There is a question that puzzle me when I apply numerical methods to principal value integral. Let me describe 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 the principal value integral 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 Lippmann-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], page: 118, it said that we can split the term in the summation into 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 singularity 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 equation: $\mathcal{P}\int_0^\infty \frac{f(x)}{k^2-k_0^2}dk = \int_0^\infty \frac{f(k)-f(k_0)}{k^2-k_0^2}dk.$
- [1] COMPUTATIONAL PHYSICS, Morten Hjorth-Jensen (NB: ~500 page PDF link)
I believe you have misread something in the derivation to arrive at this conclusion. The authors have simply re-expressed the principal value integral as the difference of two integrals: $$\mathcal{P}\int_0^\infty\frac{f(k)}{k^2-k_0^2}\,\mathrm{d}k\to\int_0^\infty\frac{f(k)-f(k_0)}{k^2-k_0^2}\,\mathrm{d}k\equiv\int_0^\infty\frac{f(k)}{k^2-k_0^2}\,\mathrm{d}k-\int_0^\infty\frac{f(k_0)}{k^2-k_0^2}\,\mathrm{d}k$$ And since $f(k_0)$ is independent of the integration variable $k$, this can be written as, $$\mathcal{P}\int_0^\infty\frac{f(k)}{k^2-k_0^2}\,\mathrm{d}k\equiv\int_0^\infty\frac{f(k)}{k^2-k_0^2}\,\mathrm{d}k-f(k_0)\int_0^\infty\frac{1}{k^2-k_0^2}\,\mathrm{d}k\tag{1}$$ From Eq (1), you can write a numeric scheme as, $$\mathcal{P}\int_0^\infty\frac{f(k)}{k^2-k_0^2}\,\mathrm{d}k\approx\sum_jw_j\frac{f(k_j)}{k_j^2-k_0^2}-f(k_0)\sum_i\frac{w_i}{k_i^2-k_0^2}\tag{2}$$ Hence, the separation of the integrals/sums only occurs after the principal value formula was used to produce the two terms $f(k),\,f(k_0)$, not before.
You may be correct in stating that the proportionality to $\mathrm{d}f/\mathrm{d}k$ is less clear when separating the two terms, but this does not mean that the principal value integral wasn't utilized to avoid the $k=k_0$ singularity (and in fact we showed that it must have been used to get the two sums as you have presented).