= Solution
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 write
$$
\Phi_{n,\rm in}=A_nr^n\cos n\theta,\qquad
\Phi_{n,\rm out}=B_nr^{-n}\cos n\theta.
$$
Continuity at $R$ gives $B_n=A_nR^{2n}$. The jump of the <normal derivative> is then
$$
-nB_nR^{-n-1}-nA_nR^{n-1}=-2nA_nR^{n-1}=-a_n.
$$
Thus $A_n=a_nR^{1-n}/(2n)$ and $B_n=a_nR^{n+1}/(2n)$. The sine mode has the same relation with $b_n$. Therefore the <potential of a charged cylindrical shell with angular Fourier data> is
$$
\boxed{\Phi(r,\theta)=\sum_{n=1}^{\infty}\frac R{2n}(a_n\cos n\theta+b_n\sin n\theta)
\begin{cases}(r/R)^n,&0\leq r\leq R,\\(R/r)^n,&r\geq R.\end{cases}}
$$
There 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.
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