionGoal -> 4] results in $8.66016$, confirming the convergence. $\endgroup$ – user64494 Feb 27 at 12:33
  • $\begingroup$ This comment is very useful to me. When I numerically-tested singular integrals, I used to isolate the singularity manually, then manually decrease the size of the isolated neighborhood. Thank you. $\endgroup$ – Medo Feb 27 at 18:57

It converges whenever $p$ and $q$ are both greater than 1. For fixed $x$, the integral against $y$ of $|y-x|^{-1/p}$ is uniformly bounded. Then, $\int_{0}^2 |1-x|^{-1/q}dx$ converges.


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