Is it possible to show this Integral identity, by assuming the hyposeses I have made?

Question

Assume there exists $p>2$ such that $B\in L^p_{\rm loc}(\mathbb{R}^2)$. Assume in addition that there exists $\tau >2$ such that \begin{equation} \label{B-decay-cond} |B(x)| \, = \, O(|x|^{-\tau}) \qquad as \quad|x| \to \infty. \end{equation} We now define \begin{equation} \label{h-eq} h(x) =- \frac{1}{2\pi} \int_{\mathbb{R}^2} B(y) \log |x-y|\, dy, \end{equation} so that the vector field given by \begin{equation} \label{A-h-def} A_h = (\partial_{x_2} h, - \partial_{x_1} h) \end{equation} satisfies $\nabla\times A_h = B$. Thanks to Assumption above we have \begin{equation} \label{A-L-infty} |A_h| \in L^\infty(\mathbb{R}^2). \end{equation} where $\alpha$ is defined as \begin{equation} \alpha= \frac{1}{2\pi} \int _{\mathbb{R}^2}B\, dx \end{equation} Notice that \begin{equation} \label{exp-h-asymp} e^{h(x)} = |x|^{-\alpha}\big(1+ O(|x|^{-1})\big) , \qquad as \quad |x|\to \infty. \end{equation} I'm struggling to show the above identity, for convenience I'll call it Integral Identity (II). I thought I could proceed in this way: $$ \exp\left\{- \frac{1}{2\pi} \int_{\mathbb{R}^2} B(y) \log |x-y|\, dy\right\} $$ where $$ - \frac{1}{2\pi} \int_{\mathbb{R}^2} B(y) \log |x-y|\, dy=- \frac{1}{2\pi} \int_{K} B(y) \log |x-y|\, dy- \frac{1}{2\pi} \int_{\mathbb{R}^2\setminus K} B(y) \log |x-y|\, dy $$ Let's call $$ I_1=\frac{1}{2\pi} \int_{K} B(y) \log |x-y|\, dy; \quad I_2=\frac{1}{2\pi} \int_{\mathbb{R}^2\setminus K} B(y) \log |x-y|\, dy $$ Either $I_1$, or $I_2$ are convergent because of the assumption. Then because of the second part of the assumption one gets

$$ I_2=\frac{1}{2\pi} \int_{\mathbb{R}^2\setminus K} B(y) \log |x-y|\, dy, \quad \text{where}\quad B(y)\log|x-y|)=O(|y|^{-\tau})\log|x-y| \quad \text{as} \quad |x|\to\infty $$ Then, I'm stuck...

Answer 1

Eventually I reckon this is a possible answer:

To show the asymptotic behavior of $ e^{h(x)} $ as $ |x| \to \infty $, we need to analyze the integral expression for $ h(x) $ given by

$$ h(x) = - \frac{1}{2\pi} \int_{\mathbb{R}^2} B(y) \log |x-y|\, dy. $$

Step-by-Step Solution:

1. Asymptotic Expansion of $ \log |x-y| $:

   For large $ |x| $, we can use the expansion: $$ \log |x-y| = \log |x| + \log \left| 1 - \frac{y}{x} \right|. $$ Since $ \Big| \dfrac{y}{x} \Big| $ is small for large $ |x| $, we can approximate: $$ \log \left| 1 - \frac{y}{x} \right| \approx - \frac{y \cdot x}{|x|^2}. $$

2. Substitute the Expansion into $ h(x) $:

   $$ h(x) = - \frac{1}{2\pi} \int_{\mathbb{R}^2} B(y) \left( \log |x| + \log \left| 1 - \frac{y}{x} \right| \right) dy. $$

   This separates into two integrals: $$ h(x) = - \frac{1}{2\pi} \log |x| \int_{\mathbb{R}^2} B(y) \, dy - \frac{1}{2\pi} \int_{\mathbb{R}^2} B(y) \log \left| 1 - \frac{y}{x} \right| dy. $$

3. Evaluate the First Integral:

   By definition, the first integral is: $$ \int_{\mathbb{R}^2} B(y) \, dy = 2\pi \alpha. $$

   Therefore, $$ - \frac{1}{2\pi} \log |x| \int_{\mathbb{R}^2} B(y) \, dy = - \frac{1}{2\pi} \log |x| \cdot 2\pi \alpha = - \alpha \log |x|. $$

4. Evaluate the Second Integral:

   For the second integral, note that $ \log \left| 1 - \frac{y}{x} \right| $ is small for large $ |x| $. Hence, the integral: $$ \int_{\mathbb{R}^2} B(y) \log \left| 1 - \frac{y}{x} \right| dy $$ will be of order $ O(|x|^{-1}) $ due to the decay condition $ |B(y)| = O(|y|^{-\tau}) $ with $ \tau > 2 $.

5. Combine the Results:

   Combining the results from the two integrals, we get: $$ h(x) = - \alpha \log |x| + O(|x|^{-1}). $$

6. Exponentiate $ h(x) $:

   Now, exponentiate $ h(x) $: $$ e^{h(x)} = e^{-\alpha \log |x| + O(|x|^{-1})}. $$

   This can be written as: $$ e^{h(x)} = e^{-\alpha \log |x|} \cdot e^{O(|x|^{-1})}. $$

   Since $ e^{-\alpha \log |x|} = |x|^{-\alpha} $, we have: $$ e^{h(x)} = |x|^{-\alpha} \cdot \left( 1 + O(|x|^{-1}) \right). $$