Trouble understanding why $ {\bf p} a_{\bf p}a_{\bf -p}$ and ${\bf p} a^\dagger_{\bf p}a^\dagger_{-\bf p}$ are odd functions of $\bf p$ Peskin and Schroeder's QFT book in equation (2.33) gives the momentum operator $\bf P$ as
$${\bf P} = - \int d^3 x \pi(x) \nabla \phi (x) = \int \frac{d^3 {\bf p}}{(2\pi)^3} {\bf p}a_{\bf p}^\dagger a_{\bf p}$$
where $\pi(x)$ is the momentum density conjugate of $\phi(x)$, the Klein-gordon field.
$$\phi(x) = \int \frac{d^3 {\bf p}}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\bf p}}}(a_{\bf p}+a_{\bf -p}^\dagger)e^{i{\bf p \cdot x}}$$
$$\pi(x) = \int \frac{d^3 {\bf p}}{(2\pi)^3}(-i)\sqrt{\frac{\omega_{\bf p}}{2}}(a_{\bf p}-a_{\bf -p}^\dagger)e^{i{\bf p \cdot x}}$$
$$\omega_{\bf p}=\sqrt{|{\bf p}|^2 + m^2}$$
From what I understand a key part of proving this is using the fact that the two functions
$$f_1({\bf p})=\frac{\bf p}{2}a_{\bf p} a_{-\bf p}\qquad f_2({\bf p})=\frac{\bf p}{2}a_{\bf p}^\dagger a_{-\bf p}^\dagger$$
are odd functions of $\bf p$ and thus
$$\int f_1({\bf p}) d^3{\bf p} = \int f_2({\bf p})d^3{\bf p}=0$$
But it's not entirely obvious to me why $f_1$ and $f_2$ should be odd. Does anyone have a hint as to why this is?
 A: First recall the relevant commutation relationships for the creation and annihilation operators:
\begin{eqnarray}
[a_{\bf p},a_{\bf q}] &=& [a_{\bf p}^\dagger,a_{\bf q}^\dagger]=0
\end{eqnarray}
Then it's obvious that $a_{\bf p} a_{\bf -p}$ is an even function of ${\bf p}$. It's also clear than ${\bf p}$ is an odd function. An even times an odd function is odd, so $f_1$ is odd. The same argument works for $f_2$.
A: $\mathbf{p}/2$ is an odd function of $\mathbf{p}$, so we can reduce your question to why $a_{\mathbf{p}} a_{-\mathbf{p}}$ (and the same with $a^\dagger$) is an even function of $\mathbf{p}$.
If we map $\mathbf{p} \rightarrow - \mathbf{p}$ we find $a_{\mathbf{p}} a_{-\mathbf{p}} \rightarrow a_{-\mathbf{p}} a_{\mathbf{p}}$, which is the same expression up to the commutation of the two annihilation operators.
Do they commute? Yes, you can show this starting from the last two relations in equation 2.20 in Peskin and Shroeder ($[\phi(x), \phi(y)] = 0$ and the same for $\pi$) following the same procedure which led you to prove that $[a_{p}, a^\dagger _{q}] = (2 \pi)^3 \delta^{(3)} (p - q)$ (equation 2.29).
