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

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 $$$$\label{B-decay-cond} |B(x)| \, = \, O(|x|^{-\tau}) \qquad as \quad|x| \to \infty.$$$$ We now define $$$$\label{h-eq} h(x) =- \frac{1}{2\pi} \int_{\mathbb{R}^2} B(y) \log |x-y|\, dy,$$$$ so that the vector field given by $$$$\label{A-h-def} A_h = (\partial_{x_2} h, - \partial_{x_1} h)$$$$ satisfies $$\nabla\times A_h = B$$. Thanks to Assumption above we have $$$$\label{A-L-infty} |A_h| \in L^\infty(\mathbb{R}^2).$$$$ where $$\alpha$$ is defined as $$$$\alpha= \frac{1}{2\pi} \int _{\mathbb{R}^2}B\, dx$$$$ Notice that $$$$\label{exp-h-asymp} e^{h(x)} = |x|^{-\alpha}\big(1+ O(|x|^{-1})\big) , \qquad as \quad |x|\to \infty.$$$$ I'm struggling to show the above identity, for convenience I'll call it Integral Identity (II). I thought I could proceed in this way: $$\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...

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