Proving $\int_{-1}^1 \frac{1}{|x-y|^{2/3}} \frac{1}{|x|^{1/3}} dx \approx -2\log y$ as $y\to 0$ I want to show that
$$\int_{-1}^1 \frac{1}{|x-y|^{2/3}} \frac{1}{|x|^{1/3}} dx \approx -2\log y$$
as $y\to 0$. Here, $f(y) \approx g(y)$ means that $\lim_{y\to 0} \frac{f(y)}{g(y)}=1$.
By computing the integral and finding asymptotic expression using Mathematica, I think the above is true. (The exact value of the integral is a very complicated function involving hypergeometric functions.) But, how can I prove this asymptotic expression in a clever way?

With $x = ty$ and $s=-t$, we find
\begin{align*}
&\int_{ - 1/y}^{1/y} {\frac{1}{{\left| {t - 1} \right|^{2/3} }}\frac{1}{{\left| t \right|^{1/3} }}dt}  = \int_1^{1/y} {\frac{1}{{(t - 1)^{2/3} }}\frac{1}{{t^{1/3} }}dt}  + \int_{ - 1/y}^0 {\frac{1}{{(1 - t)^{2/3} }}\frac{1}{{( - t)^{1/3} }}dt} \\ & \quad + \int_0^1 {\frac{1}{{(1 - t)^{2/3} }}\frac{1}{{t^{1/3} }}dt} 
\\ & = \int_1^{1/y} {\frac{1}{{(t - 1)^{2/3} }}\frac{1}{{t^{1/3} }}dt}  + \int_0^{1/y} {\frac{1}{{(1 + s)^{2/3} }}\frac{1}{{s^{1/3} }}ds}  + \frac{{2\pi }}{{\sqrt 3 }}
\\ & = \int_1^{1/y} {\frac{1}{{(t - 1)^{2/3} }}\frac{1}{{t^{1/3} }}dt}  + \int_0^1 {\frac{1}{{(1 + s)^{2/3} }}\frac{1}{{s^{1/3} }}ds}  + \int_1^{1/y} {\frac{1}{{(1 + s)^{2/3} }}\frac{1}{{s^{1/3} }}ds}  + \mathcal{O}(1)
\\ & = \int_1^{1/y} {\frac{1}{{(t - 1)^{2/3} }}\frac{1}{{t^{1/3} }}dt}  + \int_1^{1/y} {\frac{1}{{(1 + s)^{2/3} }}\frac{1}{{s^{1/3} }}ds}  + \mathcal{O}(1)
\\ & = \int_1^{1/y} {\frac{{dt}}{t}}  + \int_1^{1/y} {\frac{{ds}}{s}}  + \int_1^{1/y} {\left( {\frac{1}{{(t - 1)^{2/3} }} - \frac{1}{{t^{2/3} }}} \right)\frac{1}{{t^{1/3} }}dt} \\ &  \quad + \int_1^{1/y} {\left( {\frac{1}{{(1 + s)^{2/3} }} - \frac{1}{{s^{2/3} }}} \right)\frac{1}{{s^{1/3} }}ds}  + \mathcal{O}(1)
\\ & =  - 2\log y + \int_1^{1/y} {\left( {\frac{1}{{(t - 1)^{2/3} }} - \frac{1}{{t^{2/3} }}} \right)\frac{1}{{t^{1/3} }}dt} \\ & \quad + \int_1^{1/y} {\left( {\frac{1}{{(1 + s)^{2/3} }} - \frac{1}{{s^{2/3} }}} \right)\frac{1}{{s^{1/3} }}ds}  + \mathcal{O}(1).
\end{align*}
Now by the mean value theorem
$$
\left| {\frac{1}{{(t - 1)^{2/3} }} - \frac{1}{{t^{2/3} }}} \right| \le \frac{2}{{3(t - 1)^{5/3} }}
$$
and
$$
\left| {\frac{1}{{(1 + s)^{2/3} }} - \frac{1}{{s^{2/3} }}} \right| \le \frac{2}{{3s^{5/3} }}.
$$
Using these for $t>2$, say, and $s>1$, we see that the two integrals are $\mathcal{O}(1)$. Hence the final result is $-2\log y+ \mathcal{O}(1)$ as $y\to 0+$.
