Poisson's Equation in Cylindrical Coordinates

I have a question about a specific step in this problem that comes from ‘University of Chicago Graduate Problems in Physics’, Cronin, Greenberg, and Telegdi.

A long hollow cylindrical conductor of radius $a$ is divided into two parts by a plane through the axis, and the parts are separated by a small interval. If the two parts are kept at potentials $V_1$ and $V2$, show that the potential at any point within the cylinder is given by
$$
V=frac{left(V_{1}+V_{2}right)}{2}+2 frac{left(V_{1}-V_{2}right)}{pi} sum_{n=1}^{infty} frac{(-1)^{n-1}}{(2 n-1)}left(frac{r}{a}right)^{2 n-1} cos (2 n-1) theta
$$

When cross-referencing my solution with that of the textbook’s, I see that I have the correct approach. Using the Poisson equation,

$$
nabla^{2} V=0=frac{partial^{2} V}{partial r^{2}}+frac{1}{r} frac{partial V}{partial r}+frac{1}{r^{2}} frac{partial^{2} V}{partial theta^{2}}
$$

and separation of variables yields:

$$V=sum_{m=0}^{infty}left(frac{r}{a}right)^{m}left[A_{m} cos (m theta)+B_{m} sin (m theta)right]$$

Now

$$V(r=a, theta)=sum_{m=0}^{infty}left[A_{m} cos (m theta)+B_{m} sin (m theta)right]$$

$B_m=0$ by symmetry since $V(theta) = V(-theta)$. $A_m$ is calculated using the orthogonality of trig. functions, and the boundary conditions of $V_1$ and $V_2$,

$$
int_{-pi / 2}^{pi / 2} V_{1} cos n theta d theta+int_{pi / 2}^{3 pi / 2} V_{2} cos n theta d theta=frac{2left(V_{1}-V_{2}right)}{n} sin left(frac{n pi}{2}right)=pi A_{n}
$$

but I don’t understand how to correctly find the $A_0$ term. The solutions have the following statement:

$$
pileft(V_{2}+V_{1}right)=2 pi A_{0}
$$

I’m not sure how this comes about.