= Solution
Use units $\hbar=c=1$ and the <Minkowski metric> $g_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$, so $p\cdot x=E_pt-\mathbf p\cdot\mathbf x$. The <canonical momentum> of the <real scalar field> is $\pi=\dot\phi$, and the <Hamiltonian operator> is
$$
H=\frac12\int d^3x\,[\pi^2+(\nabla\phi)^2+m^2\phi^2].
$$
<Canonical quantization of a real scalar field> promotes $\phi$ and $\pi$ to <operator-valued distributions> satisfying the equal-time <canonical commutation relation>
$$
[\phi(t,\mathbf x),\pi(t,\mathbf y)]=i\delta^3(\mathbf x-\mathbf y),\qquad
[\phi(t,\mathbf x),\phi(t,\mathbf y)]=[\pi(t,\mathbf x),\pi(t,\mathbf y)]=0.
$$
Products and <operator commutators> can first be interpreted after spatial smearing. An additive vacuum constant in $H$ does not affect the following equations. In the <Heisenberg picture>, $\dot O=i[H,O]$. Applying the <canonical commutation relation> to the quadratic <Hamiltonian operator>, and integrating the gradient term by parts, gives
$$
\dot\phi=\pi,\qquad \dot\pi=\nabla^2\phi-m^2\phi,qquad
\boxed{(\Box+m^2)\phi=0}.
$$
Thus the <Klein-Gordon equation> is an operator equation. The spatial <Fourier transform> reduces it to $\ddot\phi_{\mathbf p}+E_p^2\phi_{\mathbf p}=0$, with $E_p=\sqrt{\mathbf p^2+m^2}$. Its two frequency branches, together with the <Hermitian adjoint> condition $\phi^\dagger=\phi$, give the <mode expansion of a free field>
$$
\phi(x)=\int d\mu(p)\,[a(\mathbf p)f_p(x)+a^\dagger(\mathbf p)f_p^*(x)],\qquad
d\mu(p)=\frac{d^3p}{(2\pi)^3 2E_p},\qquad f_p=e^{-ip\cdot x}.
$$
The measure $d\mu(p)$ is <Lorentz invariant>; its normalization fixes that of the <creation and annihilation operators> rather than imposing any extra physical assumption.
For <invariant-normalized scalar mode extraction>, take the <Klein-Gordon inner product> of $f_p$ with the field. The positive-frequency contribution has coefficient $E_p+E_q$ and spatial integral $(2\pi)^3\delta^3(\mathbf p-\mathbf q)$. The negative-frequency contribution has coefficient $E_p-E_q$ and spatial integral $(2\pi)^3\delta^3(\mathbf p+\mathbf q)$, so it vanishes because $E_{-p}=E_p$. Consequently
$$
i\int d^3x\,[f_p^*\dot\phi-\phi\dot f_p^*]
=\int d\mu(q)\,(E_p+E_q)(2\pi)^3\delta^3(\mathbf p-\mathbf q)a(\mathbf q)
=a(\mathbf p).
$$
Taking the <Hermitian adjoint>, with the order of the scalar factors immaterial, yields
$$
\boxed{a^\dagger(\mathbf p)=-i\int d^3x\,[f_p\dot\phi-\phi\dot f_p]}.
$$
Equivalently $a(\mathbf p)=\int d^3x\,f_p^*(E_p\phi+i\pi)$ and $a^\dagger(\mathbf q)=\int d^3y\,f_q(E_q\phi-i\pi)$. At a common time their <operator commutator> therefore is
$$
[a(\mathbf p),a^\dagger(\mathbf q)]
=\int d^3x\,d^3y\,f_p^*(t,\mathbf x)f_q(t,\mathbf y)(E_p+E_q)\delta^3(\mathbf x-\mathbf y)
=\boxed{(2\pi)^3 2E_p\delta^3(\mathbf p-\mathbf q)}.
$$
The same calculation gives $[a(\mathbf p),a(\mathbf q)]=[a^\dagger(\mathbf p),a^\dagger(\mathbf q)]=0$: the surviving factor is an energy difference supported at opposite spatial momenta. These are the bosonic <canonical commutation relations> in <relativistic normalization of a one-particle state>.
Choose the <Fock vacuum> by $a(\mathbf p)|0\rangle=0$ and $\langle0|0\rangle=1$. Only the annihilator-creator product contributes to the <Wightman function>, giving
$$
\langle0|\phi(x)\phi(y)|0\rangle=\int d\mu(p)\,e^{-ip\cdot(x-y)}.
$$
Using <time ordering> and the reverse product for $y^0>x^0$, the <Feynman propagator>, with the convention $D_F=i\Delta_F$ in this calculation, is
$$
\boxed{D_F(x-y)=\int d\mu(p)\,[\theta(x^0-y^0)e^{-ip\cdot(x-y)}+\theta(y^0-x^0)e^{-ip\cdot(y-x)}]}.
$$
In particular the denominator in the spatial measure is $2E_p$ in both terms.
To verify the four-dimensional <Fourier transform> representation, let $\tau=x^0-y^0$ and integrate the energy variable first:
$$
I(\tau)=\lim_{\epsilon\downarrow0}\int_{-\infty}^{\infty}\frac{dp^0}{2\pi}\,
\frac{i e^{-ip^0\tau}}{(p^0)^2-E_p^2+i\epsilon}.
$$
The <Feynman i-epsilon prescription> puts the positive-energy pole at $E_p-i0$ and the negative-energy pole at $-E_p+i0$. For $\tau>0$, the <contour integral> closes in the lower half-plane with clockwise orientation. The <residue theorem> multiplies the residue $i e^{-iE_p\tau}/(2E_p)$ by $-i$, giving $e^{-iE_p\tau}/(2E_p)$. For $\tau<0$, it closes counterclockwise above: multiplying the residue $-i e^{iE_p\tau}/(2E_p)$ by $i$ gives $e^{iE_p\tau}/(2E_p)$. Thus $I(\tau)=e^{-iE_p|\tau|}/(2E_p)$. After inserting the spatial exponential, reversing $\mathbf p$ in the negative-time term gives exactly the <time ordering> expression above. Hence
$$
\boxed{i\Delta_F(x-y)=\lim_{\epsilon\downarrow0}\int\frac{d^4p}{(2\pi)^4}\,\frac{i e^{-ip\cdot(x-y)}}{p^2-m^2+i\epsilon}}.
$$
This equality is understood as a <distribution> limit; the same prescription gives $(\Box+m^2)D_F(x-y)=-i\delta^4(x-y)$.
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