The angular momentum is conjugated to various angles in the spherical coordinates. Unlike usual coordinate $x$ those angles have limited range. So you may expect that just like in case of the Fourier transform on the limited range you will get not the integral but discrete series. However it doesn't have to look exactly like the usual Fourier series.
Indeed if we introduce the spherical coordinates,
\begin{equation}
x=r\cos\phi\sin\theta,\quad y=r\sin\phi\sin\theta,\quad z=r\cos\theta
\end{equation}
then any "good" wavefunction can be represented as a series,
\begin{equation}
\psi(r,\phi,\theta)=\sum_{l=0}^{+\infty}\sum_{m=-l}^{+l}\psi_{lm}(r)Y_{lm}(\phi,\theta)
\end{equation}
where $Y_{lm}$ are certain functions known as spherical harmonics that are orthonormal in the following sense,
\begin{equation}
\int_{0}^{2\pi} d\phi \int_{0}^{\pi} d\theta \sin\theta\, Y_{l'm'}^\ast (\phi,\theta) Y_{lm}(\phi,\theta)=\delta_{ll'}\delta_{mm'}
\end{equation}
Because of that we find that,
\begin{equation}
\psi_{lm}(r)=\int_{0}^{2\pi} d\phi \int_{0}^{\pi} d\theta \sin\theta\,Y_{lm}^\ast(\phi,\theta) \psi(r,\phi,\theta)
\end{equation}
Now if we take $\hat{L}_z=xp_x-yp_y=-i\partial_\phi$ (I'll take $\hbar=1$) then you'll get that any term in the series is an eigenfunction of this operator,
\begin{equation}
\hat{L}_z\psi_{lm}(r)Y_{lm}=m\psi_{lm}(r)Y_{lm}
\end{equation}
If you try to do the same with $\hat{L}_x$ and $\hat{L}_y$ this will not be the case. That's reflection of the fact that those operators don't commute. However if you take $\hat{L}^2=\hat{L}_x^2+\hat{L}_y^2+\hat{L}_z^2$ such operator does commute with $\hat{L}_z$ and,
\begin{equation}
\hat{L}^2\psi_{lm}(r)Y_{lm}=l\psi_{lm}(r)Y_{lm}
\end{equation}
Now you may consider $\psi_{lm}(r)$ as a new representation of the wavefunction. It's characterized by continuous parameter $r$ but also two integer parameter $l$ and $m$ with $l\geq 0,|m|\leq l$. Concerning those last two parameters the operators become simply matrices $A_{lm,l'm'}$ with $\hat{L}_z=m\delta_{mm'}$ and $\hat{L}^2=l\delta_{ll'}$. The expectation values can be written as,
\begin{equation}
\langle O\rangle = \int_{0}^{+\infty} r^2 dr\sum_{lm,l'm'} \psi_{lm}^\ast(r)\Big[\hat{O}_{lm,l'm'}\psi_{l'm'}\Big](r)
\end{equation}