Heat Kernels, Zeta Functions, and Spectral Determinants

For a nonnegative self-adjoint elliptic operator with discrete spectrum, the heat trace packages the eigenvalues into a function of positive heat time. Its Mellin transform gives the spectral zeta function in a genuine convergence half-plane. Subtracting a controlled short-time heat expansion then continues that zeta function to the origin, where

$$\log\det_\zeta'(A/\mu^2) = -\left. \frac{\mathrm d}{\mathrm ds} \zeta_{A/\mu^2}(s) \right|_{s=0}$$

defines the logarithm of the dimensionless regularized determinant. Here $\mu$ is a positive reference mass, and the prime on the determinant means that zero modes are excluded.

That chain is reliable only when the operator realization, boundary condition, zero modes, spectral sign, and reference scale are explicit. The heat coefficients determine possible zeta poles and scale dependence, while the finite part also contains global spectral information. In Euclidean QFT the resulting determinant is an input to a bosonic one-loop Gaussian integral; it is not, by itself, a choice of counterterms or a completed renormalization prescription.

Required background. Elliptic Boundary Problems and Heat Kernels supplies elliptic realizations, boundary conditions, trace-class heat evolution, and short-time expansions; Spectra, Resolvents, Spectral Measures, and Functional Calculus supplies self-adjoint spectral data, compact resolvent, and functional calculus; Mellin Transforms and Scaling Asymptotics supplies the fundamental strip, asymptotic subtraction, pole mapping, and remainder control used in the continuation.

Elliptic spectra and the operator contract

Selecting an individual Bessel or Hankel mode from boundary data is a different task from aggregating an entire operator spectrum.

Work on a smooth compact $d$-dimensional Riemannian manifold $M$, possibly with smooth boundary, and a finite-rank Hermitian bundle over it. Let $\nabla$ be a Hermitian connection and let

$$A=-g^{ij}\nabla_i\nabla_j+Q$$

be a second-order Laplace-type operator with smooth Hermitian $Q$. If $\partial M\ne\varnothing$, include a specified local, strongly elliptic boundary condition $B$. Assume that the resulting realization $A_B$ is self-adjoint, nonnegative, and has compact resolvent. Its eigenvalues, counted with multiplicity, can then be ordered as

$$0\leq \lambda_0\leq\lambda_1\leq\cdots, \qquad \lambda_n\longrightarrow\infty.$$

Write $P_0$ for the orthogonal projection onto $\ker A_B$ and $N_0=\dim\ker A_B$. A prime attached to $\operatorname{Tr}$, $\Theta_A$, or $\det_\zeta$ means that zero modes have been excluded. A prime attached to $\zeta_A(s)$ means differentiation with respect to $s$. Thus

$$\Theta_A'(t) = \operatorname{Tr}\!\left(e^{-tA_B}-P_0\right) = \sum_{\lambda_n>0}e^{-t\lambda_n}.$$

When $A_B$ is strictly positive, $P_0=0$ and the primes may be dropped. We take $A_B$ to have mass dimension two and introduce a positive reference scale $\mu$ whenever its eigenvalues occur inside a logarithm. The construction here is explicitly Euclidean and uses a positive-definite metric.

These hypotheses are not cosmetic. Compactness gives a discrete spectrum, strong ellipticity gives the standard boundary heat expansion, positivity fixes the complex-power branch, and removing the kernel makes the large-time Mellin integral converge. The same hypothesis boundary is developed in Kirsten 2000, § 1, PDF pp. 1–5 and in Vassilevich 2003, § 2.2, PDF pp. 14–17. Vassilevich’s Vassilevich 2003, Equation (2.2), PDF p. 11 uses the Laplace-type convention $D=-(g^{ij}\nabla_i\nabla_j+E)$; in the convention above, his endomorphism is $E=-Q$.

One spectrum, three global objects

It helps to keep the local kernel separate from the three global spectral objects built from it.

Object	Definition	What it records
Local heat kernel $K_A(t;x,y)$	Integral kernel of $e^{-tA_B}$	Propagation in heat time and local short-distance geometry
Primed heat trace $\Theta_A'(t)$	$\operatorname{Tr}(e^{-tA_B}-P_0)$	The positive spectrum with exponential damping
Spectral zeta function $\zeta_A(s)$	$\operatorname{Tr}'A_B^{-s}$, then analytic continuation	Power-weighted spectral data and its meromorphic structure
Zeta determinant $\det_\zeta'(A_B/\mu^2)$	$\exp[-\zeta'_{A_B/\mu^2}(0)]$	A dimensionless regularized product of positive eigenvalues

For each $t>0$, $e^{-tA_B}$ is trace class. Its short- and long-time limits do different jobs. As $t\downarrow0$, the unprimed heat trace has an asymptotic expansion

$$\Theta_A(t) \sim \sum_{j=0}^{\infty} a_j(A,B)t^{(j-d)/2}.$$

This is the same normalization as the prerequisite page’s expansion

$$\Theta_A(t) \sim (4\pi t)^{-d/2} \sum_{j=0}^{\infty}A_{j/2}(A,B)t^{j/2}, \qquad a_j=(4\pi)^{-d/2}A_{j/2}.$$

The coefficients $a_j(A,B)$ are integrals of local invariants of the metric, connection, endomorphism $Q$, and boundary data. On a closed smooth manifold, the odd-$j$ coefficients vanish for a Laplace-type operator. A boundary can produce the intervening half-powers represented by odd $j$. For every fixed $N$, the precise statement is

$$\Theta_A(t) - \sum_{j=0}^{N}a_j(A,B)t^{(j-d)/2} = O\!\left(t^{(N+1-d)/2}\right), \qquad t\downarrow0.$$

This is an asymptotic statement, not a convergent series to be integrated for all $t$. The primed trace differs by the constant $N_0$, so it is convenient to set

$$c_j=a_j(A,B)-N_0\,\delta_{jd}.$$

Then $\Theta_A'(t)\sim\sum_j c_jt^{(j-d)/2}$. At the opposite end, if $\lambda_*>0$ is the smallest positive eigenvalue, then

$$\Theta_A'(t)=O(e^{-t\lambda_*}), \qquad t\longrightarrow\infty.$$

Small heat time therefore controls the meromorphic continuation and local ultraviolet structure. Large heat time detects the kernel and the bottom of the positive spectrum.

The Mellin transform builds the zeta function

For $\lambda>0$ and $\Re s>0$,

$$\lambda^{-s} = \frac{1}{\Gamma(s)} \int_0^\infty t^{s-1}e^{-t\lambda}\,\mathrm dt.$$

Weyl growth for a second-order elliptic operator implies that $\sum_{\lambda_n>0}\lambda_n^{-s}$ converges absolutely for $\Re s>d/2$. In that half-plane, positivity and absolute convergence justify interchanging the sum and integral:

$$\begin{aligned} \zeta_A(s) &=\operatorname{Tr}'A_B^{-s} =\sum_{\lambda_n>0}\lambda_n^{-s},\\ \Gamma(s)\zeta_A(s) &=\int_0^\infty t^{s-1}\Theta_A'(t)\,\mathrm dt, \qquad \Re s>\frac d2. \end{aligned}$$

The lower endpoint gives the restriction $\Re s>d/2$: the leading heat term behaves like $t^{-d/2}$. The upper endpoint is harmless only after zero modes have been removed or when the operator is strictly positive. This separation prevents an ultraviolet subtraction from hiding an infrared failure.

Analytic continuation is finite subtraction

To reach the origin, choose a fixed reference heat time $t_0>0$, with mass dimension $-2$, and subtract only a finite number of short-time terms on $(0,t_0)$. Choose $N\geq d$ and define

$$R_N(t) = \Theta_A'(t) - \sum_{j=0}^{N}c_jt^{(j-d)/2}.$$

The remainder estimate makes the first integral below holomorphic for $\Re s>(d-N-1)/2$:

$$\begin{aligned} \zeta_A(s) =\frac{1}{\Gamma(s)}\Bigg[& \int_0^{t_0} t^{s-1}R_N(t)\,\mathrm dt +\int_{t_0}^\infty t^{s-1}\Theta_A'(t)\,\mathrm dt\\ &+\sum_{j=0}^{N} \frac{c_jt_0^{s+(j-d)/2}} {s+(j-d)/2} \Bigg]. \end{aligned}$$

This equation agrees with the original zeta series in its convergence half-plane and supplies its meromorphic continuation farther left. Increasing $N$ extends the domain without changing the continued function. Changing $t_0$ redistributes terms among the three displayed contributions but leaves their sum unchanged.

A heat term $c_jt^{(j-d)/2}$ creates a candidate simple pole at

$$s=\frac{d-j}{2}.$$

The word candidate matters. The prefactor $1/\Gamma(s)$ has zeros at $s=0,-1,-2,\ldots$ and can cancel a denominator pole. Equivalently,

$$\operatorname*{Res}_{s=(d-j)/2} \bigl[\Gamma(s)\zeta_A(s)\bigr] =c_j.$$

For the smooth Laplace-type problem specified above, the zeta function is regular at the origin. Since $1/\Gamma(s)=s+O(s^2)$, the $j=d$ term gives

$$\zeta_A(0)=c_d=a_d(A,B)-N_0.$$

Gilkey 1984, § 1.10, printed pp. 78–80 (PDF pp. 83–85) derives this heat-to-zeta continuation for positive elliptic operators and then treats a nonnegative operator by restricting away from its null space.

This identity cleanly separates a local heat coefficient from the global integer counting excluded zero modes. By contrast, $\zeta_A'(0)$ also depends on the regular integrals in the continuation formula and therefore on the full spectrum. Singular spaces, nonlocal boundary conditions, or unusual self-adjoint extensions can introduce logarithmic heat terms and a different zeta singularity structure; the formula above should not be exported to those settings without a new analysis.

The determinant must be dimensionless

For the positive spectrum of an order-two operator, define

$$\begin{aligned} \zeta_{A/\mu^2}(s) &=\operatorname{Tr}'(A_B/\mu^2)^{-s},\\ \log\det_\zeta'(A_B/\mu^2) &=-\zeta_{A/\mu^2}'(0). \end{aligned}$$

In finite dimensions this reproduces the ordinary determinant because

$$-\left.\frac{\mathrm d}{\mathrm ds} \sum_{n=1}^{N} \left(\frac{\lambda_n}{\mu^2}\right)^{-s} \right|_{s=0} = \sum_{n=1}^{N} \log\!\left(\frac{\lambda_n}{\mu^2}\right).$$

In infinite dimensions the equality is a definition through analytic continuation, not permission to rearrange a divergent product.

The reference-scale dependence follows without computing $\zeta_A'(0)$. For two positive scales $\mu_1$ and $\mu_2$,

$$\begin{aligned} \log\det_\zeta'(A_B/\mu_2^2) -\log\det_\zeta'(A_B/\mu_1^2) &=-2\zeta_A(0) \log\!\left(\frac{\mu_2}{\mu_1}\right),\\ \mu\frac{\mathrm d}{\mathrm d\mu} \log\det_\zeta'(A_B/\mu^2) &=-2\zeta_A(0). \end{aligned}$$

Thus the scale response is fixed by $a_d-N_0$. If $N_0>0$, the prime is a change of object: it defines a determinant on the orthogonal complement of the kernel. It does not integrate the corresponding flat directions or supply their collective-coordinate measure.

The positivity assumption also avoids a serious phase issue. Negative eigenvalues must be treated separately: Vassilevich 2003, § 2.2, top of PDF p. 17 describes separating their absolute values from spectral asymmetry. In a physical saddle calculation, unstable directions require a separate prescription. That information cannot be recovered from the positive determinant above.

Exact benchmark: a massive scalar on a circle

Let $x\sim x+L$ and consider

$$A_m=-\frac{\mathrm d^2}{\mathrm dx^2}+m^2, \qquad m>0,$$

on periodic functions. Its spectrum is

$$\lambda_n = \left(\frac{2\pi n}{L}\right)^2+m^2, \qquad n\in\mathbb Z.$$

The operator is positive, self-adjoint, and has no zero mode. Its heat trace is immediately

$$\Theta_m(t) = e^{-m^2t} \sum_{n\in\mathbb Z} e^{-4\pi^2n^2t/L^2}.$$

Poisson summation gives the independent winding representation

$$\Theta_m(t) = \frac{L}{\sqrt{4\pi t}}e^{-m^2t} \sum_{\ell\in\mathbb Z} e^{-\ell^2L^2/(4t)}.$$

The spectral form makes the large-$t$ decay $e^{-m^2t}$ visible. The winding form makes the short-time expansion visible:

$$\Theta_m(t) \sim \frac{L}{\sqrt{4\pi t}}e^{-m^2t}.$$

Only powers $t^{-1/2},t^{1/2},t^{3/2},\ldots$ occur. There is no $t^0$ term and no kernel, so $\zeta_{A_m}(0)=0$. The zeta determinant is therefore independent of $\mu$ in this example.

Rather than continue the Epstein zeta function from scratch, differentiate with respect to the parameter $m^2$. In one dimension $A_m^{-1}$ is trace class, so differentiating the zeta determinant gives an ordinary convergent trace:

$$\begin{aligned} \frac{\partial}{\partial m^2} \log\det_\zeta(A_m/\mu^2) &=\operatorname{Tr}A_m^{-1},\\ \operatorname{Tr}A_m^{-1} &=\sum_{n\in\mathbb Z} \frac{1}{(2\pi n/L)^2+m^2} =\frac{L}{2m}\coth\!\left(\frac{mL}{2}\right). \end{aligned}$$

The last equality is the partial-fraction expansion of $\coth z$ in NIST DLMF 2026, Equation 4.36.3, evaluated at $z=mL/2$.

Multiplying by $2m$ and integrating from $m_0>0$ to $m>0$ yields the finite, dimensionless ratio

$$\frac{\det_\zeta(A_m/\mu^2)} {\det_\zeta(A_{m_0}/\mu^2)} = \left[ \frac{\sinh(mL/2)}{\sinh(m_0L/2)} \right]^2.$$

The absolute normalization follows from the massless limit. For $A_0=-\mathrm d^2/\mathrm dx^2$, exclude the constant mode. Then

$$\zeta_{A_0/\mu^2}(s) = 2\left(\frac{\mu L}{2\pi}\right)^{2s} \zeta_{\mathrm R}(2s).$$

Using the DLMF values NIST DLMF 2026, $\zeta_{\mathrm R}(0)=-1/2$, Equation 25.6.1 and NIST DLMF 2026, $\zeta_{\mathrm R}'(0)=-\tfrac12\log(2\pi)$, Equation 25.6.11 gives

$$\det_\zeta'(A_0/\mu^2)=(\mu L)^2.$$

The nonzero-mode relative product is continuous at $m=0$. Indeed,

$$\begin{aligned} \sum_{n\ne0} \left| \frac{(2\pi n/L)^2+m^2}{(2\pi n/L)^2}-1 \right| &=m^2\sum_{n\ne0}\frac{1}{(2\pi n/L)^2}\\ &<\infty, \qquad m^2\downarrow0. \end{aligned}$$

The sum tends to zero, so the corresponding product tends to one. Only the approaching $n=0$ factor $m^2/\mu^2$ remains. Thus, as $m\downarrow0$,

$$\det_\zeta(A_m/\mu^2) \sim \frac{m^2}{\mu^2} \det_\zeta'(A_0/\mu^2) =m^2L^2.$$

This fixes the integration constant:

$$\det_\zeta(A_m/\mu^2) =4\sinh^2\!\left(\frac{mL}{2}\right).$$

The vanishing small-mass limit is not a failure of analytic continuation; it is the approaching zero eigenvalue. Boschi-Filho and Farina calculate the same periodic oscillator determinant by direct zeta continuation and obtain the bosonic partition function on Boschi-Filho and Farina 1995, PDF pp. 2–6, with their parameters mapped by $\beta=L$, $\omega=m$, periodic twist $\theta=0$, and bosonic determinant exponent $\sigma=-1/2$.

The one-loop bridge—and where it stops

For a real bosonic variable with Euclidean quadratic action

$$S_E[q] = \frac12\int_0^L q(x)A_mq(x)\,\mathrm dx,$$

the Gaussian rule gives, after fixing the same measure normalization at two positive masses,

$$\frac{Z_m}{Z_{m_0}} = \left[ \frac{\det_\zeta A_m}{\det_\zeta A_{m_0}} \right]^{-1/2} = \frac{\sinh(m_0L/2)}{\sinh(mL/2)}.$$

There is an independent canonical check. A harmonic oscillator of frequency $m$ has energies $E_r=m(r+1/2)$, and therefore

$$Z_m = \sum_{r=0}^{\infty}e^{-Lm(r+1/2)} = \frac{1}{2\sinh(mL/2)}.$$

The ratio of these operator traces agrees exactly with the determinant ratio. Equivalently, the one-loop Euclidean contribution satisfies

$$\Gamma_m^{(1)}-\Gamma_{m_0}^{(1)} = \frac12\log \frac{\det_\zeta A_m}{\det_\zeta A_{m_0}} = \log \frac{\sinh(mL/2)}{\sinh(m_0L/2)}.$$

This is the mathematical input to a one-loop calculation. In higher-dimensional QFT, $A^{-1}\partial_\alpha A$ need not be trace class, the heat coefficients identify local ultraviolet terms, and a physical answer still needs a measure, counterterms, normalization conditions, and a renormalization scheme. Gauge systems also require gauge fixing and ghost determinants; fermions and unstable saddles introduce signs or phases. Integrating Out Heavy Fields places determinant contributions inside matching and the local low-energy expansion.

A reliable spectral-determinant workflow

1. Specify the realization. Record the manifold, bundle, operator order, coefficients, domain, and boundary condition. Stop if the differential expression has not yet been turned into an operator.
2. Classify the spectrum. Verify self-adjointness, strong ellipticity when there is a boundary, compact resolvent, the kernel dimension, and the sign of every exceptional mode. Stop before using the positive zeta formula if negative or complex spectrum remains untreated.
3. Control both heat-time limits. Obtain enough short-time coefficients to reach the desired $s$-value, and prove decay of the primed trace at large time. A local expansion alone does not control zero modes.
4. Start inside the true Mellin half-plane. Define the zeta series where it converges before analytically continuing it. This anchors every later formula to a unique function.
5. Subtract finitely and locally. Split the Mellin integral, subtract a declared number of heat terms only near $t=0$, and retain the remainder integral. Increasing the subtraction order should leave the continued answer unchanged.
6. Make the determinant dimensionless. State $\mu$, or compute a ratio in which it cancels. Use a prime only after the kernel and its separate treatment have been identified.
7. Run independent checks. Compare with an exact spectrum, a Poisson or resolvent identity, a parameter derivative when it is trace class, the predicted $\mu$ dependence, and a controlled zero-mode limit.

The outputs are a meromorphic zeta function in a stated domain, its residues, $\zeta_A(0)$, $\zeta_A'(0)$, a primed or unprimed dimensionless determinant, and its scale response. The main cost is not the formal derivative at zero; it is obtaining trustworthy heat coefficients and the global finite part.

For a noncompact space, the ordinary heat trace usually contains an infinite volume. One then needs a relative trace, density, reference subtraction, or another problem-specific construction. Likewise, singular geometry, nonlocal boundary data, or a failed strong-ellipticity test is a stop condition for the standard expansion used here.

Common pitfalls

Confusing the heat kernel with its trace. $K_A(t;x,y)$ is a local integral kernel. $\Theta_A(t)$ integrates its diagonal and traces bundle indices, losing location data while retaining the eigenvalue sum.

Integrating the short-time series over all heat time. The heat expansion is asymptotic near $t=0$. Extending it to infinity can manufacture an infrared divergence or erase global information.

Calling every heat coefficient a zeta pole. The Mellin denominator gives a candidate pole, but $1/\Gamma(s)$ cancels candidates at nonpositive integers in the standard smooth problem.

Hiding a zero mode with a prime. A primed determinant is defined on the positive spectral complement. The missing mode still requires a separate measure or collective-coordinate treatment in a physical integral.

Taking the logarithm of a dimensionful eigenvalue. The determinant must be formed from $A/\mu^2$ or as a compatible ratio. Its scale response is part of the answer whenever $\zeta_A(0)\ne0$.

Treating analytic continuation as renormalization. Zeta continuation assigns a spectral invariant under stated hypotheses. It does not choose the physical counterterms, scheme, normalization conditions, or matching scale.

Using the positive formula for an unstable spectrum. Negative modes and complex eigenvalues carry phase information. Taking absolute values silently changes the problem.

Assuming all regularized determinants multiply. Zeta determinants need not satisfy $\det_\zeta(AB)=\det_\zeta(A)\det_\zeta(B)$ without additional hypotheses; Vassilevich 2003, § 7, PDF p. 65 records this multiplicative anomaly. They are also distinct from the Fredholm determinants treated in Trace Ideals and Fredholm Determinants.

Check your understanding

1. Retrieval: reconstruct the chain

For a strictly positive order-two elliptic operator on a closed compact $d$-manifold, state the heat trace, spectral zeta function, their Mellin relation, the initial convergence half-plane, and the dimensionless zeta determinant.

Solution

With eigenvalues $\lambda_n>0$,

$$\begin{aligned} \Theta_A(t)&=\sum_ne^{-t\lambda_n},\\ \zeta_A(s)&=\sum_n\lambda_n^{-s},\\ \Gamma(s)\zeta_A(s) &=\int_0^\infty t^{s-1}\Theta_A(t)\,\mathrm dt, \qquad \Re s>\frac d2. \end{aligned}$$

Finite short-time subtraction continues $\zeta_A$ to $s=0$. With an order-two operator and reference mass $\mu$,

$$\log\det_\zeta(A/\mu^2) =-\zeta_{A/\mu^2}'(0).$$

The half-plane comes from the $t^{-d/2}$ leading heat behavior; strict positivity controls the large-$t$ endpoint.

2. Hypothesis check: expose the circle zero mode

Set $m=0$ in the periodic circle operator. What fails in the unprimed construction, and what does the prime repair?

Solution

The constant function has eigenvalue $\lambda_0=0$. Hence $\Theta_{A_0}(t)\to1$ as $t\to\infty$, the expression $0^{-s}$ makes the unprimed zeta sum meaningless, and the quadratic Gaussian has a flat direction. Define instead

$$\Theta_{A_0}'(t) = \Theta_{A_0}(t)-1, \qquad \zeta_{A_0}(s) = \sum_{n\ne0} \left(\frac{2\pi n}{L}\right)^{-2s}.$$

This produces $\det_\zeta'(A_0/\mu^2)$ on the orthogonal complement of the constant mode. It does not integrate that mode or determine its physical measure.

3. Calculation: map a heat power to the zeta plane

Suppose the primed heat trace contains $c_jt^{(j-d)/2}$. Find its contribution to the split Mellin integral, locate the candidate zeta pole, and explain the special role of nonpositive integers.

Solution

On $(0,t_0)$,

$$\int_0^{t_0} t^{s-1}c_jt^{(j-d)/2}\,\mathrm dt = \frac{c_jt_0^{s+(j-d)/2}} {s+(j-d)/2}.$$

Thus $\Gamma(s)\zeta_A(s)$ has a candidate pole at $s=(d-j)/2$ with residue $c_j$. The zeta function itself includes the factor $1/\Gamma(s)$. Because this factor vanishes at $s=0,-1,-2,\ldots$, it can cancel the candidate there. In particular, the $j=d$ denominator is $c_d/s$, and multiplication by $1/\Gamma(s)=s+O(s^2)$ gives the finite value $\zeta_A(0)=c_d$.

4. QFT transfer: integrate the circle determinant

Use

$$2m\sum_{n\in\mathbb Z} \frac{1}{(2\pi n/L)^2+m^2} =L\coth(mL/2)$$

to compare masses $m$ and $m_0$. What is the determinant ratio, what is the bosonic one-loop difference, and what information is still absent from a higher-dimensional QFT calculation?

Solution

Since the left side is $\mathrm d\log\det_\zeta A_m/\mathrm dm$, integration gives

$$\frac{\det_\zeta A_m}{\det_\zeta A_{m_0}} = \left[ \frac{\sinh(mL/2)}{\sinh(m_0L/2)} \right]^2.$$

For one real positive Euclidean boson,

$$\Gamma_m^{(1)}-\Gamma_{m_0}^{(1)} = \log \frac{\sinh(mL/2)}{\sinh(m_0L/2)}.$$

In a higher-dimensional QFT this spectral result does not yet specify the functional measure, local counterterms, subtraction scheme, normalization conditions, matching prescription, gauge and ghost factors, or treatment of zero and negative modes. Those data belong to the physical calculation.

Synthesis and continuations

The heat trace turns a discrete positive spectrum into a function whose two time limits separate local ultraviolet structure from infrared spectral data. Its Mellin transform defines $\zeta_A(s)$ where the series converges; finite short-time subtraction continues it to the origin. The value $\zeta_A(0)=a_d-N_0$ fixes scale dependence, while $-\zeta_A'(0)$ supplies the global finite part of the zeta determinant. The circle benchmark verifies the whole chain against both Poisson summation and an operator-trace partition function.

Mellin Transforms and Scaling Asymptotics gives the general pole and remainder machinery used in the continuation. Special Functions from Equations and Boundary Data solves the complementary problem of selecting individual modes. Integrating Out Heavy Fields continues from a mathematical determinant to a physical one-loop matching calculation with counterterms and scale separation.

References

• H. Boschi-Filho and C. Farina, Generalized Thermal Zeta-Functions, Physics Letters A 205 (1995) 255–260, arXiv:hep-th/9505154v1, PDF pp. 2–6. Periodic oscillator spectrum, analytic zeta continuation, determinant, and the bosonic partition function $[2\sinh(mL/2)]^{-1}$.

• Peter B. Gilkey, Invariance Theory, the Heat Equation, and the Atiyah–Singer Index Theorem, PDF, first edition, Publish or Perish (1984), electronic reprint (1996), § 1.10, printed pp. 78–80 (PDF pp. 83–85). Heat asymptotics, the Mellin construction of the zeta function, regularity at the origin, and removal of the null space.

• Klaus Kirsten, Spectral Functions in Mathematics and Physics, arXiv:hep-th/0005133v1 (2000), § 1, PDF pp. 1–5. Elliptic spectral zeta functions, their heat-kernel relation, regularity at the origin under standard hypotheses, dimensionless determinants, and QFT scale ambiguity.

• NIST Digital Library of Mathematical Functions, version 1.2.7, released June 15, 2026, National Institute of Standards and Technology, Equation 4.36.3 and Equations 25.6.1 and 25.6.11. The partial fraction expansion of $\coth z$ and the required Riemann-zeta values at the origin.

• D. V. Vassilevich, Heat Kernel Expansion: User’s Manual, Physics Reports 388 (2003) 279–360, arXiv:hep-th/0306138v3, Equation (2.2), PDF p. 11; § 2.2, PDF pp. 14–17; § 7, PDF p. 65. Laplace-type convention, trace-class heat evolution, local heat coefficients, the heat–zeta transform, determinant scale dependence, zero- and negative-mode qualifications, and the multiplicative anomaly.