# Asymptotic solution of an integral equation

Consider an integral equation of the form

$$\sigma_B(\Lambda)+\int_{-B}^{B} K\left(\Lambda-\Lambda^{\prime}\right) \sigma_B\left(\Lambda^{\prime}\right) d \Lambda^{\prime}=f(\Lambda)$$

where the Kernel is $$K(x)=\frac{1}{\pi}\frac{1}{1+x^2}$$ and the driving term is some function like $$f(x)=\frac{1}{\pi} \frac{c^2}{c^2+x^2}$$ where $$c$$ is a real number.

Now, let's say one is interested in the integral only in the limits $$B\to 0$$ and $$B\to \infty$$.

$$B\to 0$$ limit is quite easy to solve as one can then expand the unknown function around $$\Lambda\to 0$$ and work order by order.

But how does one systematically look at $$B\to \infty$$ limit?

When $$B=\infty$$, the equation can be easily be solved in the Fourier space. But what would be the way to look at the corrections for $$B$$ approaching $$\infty$$.

Too long for a comment

Let's rewrite the integral equation as $$\sigma_B(\Lambda)+\int_{-\infty}^{\infty} K(\Lambda-\Lambda') \sigma_B(\Lambda')\,d \Lambda'=f(\Lambda)+\int_{{\mathbb R}\setminus[-B,B]}K(\Lambda-\Lambda') \sigma_B(\Lambda')\,d \Lambda'. \tag{1}$$ Now let's define the sequence $$(\sigma_n(\Lambda))_{n\in\mathbb{N}}$$ as the solutions to $$\sigma_n(\Lambda)+\int_{-\infty}^{\infty} K(\Lambda-\Lambda') \sigma_n(\Lambda')\,d \Lambda'=f_n(\Lambda), \tag{2}$$ where $$f_0(\Lambda):=f(\Lambda)$$ and $$f_n(\Lambda):=f(\Lambda)+\int_{{\mathbb R}\setminus[-B,B]}K(\Lambda-\Lambda') \sigma_{n-1}(\Lambda')\,d \Lambda'\qquad(n\geq 1). \tag{3}$$ Notice that

1. $$\sigma_0(\Lambda)=\lim_{B\to\infty}\sigma_B(\Lambda)$$;
2. Given $$\sigma_{n-1}(\Lambda)$$, one may, in principle, compute $$f_n(\Lambda)$$ using $$(3)$$, then use Fourier transform to solve $$(2)$$ for $$\sigma_n(\Lambda)$$;
3. Hopefully, under suitable conditions the sequence $$(\sigma_n(\Lambda))_{n\in\mathbb N}$$ converges to $$\sigma_B(\Lambda)$$ as $$n\to\infty$$;
4. It follows from $$(2)$$ and $$(3)$$ that $$\Delta\sigma_n(\Lambda)+\int_{-\infty}^{\infty} K(\Lambda-\Lambda') \Delta\sigma_n(\Lambda')\,d \Lambda'= \int_{{\mathbb R}\setminus[-B,B]}K(\Lambda-\Lambda') \Delta\sigma_{n-1}(\Lambda')\,d \Lambda', \tag{4}$$ where $$\Delta\sigma_n:=\sigma_n-\sigma_{n-1}$$. Because of the range of integration in the latter integral, I suspect that $$\frac{\Delta\sigma_n}{\Delta\sigma_{n-1}}\to 0$$ as $$B\to\infty$$, so the sequence $$(\sigma_n(\Lambda))_{n\in\mathbb N}$$ also plays the role of an asymptotic expansion of $$\sigma_B(\Lambda)$$ for $$B\to\infty$$.