Suppose that you're doing electrostatics in two dimensions (or, equivalently, in 3D with translation invariance along $z$) and you're studying the response to a point charge in an angle sector of opening angle $\alpha$, i.e. in the domain $$ \mathcal D=\{(r\cos(\theta),r\sin(\theta))\in\mathbb R^2:0<\theta<\alpha\}, $$ under Dirichlet boundary conditions, i.e. between two conducting planes at dihedral angle $\alpha$.

Now, if you're actually faced with such a problem, the natural response is to say "ah, yeah, you can probably get the Green's function of the problem in terms of some expansion in terms of fractional-order Bessel functions something something something", and nobody can really remember the details, but luckily we have Jackson's Classical Electrodynamics, which informs us in §2.11 and in problem 2.25 that the Green's function for the problem is \begin{align} G(\mathbf r,\mathbf r') = 4 \sum_{n=1}^\infty \frac1n \frac{r_<^{n\pi/\alpha}}{r_>^{n\pi/\alpha}} \sin\left(\frac{n\pi}{\alpha}\theta \right)\sin\left(\frac{n\pi}{\alpha}\theta' \right), \tag 1 \end{align} satisfying $\nabla^2_\mathbf r G(\mathbf r,\mathbf r') = \delta(\mathbf r-\mathbf r')$ (up to signs and factors of $\pi$), where the arbitrary-real-number nature of the corner angle $\alpha$ manifests itself in the non-integer powers of the radii involved in the expansion.

On the other hand, for some special angles $\alpha$ of the form $\alpha = \pi/k$, where $k$ is an integer, one can just ditch all of that and simply use the method of image charges, by judiciously planting copies of our source charge at angles between $\alpha$ and $2\pi$ in such a way that the potential vanishes at the boundary of $\partial \mathcal D$ like it needs to, giving us the Green function in the form $$ G(\mathbf r,\mathbf r') =\sum_{i}q_i\ln(\|\mathbf r- \mathbf r_i\|) \tag 2 $$ where one of the $\mathbf r_i$ is $\mathbf r'$ and the rest are image charges.

OK, so after all that setup, here's my question: is it possible to directly reduce the general result in $(1)$ to the method-of-images result in $(2)$ through some means or another? Intuitively, it feels like that should be the case (though there's no promises that it'd be pretty), because $(1)$ contains all the information about the problem's solution and it should be possible to whittle it down to simpler versions where those exist; more prosaically, when $\alpha=\pi/k$ those exponents now read $r^{n\pi/\alpha} = r^{nk}$, which looks rather more manageable.

Along these lines, Jackson does add one tack to that problem 2.25, saying

(b) By means of complex-variable techniques or other means, show that the series can be summed to give a closed form, $$ G(\mathbf r,\mathbf r') = \ln\left[\frac{ r^{2\pi/\alpha}+r'^{2\pi/\alpha}-2rr'^{\pi/\alpha}\cos(\pi(\theta+\theta')/\alpha) }{ r^{2\pi/\alpha}+r'^{2\pi/\alpha}-2rr'^{\pi/\alpha}\cos(\pi(\theta-\theta')/\alpha) }\right], $$

which looks like a step in the right direction, but (i) I can't fully see how you'd sum that series, and more importantly (ii) I'm not fully sure how you'd reduce that simplified series to the method-of-images solution.

So: is this possible? and if so, how? Also, a bit more ambitiously: does that still hold in 3D, where the Green's function goes up into Bessel expansions but the method of images still works?

Suppose that you're doing electrostatics in two dimensions (or, equivalently, in 3D with translation invariance along $z$) and you're studying the response to a point charge in an angle sector of opening angle $\alpha$, i.e. in the domain $$ \mathcal D=\{(r\cos(\theta),r\sin(\theta))\in\mathbb R^2:0<\theta<\alpha\}, $$ under Dirichlet boundary conditions, i.e. between two conducting planes at dihedral angle $\alpha$.

Now, if you're actually faced with such a problem, the natural response is to say "ah, yeah, you can probably get the Green's function of the problem in terms of some expansion in terms of fractional-order Bessel functions something something something", and nobody can really remember the details, but luckily we have Jackson's Classical Electrodynamics, which informs us in §2.11 and in problem 2.25 that the Green's function for the problem is \begin{align} G(\mathbf r,\mathbf r') = 4 \sum_{n=1}^\infty \frac1n \frac{r_<^{n\pi/\alpha}}{r_>^{n\pi/\alpha}} \sin\left(\frac{n\pi}{\alpha}\theta \right)\sin\left(\frac{n\pi}{\alpha}\theta' \right), \tag 1 \end{align} satisfying $\nabla^2_\mathbf r G(\mathbf r,\mathbf r') = \delta(\mathbf r-\mathbf r')$ (up to signs and factors of $\pi$), where the arbitrary-real-number nature of the corner angle $\alpha$ manifests itself in the non-integer powers of the radii involved in the expansion.

On the other hand, for some special angles $\alpha$ of the form $\alpha = \pi/k$, where $k$ is an integer, one can just ditch all of that and simply use the method of image charges, by judiciously planting copies of our source charge at angles between $\alpha$ and $2\pi$ in such a way that the potential vanishes at the boundary of $\partial \mathcal D$ like it needs to, giving us the Green function in the form $$ G(\mathbf r,\mathbf r') =\sum_{i}q_i\ln(\|\mathbf r- \mathbf r_i\|) \tag 2 $$ where one of the $\mathbf r_i$ is $\mathbf r'$ and the rest are image charges.

OK, so after all that setup, here's my question: is it possible to directly reduce the general result in $(1)$ to the method-of-images result in $(2)$ through some means or another? Intuitively, it feels like that should be the case (though there's no promises that it'd be pretty), because $(1)$ contains all the information about the problem's solution and it should be possible to whittle it down to simpler versions where those exist; more prosaically, when $\alpha=\pi/k$ those exponents now read $r^{n\pi/\alpha} = r^{nk}$, which looks rather more manageable.

Along these lines, Jackson does add one tack to that problem 2.25, saying

(b) By means of complex-variable techniques or other means, show that the series can be summed to give a closed form, $$ G(\mathbf{r},\mathbf{r}') = \ln\left[\frac{ r^{2\pi/\alpha}+r'^{2\pi/\alpha}-2(rr')^{\pi/\alpha}\cos(\pi(\theta+\theta')/\alpha) }{ r^{2\pi/\alpha}+r'^{2\pi/\alpha}-2(rr')^{\pi/\alpha}\cos(\pi(\theta-\theta')/\alpha) }\right], $$

which looks like a step in the right direction, but (i) I can't fully see how you'd sum that series, and more importantly (ii) I'm not fully sure how you'd reduce that simplified series to the method-of-images solution.

So: is this possible? and if so, how? Also, a bit more ambitiously: does that still hold in 3D, where the Green's function goes up into Bessel expansions but the method of images still works?