Numerical method to Lippman-Schwinger equation

[HadronPhysics]

There is a question that puzzle me when I apply numerical method to principal value integral. Let me descibe it. We make use of the fact that the integral $\int_0^\infty \frac{dk}{k^2-k_0^2}$ vanishes, namely,
$$
\int_0^\infty \frac{dk}{k^2-k_0^2} = 0 .
$$
We use this formula to express a principal value intergral as
$$ \mathcal{P}\int_0^\infty \frac{f(x)}{k^2-k_0^2}dk = \int_0^\infty \frac{f(k)-f(k_0)}{k^2-k_0^2}dk .$$
Now, the right hand side is no longer singular at $k=k_0$ because it is proportional to the derivative $df/dx$. We can approximate this integral numerically, i.e.,
$$ \int_0^\infty \frac{f(k)-f(k_0)}{k^2-k_0^2}dk \approx \sum_i^N w_i \frac{f(k_i)-f(k_0)}{k_i^2-k_0^2} ,$$
where we adopt the Gaussian quadrature method.
Next, we change to Lippman-Schwinger equation where the principal integral occurs. That is
$$ R(k', k) = V(k', k) + \frac{2}{\pi} \mathcal{P}\int_0^\infty dp \frac{p^2V(k', p)R(p, k)}{(k_0^2-p^2)/2\mu} .$$
Then, we can evaluate this equation by the method that we have mentioned. we get
$$ R(k, k_0) = V(k, k_0) + \frac{2}{\pi} \sum_i^N \frac{p_i^2V(k', p_i)R(p_i, k_0)-k_0^2V(k', k_0)R(k_0, k_0)}{(k_0^2-p_i^2)/2\mu} w_i ,$$
where we let $k$ be $k_0$.
At present, everything is ok. the question that puzzles me will occur at the next step. In some computational physics books, for example, you can refer to [[1]](#id1), page: 118, it said that we can split term in summation to two part, namely,
$$ R(k, k_0) = V(k, k_0) \frac{2}{\pi} \left[ \sum_i^N \frac{k_i^2V(k, k_i)R(k_i, k_0)w_i}{(k_0^2-k_i^2)/2\mu} - k_0^2V(k, k_0)R(k_0, k_0)\sum_j^N\frac{w_j}{(k_0^2-k_j^2)/2\mu} \right] .$$
In the previous discussion, we constructed the term $\frac{f(k)-f(k_0)}{k^2-k_0^2}$ to avoid the singular at $k=k_0$. But here, we split the summation into two part. If $k_j\to k_0$, or $k_i\to k_0$, we can not see the term that is proportional to $df/dk$. I can not understand this step, because I think it contradicts the eqaution: $\mathcal{P}\int_0^\infty \frac{f(x)}{k^2-k_0^2}dk = \int_0^\infty \frac{f(k)-f(k_0)}{k^2-k_0^2}dk.$<div id="id1"></div>
- [1] [COMPUTATIONAL PHYSICS](https://courses.physics.ucsd.edu/2017/Spring/physics142/Lectures/Lecture18/Hjorth-JensenLectures2010.pdf)

 

[pasmith]

You have approximated the integral by a finite sum of bounded quantities. Terms of finite sums can always be re-ordered as you find convenient without affecting the result, and doing so in a context where you are using floating-point arithmetic might actually increase the accuracy of the result.

You are not then taking the limit N \to \infty and trying to sum an infinite series, where I agree that any re-ordering of the terms would require rigorous jusitification that the limit is not thereby affected. (It is more complicated here, since the values of the summands are themselves dependent on N.)