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On $[0,T]$ with vanishing endpoint fluctuations, the sine-mode eigenvalues are $(\pi n/T{)}^2-{\omega }^2$. Divide by the free eigenvalues and use the sine infinite product. The ratio determines the harmonic oscillator transition kernel once the free configuration-space path integral is normalized. Zeros correspond to caustics where the fluctuation operator has a zero mode.

Write $z=z_c+\eta$, with $\eta (0)=\eta (T)=0$. Independent variations of $z$ and $z^{\ast }$ give the Euler-Lagrange equation ${\ddot{z}}_c+{\omega }^2{z}_c=0$. Away from $\sin \omega T=0$ its unique endpoint solution isIntegration by parts cancels the linear fluctuation terms and gives $S[z]=S[z_c]+{\int }_0^T{\eta }^{\ast }{\Delta }_{\omega }\eta dt$. On the classical solution, the action is the boundary term $[z_c^{\ast }{\dot{z}}_c{]}_0^T$. Substituting the endpoint derivatives therefore givesThe remaining Gaussian path integral contains two real fluctuation coordinates per mode, so it contributes an inverse functional determinant, rather than its inverse square root. Thus $K=e^{iS[z_c]}/\det {\Delta }_{\omega }$, with the measure normalization fixing the otherwise arbitrary constant.

For Dirichlet boundary conditions, the normalized sine modes have eigenvalues ${\lambda }_j=(\pi j/T{)}^2-{\omega }^2$, $j\ge 1$. Their determinant ratio is the convergent Dirichlet oscillator determinant ratioThe last equality is the sine infinite product. With the prescribed free determinant this givesThe kernel uses the Feynman i-epsilon prescription; at a caustic it is a distributional limit, not an ordinary finite function. Its $\omega \to 0$ limit is $(\pi iT{)}^{-1}\exp (i|z_f-z_i{|}^2/T)$.

There is a sign error in the printed complex-integral hint. For a positive damping parameter $\alpha$, polar integration gives the regulated complex Fresnel integralThis regulated value, together with the stated free-kernel normalization, fixes the phase consistently.

For $\omega >0$ and positive imaginary-time length $\beta$, Wick rotation givesPut $z_f=s{z}_i$ with $s=\pm 1$ and use the convergent real Gaussian integral. Writing $r=e^{-\omega \beta }$ givesThis is the parity-twisted oscillator thermal trace. The complex coordinate describes two independent real quantum harmonic oscillators, each with mass two in these units. Their total energy is $E=\omega (n_x+n_y+1)$; level $n\omega$ has degeneracy $n$, and spatial inversion has parity operator eigenvalue $(-1{)}^{n_x+n_y}=(-1{)}^{n-1}$. The plus sign is $\mathrm{Tr}{e}^{-\beta H}$; the minus sign is $\mathrm{Tr}(P{e}^{-\beta H})$. The alternating trace inserts parity into a bosonic system; it does not change the oscillators into fermions.