= Solution
For unnormalized <coherent states>, $|z\rangle=e^{za^\dagger}|0\rangle$, $\langle w|z\rangle=e^{\bar wz}$ and $1=\int d^2z\,e^{-|z|^2}|z\rangle\langle z|/\pi$. Insert this identity between short imaginary-time evolutions. The overlaps generate $\bar\phi\partial_\tau\phi$, while the normal-ordered symbol of $H=\hbar\omega(a^\dagger a+1/2)$ gives
$$
\boxed{Z=e^{-\beta\hbar\omega/2}\int_{\phi(\beta)=\phi(0)}\mathcal D\phi\,\exp\!\left[-\int_0^\beta\bar\phi(\partial_\tau+\hbar\omega)\phi\,d\tau\right]_{\rm normal}.}
$$
Here $\tau=\tau_E/\hbar$ runs from zero to $\beta$; the exponent is dimensionless. The bosonic trace makes the complex fields \b[periodic], not antiperiodic.
The printed expression omits the explicit vacuum factor. It represents the full oscillator only if a symmetric/Weyl determinant prescription implicitly includes that factor. With the ordinary normal-ordered time-sliced coherent-state measure, it instead represents $H=\hbar\omega a^\dagger a$. The continuum formula alone does not specify the energy-zero convention.
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