I asked this question on math.stackexchange:


[Does this integral converge when $\frac{1}{p}+\frac{1}{q}\ge1$?](https://math.stackexchange.com/questions/4041263/does-this-integral-converge-when-frac1p-frac1q-geq1)


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.



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


