= Solution
In the incoming <Bunch-Davies vacuum>, only $\langle a_{\mathbf k}a_{\mathbf q}^{\dagger}\rangle=\delta_{\mathbf k\mathbf q}$ contributes to the coincident <vacuum expectation value>. Thus
$$
\langle\delta\hat\phi^2\rangle=\sum_{\mathbf k}|\delta\phi_k^{(+)}|^2=\sum_{\mathbf k}\frac1{2k\mathcal V}\left(\frac1{a^2}+\frac{H^2}{k^2}\right).
$$
Replacing the box sum by $\mathcal V\int d^3k/(2\pi)^3$ gives
$$
\boxed{\langle\delta\hat\phi^2\rangle=\int\frac{d^3k}{(2\pi)^3}\frac1{2k}\left(\frac1{a^2}+\frac{H^2}{k^2}\right)=\frac1{4\pi^2}\int_0^\infty dk\left(\frac{k}{a^2}+\frac{H^2}{k}\right).}
$$
The <subhorizon vacuum contribution> is the flat-space zero-point term; at fixed comoving $k$ it redshifts as $a^{-2}$. The other term gives <frozen scalar power> on <superhorizon scales>. Specifically, the <dimensionless cosmological power spectrum> per logarithmic interval is
$$
\Delta_{\delta\phi}^2(k,\tau)=\frac{k^2/a^2+H^2}{4\pi^2}\ \longrightarrow\ \boxed{\left(\frac H{2\pi}\right)^2}.
$$
Its late-time value is independent of $k$, producing a <scale-invariant inflationary power spectrum>. Each fixed <Fourier mode> freezes separately as $\tau\to0$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-62-scalar-freezing.png]
{title=Vacuum redshifting and frozen power of massless scalar modes during de Sitter expansion}
The coincident integral is formal: the vacuum part diverges at large $k$, and an unlimited scale-invariant part diverges at small $k$. The conclusion concerns the power in each finite logarithmic band. A finite inflationary history supplies a largest generated wavelength, as described by the <infrared qualification of a scale-invariant covariance>; taking the late-time limit inside an unregulated integral would not be justified.
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