In chapter 2 of Quantum Field Theory and the Standard Model, Schwartz derives the equal-time commutation relations of the second-quantised field. Using $$ \phi(\vec{x}) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \left(a_p e^{i\vec{p}\cdot\vec{x}} + a_p^{\dagger} e^{-i\vec{p}\cdot\vec{x}}\right),\tag{2.75} $$ he arrives at $$ [\phi(\vec{x}), \phi(\vec{y})] = \int\frac{d^3p}{(2\pi)^3} \frac{1}{2\omega_p} \left(e^{i\vec{p}\cdot(\vec{x}-\vec{y})} - e^{-i\vec{p}\cdot(\vec{x}-\vec{y})} \right).\tag{2.89} $$ Now he argues that the integral measure and $\omega_p$ are symmetric under $\vec{p} \rightarrow -\vec{p}$ and hence the commutator vanishes. I cannot see why $d^3 p$ is symmetric. From my understanding, it is just the 3-momentum and should be $-d^3p$ under the substitution. Am I missing something here?
Also, in an earlier section he defines, $$ \phi_0(\vec{x},t) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \left(a_p e^{-ipx} + a_p^{\dagger} e^{ipx}\right).\tag{2.78} $$ The $p$'s and $x$'s here are 4-vectors. Since the integral is over only the 3-momentum, is this Lorentz invariant? (I can see that the integrand is Lorentz invariant).
References:
Section 2.3.2 & 2.3.3, Quantum Field Theory and the Standard Model, M.D. Schwartz