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In Jackson's "Classical Electrodynamics" third edition in section 3.9 page 121 while explaining a Green function expansion the following equation is attained:

$ \frac{1}{r}\frac{d^2}{dr^2}(rg_l(r,r')) - \frac{l(l+1)}{r^2}g_l(r,r')=-4π\delta(r-r') $ $(eq.1)$

which for the relevant boundary conditions has as the solution

$C(r_<^l-\frac{a^{2l+1}}{r_<^{l+1}})(\frac{1}{r_>^{l+1}}-\frac{r_>^{l}}{b^{2l+1}}) $ $(eq2)$

then to obtain the constant $C$ he multiplies $eq1$ by $r$ and integrates from $r'-\epsilon$ to $r'+\epsilon$ but for some reason completely leaves out the $2$nd term as if it integrates to zero or something... This has been bothering me for a while since I had to use this method a couple of times in the exercises and often I had to ignore some term but I don't understand why this is the case...

something i thought of while writing: maybe to get the constant I just have to consider the $l=0$ case or something? but then why are there $l$s in the solution?

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The second term is bounded in the region of integration. As we are integrating over a region of length $2\epsilon$ it is less in magnitude than $2M\epsilon$, where $M$ is the bound. This becomes negligible as $\epsilon\to 0$. The other terms are are singular and give contributions that do not go to zero with epsilon.

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  • $\begingroup$ First of all thanks for the answer... but I still don't understand. What do you mean with the bound $M$? $\endgroup$
    – ctsmd
    Nov 11, 2019 at 14:23
  • $\begingroup$ I mean that there exists some positive number $M$ such that everywhere in the interval $r'-\epsilon$ to $r'+\epsilon$ we have $|(l(l+1)/r^2) g_l(r,r')|<M$. Hence $$\int_{r'-\epsilon}^{r'+\epsilon}(l(l+1)/r^2) g_l(r,r')dr<2M\epsilon$$. $\endgroup$
    – mike stone
    Nov 11, 2019 at 19:31
  • $\begingroup$ Oh, ok, and the first term is infinite because there is a discontinuity in $g(r,r')$ and thus cancels the $\epsilon$? $\endgroup$
    – ctsmd
    Nov 11, 2019 at 19:44
  • $\begingroup$ Yes. Just use the fundamental theorem of calculus to do the integral. Of course you should say something about distributions and test functions to be rigorous in doing this. $\endgroup$
    – mike stone
    Nov 11, 2019 at 19:47

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