This solution is incorrect
This problem is not quite the same as that in [1], however, under fair use, I have used the language therein as a template for solving this problem.
In the region of interest there is no charge.
Our solution of Laplace's equation in circular cylindrical coordinates is a linear combination of
\begin{align*}
\psi_{k,m}(r,\phi,z)
&=
\left[A_{k,m}\,J_{m}{(kr)} + B_{k,m}\,N_{m}{(kr)}\right]
\\
&\qquad \left[ C_{k,m}\,\cos{(m\,\phi)} + D_{k,m}\,\sin{(m\phi)} \right]\left[ E_{k,m}\,e^{-k\,z} + F_{m,k}\, e^{+k\,z} \right].
\end{align*}
The particular linear combination is determined by the boundary conditions to be satisfied. Our cylinder here has a radius $a$ and on its entire surface the potential is zero. The problem is to find the electric potential
$$
\Psi(r,\phi,z) = \sum_{k,m}\psi_{k,m}(r,\phi,z)
$$
everywhere in the exterior of the hollow cynlinder and above the $z=0$ plane.
As one boundary conditions is $\lim_{z\to\infty}V{(r,\phi,z)} = 0$, we must take $F_{k,m}=0$. The $z$ dependence becomes $e^{-k\,z} $. The requirement that $\Psi = 0$ on the cylindrical sides is met by requiring the separation constant $k$ to be
$$k=k_{m,n} = \frac{\alpha_{m,n}}{a},$$
where the first subscript, $m$, gives the index of the Bessel function, whereas the second subscript identifies the particular zero of $J_m$. Consequently, we must take $B_{k,m}=0$. The electrostatic potential becomes
\begin{equation}
\begin{aligned}
\Psi(r,\phi,z)
=
\sum_{m=0}^\infty \, \sum_{n=1}^\infty J_{m}{\left(\frac{\alpha_{m,n}\,r}{a}\right)}
\,\left[ C_{n,m}\,\cos{(m\,\phi)} + D_{n,m}\,\sin{(m\phi)} \right]\, \exp{\left(- \frac{\alpha_{m,n}\,z}{a} \right) }.
\end{aligned}
\tag{1}
\end{equation}
Equation (1) is a double series: A piece-wise Bessel series in $r$ and a Fourier series in $\phi$. At $z=0$, $\Psi= \Psi_o{(r,\phi)}$ (i.e., $\Psi_o$ is a known function of $r$ and $\phi$). Therefore,
\begin{equation*}
\begin{aligned}
&\Psi(r,\phi,0)
= \sum_{m=0}^\infty \, \sum_{n=1}^\infty J_{m}{\left(\frac{\alpha_{m,n}\,r}{a}\right)}
\,\left[ C_{n,m}\,\cos{(m\,\phi)} + D_{n,m}\,\sin{(m\phi)} \right].
\end{aligned}
\end{equation*}
The constants $C_{k,m}$ and $D_{k,m}$ are evaluated by using orthogonality rules of the harmonic functions and the closure relation of the Bessel functions. First, I multiply and integrate as
\begin{equation*}
\begin{aligned}
& \int_{\phi=0}^{2\pi} \, \cos{(m'\,\phi)}\,\Psi_o(r,\phi )\,d\phi
\\
&=\sum_{m=0}^\infty \, \sum_{n=1}^\infty
J_{m}{\left(\frac{\alpha_{m,n}\,r}{a}\right)}
\, \int_{\phi=0}^{2\pi}\left[ C_{n,m}\,\cos{(m\,\phi)} + D_{n,m}\,\sin{(m\phi)} \right]\,\cos{(m'\,\phi)} \, dk\,d\phi.
\\
&=\sum_{m=0}^\infty \, \sum_{n=1}^\infty
C_{n,m}\,J_{m}{\left(\frac{\alpha_{m,n}\,r}{a}\right)}
\, \int_{\phi=0}^{2\pi} \cos{(m\,\phi)} \,\cos{(m'\,\phi)} \, d\phi.
\\
&=\sum_{m=0}^\infty \, \sum_{n=1}^\infty
C_{n,m}\,J_{m}{\left(\frac{\alpha_{m,n}\,r}{a}\right)}
\,\begin{cases}
\pi\,\delta_{m,m'},&(m\neq 0),
\\
2\,\pi,&(m=m'= 0).
\end{cases}
\\
&=
\pi\,\left[
1+ \delta_{m',0}
\right]
\,
\sum_{n=1}^\infty
C_{n,m'}\, J_{m'}{\left(\frac{\alpha_{m',n}\,r}{a}\right)}
\end{aligned}
\end{equation*}
Second, I multiply and integrate as
\begin{equation*}
\begin{aligned}
&\int_{r=0}^{\infty} \int_{\phi=0}^{2\pi}
r\,J_{m'}{\left(\frac{\alpha_{m',n'}\,r}{a}\right)} \, \cos{(m'\,\phi)}\,\Psi_o(r,\phi )\,d\phi\,dr
\\
&=
\pi\,\left[
1+ \delta_{m',0}
\right]
\,
\sum_{n=1}^\infty
C_{n,m'}\,
\int_{r=0}^{\infty}r\, J_{m'}{\left(\frac{\alpha_{m',n}\,r}{a}\right)}
\,
J_{m'}{\left(\frac{\alpha_{m',n'}\,r}{a}\right)} \,dr \\
&=
\pi\,\left[
1+ \delta_{m',0}
\right]
\,
\sum_{n=1}^\infty
C_{n,m'}\,
\frac{1}{\frac{\alpha_{m',n'}}{a}}\,\delta_{n,n'}
\\
&=
\pi\,\left[
1+ \delta_{m',0}
\right]
\,
C_{n',m'}\,
\frac{a} {\alpha_{m',n'}}
\end{aligned}
\end{equation*}
I find that the coefficient
\begin{equation*}
C_{n,m}
=
\frac{ \alpha_{m,n} }
{\pi\,a}\,
\int_{r=0}^{\infty} \int_{\phi=0}^{2\pi}
J_{m}{\left(\frac{\alpha_{m,n}\,r}{a}\right)} \, \cos{(m \phi)}\,\Psi_o(r,\phi )\,d\phi\,dr \,
\left[
1+ \delta_{m,0}\right].
\end{equation*}
By a similar process, I also find that the coefficient
\begin{equation*}
D_{n,m}
=
\frac{ \alpha_{m,n} }
{\pi\,a}\,
\int_{r=0}^{\infty} \int_{\phi=0}^{2\pi}
J_{m}{\left(\frac{\alpha_{m,n}\,r}{a}\right)} \, \sin{(m \phi)}\,\Psi_o(r,\phi )\,d\phi\,dr \,
\left[
1- \delta_{m,0}\right].
\end{equation*}
Now, in this particular case, $\Psi_o(r,\phi ) = 0$ for $r<a$ and for $b<r$; thus,
\begin{align*}
C_{n,m}
&=
\frac{ \alpha_{m,n} }
{\pi\,a}\,
\int_{r=a}^{b} \int_{\phi=0}^{2\pi}
J_{m}{\left(\frac{\alpha_{m,n}\,r}{a}\right)} \, \cos{(m \phi)}\,\Psi_o(r,\phi )\,d\phi\,dr \,
\left[
1+ \delta_{m,0}\right].
\\
D_{n,m}
&=
\frac{ \alpha_{m,n} }
{\pi\,a}\,
\int_{r=a}^{b} \int_{\phi=0}^{2\pi}
J_{m}{\left(\frac{\alpha_{m,n}\,r}{a}\right)} \, \sin{(m \phi)}\,\Psi_o(r,\phi )\,d\phi\,dr \,
\left[
1- \delta_{m,0}\right].
\end{align*}
These are definite integrals, that is these are numbers. Upon substitution of these coefficients back into Equation (1), the series is specified and the potential $\Psi{(r,\phi,z)}$ is determined.
[1] Arfken and Weber, "Mathematical Methods for Physicists," 5th edition, page 690-691.
