Parity of annihilation and creation operator - Real Klein-Gordon field In calculating the total-momentum operator of the real Klein-Gordon field, I end up with an equation like 
$$ \vec P = \frac{1}{(2\pi)^3}\int d^3p \mspace{9mu} \vec p\Big(a_pa_{-p} + a_pa^\dagger_{p}-a_{-p}^\dagger a_{-p} - a_{-p}^\dagger a_p^\dagger\Big)  $$
and it seems the first and the last integrals vanish because they are antisymmetric w.r.t $ \vec p \rightarrow -\vec p $. I am unable to see this anti-symmetry. Also I am not able to see how to relate $ a_{-p} $ to $ a_p $ or $ a_p^\dagger $ other than some implicit integral relations like(which also am not very sure how to prove),
$$ \int d^3p a_p^\dagger e^{-i\vec p.\vec x}  = \int d^3p a_p e^{i\vec p.\vec x}$$
Thanks for any inputs.
 A: I am not a 100% sure, but I think they mean the following:
\begin{equation}
\begin{aligned}
\mathbf{P} & \sim \int \mathrm{d}^3 p \; \mathbf{p}a_p a_{-p} \\&
= \int \mathrm{d}^3 p \; \left(\frac{1}{2} \mathbf{p}a_p a_{-p} + \frac{1}{2} \mathbf{p}a_p a_{-p} \right) \\&
= \int \mathrm{d}^3 p \; \left(\frac{1}{2} \mathbf{p}a_p a_{-p} - \frac{1}{2} \mathbf{p}a_{-p} a_{p} \right) \\&
= \int \mathrm{d}^3 p \; \frac{1}{2} [a_p, a_{-p}] \\&
= 0
\end{aligned}
\end{equation}
where we performed a change of integration variable $\mathbf{p} \rightarrow - \mathbf{p}$ to get to the third line. In order to understand this, note that for $\mathbf{p} \rightarrow - \mathbf{p}$ we get:
\begin{equation}
\int\limits_{-\infty}^{\infty} \mathrm{d}p \rightarrow \int\limits_{\infty}^{-\infty} \left(-\mathrm{d}p\right)= - \int\limits_{-\infty}^{\infty} \left(-\mathrm{d}p\right)=\int\limits_{-\infty}^{\infty} \mathrm{d}p
\end{equation}
I think this may also explain your second question: $a_p$ is related to $a_{-p}$ through this change of integration variable.
