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 $N$.)