= Solution
With the mode normalization given, the oscillator <canonical commutation relations> are $[a_{\mathbf p},a_{\mathbf q}^\dagger]=(2\pi)^3\delta^3(\mathbf p-\mathbf q)$, with the other two <commutators> zero. Let $E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}$ and $p^0=E_{\mathbf p}$. The convention compatible with the requested numerator is
$$
\Delta_F(x-y)=\langle0|T\{\phi(x)\phi(y)\}|0\rangle,
$$
without an additional factor of $i$ outside the <vacuum expectation value>. Put $\tau=x^0-y^0$ and $\mathbf r=\mathbf x-\mathbf y$. The <Fock vacuum> is annihilated by $a_{\mathbf p}$, so only $aa^\dagger$ contributes to the two <Wightman functions>. Consequently,
$$
\Delta_F(\tau,\mathbf r)=\int\frac{d^3p}{(2\pi)^3}\frac{e^{i\mathbf p\cdot\mathbf r}}{2E_{\mathbf p}}\left(\theta(\tau)e^{-iE_{\mathbf p}\tau}+\theta(-\tau)e^{iE_{\mathbf p}\tau}\right).
$$
In the second term we changed $\mathbf p\mapsto-\mathbf p$ to give the same spatial exponential.
Now perform the <energy> <contour integral>
$$
g_E(\tau)=\lim_{\epsilon\downarrow0}\int\frac{dp^0}{2\pi}\frac{i\,e^{-ip^0\tau}}{(p^0)^2-E^2+i\epsilon}.
$$
Here the residue step uses $E>0$; the massless zero-momentum point is interpreted through the smeared <distribution> limit, not as an isolated normalized oscillator. The <Feynman i-epsilon prescription> puts the positive-energy pole below the real axis and the negative-energy pole above it: $p^0=E-i0$ and $p^0=-E+i0$. For $\tau>0$, close in the lower half-plane, clockwise; the <residue theorem> gives $-i$ times the residue $ie^{-iE\tau}/(2E)$, hence $e^{-iE\tau}/(2E)$. For $\tau<0$, close in the upper half-plane, counterclockwise; the negative-energy residue is $-ie^{iE\tau}/(2E)$, again giving a positive $e^{iE\tau}/(2E)$. These are exactly the two time-ordered terms. Restoring the spatial integral proves the <scalar Feynman propagator pole prescription>:
$$
\boxed{\Delta_F(x-y)=\lim_{\epsilon\downarrow0}\int\frac{d^4p}{(2\pi)^4}\frac{i\,e^{-ip\cdot(x-y)}}{p^2-m^2+i\epsilon}.}
$$
The prescription in this formula supplies the pole convention left unspecified in the printed display; an unprescribed ordinary real-axis integral would not be well-defined. It is a <distribution> limit after smearing, not an absolutely convergent four-dimensional integral. With this normalization the <derivative jump of a free scalar time-ordered two-point function> gives $(\Box+m^2)\Delta_F=-i\delta^4(x-y)$, an independent check of both the numerator and the sign.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-301-feynman-contours.png]
{title=<Feynman propagator> integration contours and displaced poles}
{description=Clockwise and counterclockwise <contour integrals> for the <Feynman i-epsilon prescription>.}
The pole displacement is exaggerated in this original schematic. The contour orientation and selected pole reproduce the <time ordering> of the <Feynman propagator>.
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