I am reading the original work of Casimir: On the attraction between two perfectly conducting plates (1948)
In general, the zero-point energy is
$$U_0 =\frac12 \hbar c \sum_{k_x=0}^{\infty} \sum_{k_y=0}^{\infty} \sum_{k_z=0}^{\infty} \sqrt{k_x^2 + k_y^2 + k_z^2} \, g(\omega_k)$$
where $g$ is the degeneracy of the mode.
The zero-point energy $E_0$ when the plates are separated by some distance $a$ is calculated by going to the continuum in the plane $k_x \, k_y$ as $L\to\infty$ and taking into account that $g = 1$ for modes with $k_z = 0$
$$E_0 = \frac12 \sum_{\vec{k}}\hbar \omega_k = \hbar c \frac{L^2}{\pi^2} \frac{\pi}{2} \int_{0}^{\infty} \rho \, d\rho \left\{ \sum_{n = 1}^{\infty} \sqrt{\rho^2 + \left( \frac{\pi n}{a} \right)^2} + \frac{\rho}{2} \right\} $$
The next step is to find the zero point energy $E_0'$ when $a\to\infty$. Now, we also go to the continuum in $k_z$ as $a$ is comparable to $L$. It is easy to show that
$$E_0' = \hbar c \frac{L^2}{\pi^2} \frac{\pi}{2} \frac{a}{\pi} \int_{0}^{\infty} \int_{0}^{\infty} \rho \, d\rho dk_z \sqrt{\rho^2 +k_z^2}$$
The zero-point energy diference is just $\delta E = E_0 - E_0'$
My question is... is it right to use the same variable '$a$' for both configurations? We could write $\delta E(a) = E_0(a) - E_0(a\to\infty)$ In that case, doesn't it only hold for large enough values of '$a$'?
For the next steps, he mixes the '$a$' from the first configuration (which is finite) with the '$a$' from the second configuration (which is "infinite")
