Casimir effect when the zeta function has a pole

let us suppose the Casimir force/ Vaccumm energy of a certain system

$$E_{casimir}= \sum _{n} \omega _{n}$$

which is formally equal to $E_{casimir}= \zeta _{spec}(-1)$ with $\zeta (s)_{spectral}=\sum_{n}\omega _{n} ^{-s}$

however what would happen if the spectral function has a pole at $s=1$ what would be then the value of $E_{casimir}$ can this pole problem be overcome by regularization ??

perhaps in this case we could take the finite part

$FP \zeta (-1)_{spec}= \sum_{n}\omega_{n}-\int_{0}^{\infty}\omega (x)dn(x)$

the secon integral would be the SEMICLASSICAL approximation ot the frecuencies energies $E_{n}= \hbar \omega _{n}$

The zeta function always has poles, depending on the number of dimensions you're working in. For $d=4$ you will generically have poles at $s=2$ and $s=1$ (but only there). The sum $\sum_n \omega_n^{-s}$ only converges if $\text{Re } s > 2$ (staying in $4d$ for simplicity). So in every case you'll encounter, you will need to use 'analytic continuation'.
This is part of the theory of complex functions. You can expand the zeta function around $s=3$, for example, and this expansion will have radius of convergence one. Now pick a point $w$ near the boundary of that sphere (far away from $s=2$ though) and you can again expand the zeta function, with radius $|w-2|.$ This expansion is well-defined (unique). Now you can play this game several times, until you reach any point you like (such as $s=-1$). There's a useful picture of these overlapping domains of convergence on Wikipedia, here http://upload.wikimedia.org/wikipedia/commons/thumb/4/4d/Imaginary_log_analytic_continuation.png/316px-Imaginary_log_analytic_continuation.png.
This recipe always works, in theory. But in practice it can be very painful to use. Often in physics we recycle existing analytic continuations. Mathematica knows how to calculate the Riemann zeta function at any $s \neq 1$ for example. There is a useful theorem that tells you that analytic continuations are unique, so it doesn't matter how you find it. But how you precisely find those analytic continuations is not so easy.
But forget about other 'regularisations'. The zeta function is already a regularisation and there is a mathematically well-defined procedure to find its value at negative $s$. If you start subtracting things, you'll get in trouble.