Is the phase of a digital filter always an odd function from $[-\pi, \pi]$ Trying to prove something about the integral of the phase of an FIR from, does this hold true at all? Or maybe for some?
To expand on RBJ's comment from above, your statement is true iff the impulse response of the filter is real. Recall that the frequency response of a filter is the discrete time Fourier transform of its impulse response. That is:

$$
X(\omega) = \sum_{n=-\infty}^{\infty} x[n] e^{j\omega n}
$$

Look at the relationship between $X(\omega)$ and $X(-\omega)$:

$$
X(-\omega) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n}
$$
$$
X(-\omega) = \sum_{n=-\infty}^{\infty} x[n] \left(e^{j\omega n}\right)^*
$$

If $x[n]$ is real, then we can absorb it into the conjugate:

$$
X(-\omega) = \sum_{n=-\infty}^{\infty} \left(x[n] e^{j\omega n}\right)^*
$$

And since conjugation is a linear operation, we can pull it around the sum as well:

$$
X(-\omega) = \left(\sum_{n=-\infty}^{\infty} x[n] e^{j\omega n}\right)^*
$$
$$
X(-\omega) = \left(X(\omega)\right)^*
$$

So $X(\omega)$ exhibits Hermitian symmetry; its phase is oddly symmetric about $\omega = 0$, while its magnitude is evenly symmetric.