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Boundedness at the origin allows only $r^n$ for nonconstant interior modes; decay at infinity allows only $r^{-n}$ outside. Write the common surface potential in its Fourier series as ${\sum }_{n\ge 1}{v}_n(\theta )$, where $v_n=A_n\cos n\theta +B_n\sin n\theta$. Continuity then makes the corresponding interior and exterior factors $(r/R{)}^n$ and $(R/r{)}^n$.

Their outward radial derivatives at the surface are $n{v}_n/R$ inside and $-n{v}_n/R$ outside. Thus the jump is $-2n{v}_n/R$. Equating it to minus the prescribed surface-charge mode givesTherefore the potential of a charged cylindrical shell with angular Fourier data isEither branch gives the same value at $R$. The mean angular mode is zero: outside, neither a constant nor a logarithm can survive the condition at infinity; continuity then forces the interior constant to zero. Consequently $\Phi (0)=0$.

The absence of a mean term in the surface charge density is essential. A nonzero mean would require a logarithmic exterior potential and would be incompatible with decay at infinity in this two-dimensional reduction. The normalization follows the stated derivative jump, with no additional permittivity factor introduced.

Treat each angular Fourier mode independently. Boundedness at the axis excludes the negative radial powers and the logarithm inside; decay at infinity excludes the positive powers, constant and logarithm outside. For the cosine mode writeContinuity at $R$ gives $B_n=A_n{R}^{2n}$. The jump of the normal derivative is thenThus $A_n=a_n{R}^{1-n}/(2n)$ and $B_n=a_n{R}^{n+1}/(2n)$. The sine mode has the same relation with $b_n$. Therefore the potential of a charged cylindrical shell with angular Fourier data isThere is no angular mean term: the given surface charge density has zero average. The interior constant is zero because its boundary value must match the vanishing exterior mean. A nonzero mean charge per unit length would instead produce a logarithmic electrostatic potential, incompatible with the specified decay at infinity.