Commutation relations for creation, annihilator operators $a_\mathbf{p}^\dagger, a_\mathbf{p}$ Write the field $\phi$ and momentum $\pi$ in terms of creation and annihilation operators $a_\mathbf{p}^\dagger, a_\mathbf{p}$
$$
\phi(\mathbf{x}) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_\mathbf{p}}} [a_\mathbf{p}e^{i\mathbf{p}\cdot\mathbf{x}} + a_\mathbf{p}^\dagger e^{-i\mathbf{p}\cdot\mathbf{x}}],
$$
$$
\pi(\mathbf{x}) = \int \frac{d^3p}{(2\pi)^3} (-i) \sqrt{\frac{\omega_\mathbf{p}}{2}} [a_\mathbf{p}e^{i\mathbf{p}\cdot\mathbf{x}} - a_\mathbf{p}^\dagger e^{-i\mathbf{p}\cdot\mathbf{x}}].
$$
The goal is to show that
$$
[a_\mathbf{p}, a_\mathbf{q}] = [a_\mathbf{p}^\dagger, a_\mathbf{q}^\dagger] = 0,
$$
$$
[a_\mathbf{p}, a_\mathbf{q}^\dagger] = (2\pi)^3 \delta^{(3)}(\mathbf{p}-\mathbf{q}).
$$
I have no luck in arriving at the commutation relations.
Take inverse Fourier transform,
$$
\tilde{\phi}(\mathbf{p}) = \int d^3x\ \phi(\mathbf{x}) e^{-i\mathbf{p}\cdot\mathbf{x}} = \frac{1}{\sqrt{2\omega_\mathbf{p}}} a_\mathbf{p} + \frac{1}{\sqrt{2\omega_\mathbf{-p}}} a_\mathbf{-p}^\dagger,
$$
$$
\tilde{\pi}(\mathbf{p}) = \int d^3x\ \pi(\mathbf{x}) e^{-i\mathbf{p}\cdot\mathbf{x}} = (-i)\Bigg(\sqrt{\frac{\omega_\mathbf{p}}{2}} a_\mathbf{p} - \sqrt{\frac{\omega_\mathbf{-p}}{2}} a_\mathbf{-p}^\dagger\Bigg).
$$
Then
$$
a_\mathbf{p} = \frac{1}{2} \Bigg(\sqrt{2\omega_\mathbf{p}} \tilde{\phi}(\mathbf{p}) + i\sqrt{\frac{2}{\omega_\mathbf{p}}} \tilde{\pi}(\mathbf{p}) \Bigg),
$$
$$
a_\mathbf{-p}^\dagger = \frac{1}{2} \Bigg(\sqrt{2\omega_\mathbf{p}} \tilde{\phi}(\mathbf{p}) - i\sqrt{\frac{2}{\omega_\mathbf{p}}} \tilde{\pi}(\mathbf{p}) \Bigg).
$$
Using $[\phi(\mathbf{x}), \phi(\mathbf{y})] = [\pi(\mathbf{x}), \pi(\mathbf{y})] = 0$, $[\phi(\mathbf{x}), \pi(\mathbf{y})] = i\delta^{(3)}(\mathbf{x}-\mathbf{y})$,
\begin{align}
[a_\mathbf{p}, a_\mathbf{q}]
&=
\frac{1}{4} \int d^3x d^3y\ 2i\sqrt{\frac{\omega_\mathbf{p}}{\omega_\mathbf{q}}} [\phi(\mathbf{x}), \pi(\mathbf{y})] e^{-i\mathbf{x}\cdot\mathbf{p}} e^{-i\mathbf{y}\cdot\mathbf{q}} + 2i\sqrt{\frac{\omega_\mathbf{q}}{\omega_\mathbf{p}}} [\pi(\mathbf{x}), \phi(\mathbf{y})] e^{-i\mathbf{x}\cdot\mathbf{p}} e^{-i\mathbf{y}\cdot\mathbf{q}} \\
&=
\frac{1}{4} \int d^3x d^3y\ 2i\sqrt{\frac{\omega_\mathbf{p}}{\omega_\mathbf{q}}} i\delta^{(3)}(\mathbf{x}-\mathbf{y}) e^{-i\mathbf{x}\cdot\mathbf{p}} e^{-i\mathbf{y}\cdot\mathbf{q}} + 2i\sqrt{\frac{\omega_\mathbf{q}}{\omega_\mathbf{p}}} (-i)\delta^{(3)}(\mathbf{y}-\mathbf{x}) e^{-i\mathbf{x}\cdot\mathbf{p}} e^{-i\mathbf{y}\cdot\mathbf{q}} \\
&=
-\frac{i}{2} \int d^3x \Bigg(\sqrt{\frac{\omega_\mathbf{p}}{\omega_\mathbf{q}}} 
 - \sqrt{\frac{\omega_\mathbf{q}}{\omega_\mathbf{p}}} \Bigg) e^{-i\mathbf{x}\cdot(\mathbf{p} + \mathbf{q})}
\end{align}
Why is this equal to $0$? It's not true that $\omega_\mathbf{p} = \omega_\mathbf{q}$, is it?

The only dependence on $\mathbf{x}$ that remains is in the exponent factor. Integrating it we get $\delta$-function,
\begin{equation}
\int d^3x\, e^{-i\mathbf{x}\cdot(\mathbf{p}+\mathbf{q})}=(2\pi)^3\delta^{(3)}(\mathbf{p}+\mathbf{q})
\end{equation}
That means that we can replace $\omega_\mathbf{q}$ with $\omega_{-\mathbf{p}}=\omega_\mathbf{p}$. That results in cancellation.
