How to solve the two dimensional Laplace's equation for certain cases? Had a doubt regarding Laplace's equation.
In many textbooks, the general solution to the two dimensional Laplace equation is mentioned as:
$$\Phi(\rho,\phi) = A_{0} + B_{0}\ln(\rho) + \sum_{n=1}^{\infty}\rho^n(A_n\cos(n\phi) + B_n \sin(n\phi)) + \sum_{n=1}^{\infty}\rho^{-n}(C_n \cos(n\phi) + D_n \sin(n\phi))$$
in polar coordinates.
For convenience, I will name the two summation terms as:
$$T_1 = \sum_{n=1}^{\infty}\rho^n(A_n\cos(n\phi) + B_n \sin(n\phi))$$
$$T_2 =  \sum_{n=1}^{\infty}\rho^{-n}(C_n \cos(n\phi) + D_n \sin(n\phi))$$
Not much has really been mentioned on whether these two series, $T_1$ and $T_2$ converge or not. When the solution space does not include either zero or infinity it is generally implicitly assumed that they converge. When the solution space includes the origin, generally the procedure has been to equate the coefficients of the second series, $T_2$, to zero, in other words, to exclude the second series from the solution. 
In many physical problems of interest however, the solution space includes both the origin of the coordinate system and is an unbounded region in all directions. My question is, how can we use the above expansions to handle such cases? Especially when there is a boundary condition of zero (or $ k ln(\rho) $) at infinity. The radius of convergence doesn't seem to be known a priori as the coefficients are unknown. 
For example, if you consider the case of an infinitely long conducting cable of arbitrary but uniform cross-section, (Arbitrary meaning that it need not be something geometrical or symmetric like a circle or a square, and could even be concave; and uniform meaning that the shape and area of the cross-section does not change in the $z$ direction so it can be reduced to a two dimensional problem.) maintained at a constant potential $V$, and where the potential is taken to approach $ k ln(\rho) $ at infinity. It may often be convenient to choose the origin outside the cable. In this case, if I exclude the term $T_2$ (for finiteness at the origin), again the boundary condition $\Phi\to k \ln(\rho)$ as $\rho\to\infty$ will imply that the coefficients in the term $T_1$ become zero as well. So the solution appears to be trivially $\Phi = A_0$ (obviously wrong!) even before I apply the boundary condition $\Phi = V$ at the surface of the cable. 
Can you let me know what I am doing wrong? Can you throw some light on this?
Thanks.
 A: HiPragabhava, Thank you for your assistance. If I proceed along those lines, this is what I get:
$$\phi_\text{in} = A_0 + \sum_{n=1}^\infty\rho^n\Big(A_n\sin(n\phi) + B_n\cos(n\phi)\Big) $$ for $\rho\leq a$ .
and
$$\phi_\text{out} = A_0 + B_0\ln(\rho) + \sum_{n=1}^\infty\rho^{-n}\Big(C_n\sin(n\phi) + D_n\cos(n\phi)\Big) $$ for $\rho>a$ .
Now if I assume symmetry about x axis, the sine terms will vanish. Also, 'glueing' the two solutions will give me:
$$D_n = a^{2n}B_n$$ .
So, now the solutions become :
$$\phi_\text{in} = A_0 + \sum_{n=1}^\infty\rho^nB_n\cos(n\phi) $$ for $\rho\leq a$
and
$$\phi_\text{out} = A_0 + B_0\ln(\rho) + \sum_{n=1}^\infty\rho^{-n}a^{2n}B_n\cos(n\phi) $$ for $\rho>a$ .
Now, the constants $B_n$ are yet to be determined. Though we don't expect the electric field to be continuous on the (incomplete) cylinder, we would expect it to be continuous on the slit. In other words, the $\rho$ component of the electric field would be continuous at $\rho = a$, $-\alpha<\phi<\alpha$. This means that:
$$ \frac{\partial\phi_\text{in}}{\partial\rho} = \frac{\partial\phi_\text{out}}{\partial\rho}$$ for $\rho = a$, and $-\alpha<\phi<\alpha$.
After differentiation and simplification, this is what I get:
$$ B_0 = 2\sum_{n=1}^\infty na^nB_n\cos n\phi $$
for $-\alpha<\phi<\alpha$ .
Since this relation is valid only for the range between $-\alpha$ and $\alpha$ I would not be able to get the coefficients by multiplying both sides by cosines and integrating between 0 to 2$\pi$.  So how to get the coefficients $B_n$ ?
Even if I could get the coefficients something looks fishy here as the L.H.S. is a constant while the R.H.S. depends on the angle $\phi$.
So how can I proceed from here on?
Thanks.
