# A question on a simple integral with a singular kernel?

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 $$1, $$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.

• 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. – user64494 Feb 27 at 12:33
• 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. – 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.