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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 $\infty$.

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1 Answer 1

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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 $B\to\infty$.
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