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For action $\frac{m}{2}\int ({\dot{q}}^2-{\omega }^2{q}^2)dt$, the kernel is $\sqrt{m\omega /(2\pi i\sin \omega T)}\exp \{im\omega [(q_f^2+q_i^2)\cos \omega T-2q_f{q}_i]/(2\sin \omega T)\}$. The Dirichlet oscillator determinant ratio supplies its prefactor, while the classical boundary action supplies its exponent. The Feynman i-epsilon prescription specifies its continuation through caustics.

Work in units $\hslash =1$ and with the particle mass absorbed into the stated kinetic term. Divide the interval into $N$ pieces of length $\Delta =T/N$. For a suitable self-adjoint Hamiltonian operator, the Trotter product formula gives the regulated limitInsert $1=\int d{q}_j|q_j⟩⟨q_j|$ between neighboring factors. Because $V(\hat{q})$ is diagonal in the position operator basis, the given free-particle propagator yieldsThe exponent is the time-sliced action. Its continuum notation isThe measure is defined by the displayed finite-dimensional limits and their normalization, not by a translation-invariant flat measure on an infinite-dimensional space. The oscillatory limit needs a damping/analytic-continuation prescription. This is the time-sliced configuration-space path integral derived from the operator kernel.

For a quadratic potential energy, write $q=q_c+\eta$, where the classical path satisfies ${\ddot{q}}_c+V^{\prime }(q_c)=0$ and the endpoint conditions, and $\eta (0)=\eta (T)=0$. Expansion is exact:Integration by parts makes the linear term equal to $-[{\int }_0^T({\ddot{q}}_c+V^{\prime }(q_c))\eta dt]$, since the endpoint term vanishes. It is zero by the Euler-Lagrange equation. Also $V^{\prime \prime }$ is constant, independent of the endpoints. Thus classical-path factorization of a quadratic path integral is exact:Only the classical action carries the endpoint dependence.

Let $D={\partial }_t^2+V^{\prime \prime }$ on homogeneous Dirichlet boundary conditions. Expand $\eta$ in its normalized eigenfunctions, with eigenvalues ${\lambda }_n$. The fluctuation integral becomes a product of regulated integrals $\int d{a}_n\exp (-i{\lambda }_n{a}_n^2/2)$. Each is proportional to ${\lambda }_n^{-1/2}$, with a prescribed complex phase. Relative to $D_0={\partial }_t^2$, the normalization is thereforeThe subscript denotes the Dirichlet operator domain, and the square-root branch is continued from the free kernel with its convergence prescription. Equivalently one can use determinants of $-D$ and $-D_0$. For $V^{\prime \prime }={\omega }^2$, the Dirichlet oscillator determinant ratio isThis also checks the free limit. At conjugate times, the fluctuation operator has a zero eigenvalue and the ordinary finite-prefactor expression fails; the kernel is obtained by distributional continuation. Away from these times the classical boundary problem is unique and the stated factorization applies directly.

For the infinite-time oscillator with an external source, use vacuum boundary conditions and the Feynman i-epsilon prescription. For a smooth source of compact support, defineThe imaginary term damps the real-field integral. Integration by parts gives $S_{ϵ}[q]=\frac{1}{2}q{A}_{ϵ}q+Jq$, with the pairings understood as time integrals. Completing the square givesTranslate the integration variable. The source-independent Gaussian functional integral is $N$, soOn Fourier modes $e^{i\omega t}$, $A_{ϵ}$ has eigenvalue ${\omega }^2-m^2+iϵ$, henceFor $m>0$, closing the contour above for $t>0$ and below for $t<0$ gives $G(t)=-i{e}^{-im|t|}/(2m)$. The physical time-ordered oscillator quantum field theory propagator is $iG$, not $G$ itself. The shifted stationary path is $q_c=-GJ$ and obeys ${\ddot{q}}_c+m^2{q}_c=J$ with the same vacuum prescription. A retarded inverse would define a different boundary problem. Dividing by the zero-source integral removes $N$ and gives the vacuum source functional of a quantum oscillator. At zero mass an additional infrared prescription is needed for the inverse on constant modes.

Set $\Delta =T/M$, $q_0=q$, and $q_M=q^{\prime }$. The Trotter product formula separates the kinetic and potential energy evolution to first order in $\Delta$. Inserting $M-1$ resolutions of the identity in the position operator basis and using the free short-time kernel givesThe exponent converges formally to the action along the path. The prefactors and intermediate integrations define the time-sliced configuration-space path integral measure, soA real-time path integral is an oscillatory limit with the Feynman i-epsilon prescription, rather than an ordinary probability integral. The operator derivation assumes the usual self-adjointness and product-formula hypotheses for the Hamiltonian.

For a quadratic potential energy write $q=q_c+\eta$, with $\eta (0)=\eta (T)=0$ and ${\ddot{q}}_c+V^{\prime }(q_c)=0$. The first variation vanishes by the Euler-Lagrange equation and the fixed endpoints. Because $V$ is quadratic, the remaining expansion is exact:Here integration by parts has no boundary contribution. Translation of the integration variables leaves the time-sliced measure unchanged. Thus the Gaussian path integral over $\eta$ is independent of the endpoint values, and the entire endpoint dependence lies in $e^{iS[q_c]}$.

Expanding $\eta$ in eigenfunctions satisfying Dirichlet boundary conditions, each real mode contributes an inverse square root of its quadratic eigenvalue, with the oscillatory phase fixed by continuation. Normalizing against $A_0={\partial }_t^2$ gives a precise determinant-ratio form:The determinant ratio is meaningful after a common regulator; the phase is part of the prescription. It is equivalent to writing $N=C(T)({\det }_{D}A{)}^{-1/2}$ with an endpoint-independent normalization.

For example, if $V^{\prime \prime }={\omega }^2$, the Dirichlet oscillator determinant ratio isA linear or constant term in $V$ changes the action but not this prefactor. The formula holds away from conjugate times, when the boundary problem and determinant are nonsingular. At $\sin \omega T=0$, it must be continued as a distribution with its Maslov index phase; the assertion of an ordinary finite prefactor cannot be used literally there. For the unshifted oscillator, at $T=r\pi /\omega$ the kernel is $e^{-ir\pi /2}\delta (q^{\prime }-(-1{)}^{r}q)$.

For the full-line source integral, impose vacuum boundary conditions by adiabatic damping. Take the source initially as a test function, so its pairings with the Green function are defined. After integration by parts, writeWith the Fourier transform convention $G(t)=\int d\omega e^{i\omega t}\tilde{G}(\omega )/(2\pi )$, the inverse isSet $q=\eta -GJ$. Completing the square givesand consequentlyThe Gaussian functional integral $Z[0]$ is source independent and contains the regulated determinant. This is most naturally read as a normalized vacuum generating functional. For $m>0$, closing the frequency contour gives $G(t)=-i{e}^{-im|t|}/(2m)$; the time-ordered oscillator two-point function is $iG(t)$. This checks the sign of the source exponent and identifies the required vacuum prescription.

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.

The paired zero factors converge locally uniformly because $\sum n^{-2}$ converges. Hadamard factorization for the order-one entire sine function gives a possible exponential multiplier; oddness removes its linear exponent and the derivative at zero fixes its constant. This product turns a Dirichlet oscillator determinant ratio into an elementary sine ratio.