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Interpret $L=-\square$ as the minimally coupled scalar positive Laplace-Beltrami operator on the smooth closed manifold. Writing $dV$ for its metric volume measure (the Riemannian volume form on an oriented chart), integration by parts gives $⟨u,Lu⟩={\int }_M|\nabla u{|}^{2}dV\ge 0$. The compact elliptic spectral theorem supplies an orthonormal basis $u_j$ with $L{u}_j={\lambda }_j{u}_j$, real nonnegative eigenvalues tending to infinity and finite multiplicities. Solve this eigenvalue problem to obtain the spectral zeta functionThe zero eigenvalues are excluded, not assigned inverse powers. Their multiplicity $n_0$ is the number of connected components: $Lu=0$ implies $\int |\nabla u{|}^{2}dV=0$, so $u$ is constant on each component.

An equivalent calculation uses the spectral expansion of the Riemannian heat kernel,Since ${\int }_0^{\infty }{t}^{s-1}{e}^{-t\lambda }dt=\Gamma (s){\lambda }^{-s}$ for $\lambda >0$, the Mellin transform givesThe reduced trace decays exponentially at large $t$. At small $t$, subtract a finite number of terms of the heat kernel expansion, integrate those terms explicitly, and then integrate the remainder. Increasing the number of subtractions extends the formula successively to further left half-planes, producing the meromorphic continuation of ${\zeta }_L$ rather than substituting $s=0$ in its divergent defining series.

For completeness, the needed curvature coefficient comes from the heat-kernel transport equations. In normal coordinates $\xi$ centered at $y$, the Riemannian volume form has density $J(\xi )=1-\frac{1}{6}{R}_{ab}(y){\xi }^a{\xi }^b+O(|\xi {|}^3)$. The leading Gaussian amplitude is thereforeSubstitution of the Gaussian amplitude series into the heat equation gives the first diagonal transport relation $u_1(y,y)={\square }_x{u}_0(x,y){|}_{x=y}=R(y)/6$. The curvature coefficient of the scalar heat kernel is consequentlyIn dimension two this impliesSet $A_0=\mathrm{Area}(M)/(4\pi )$ and $A_1={\int }_{M}RdV/(24\pi )$. Splitting the Mellin integral at $t=1$ and subtracting these two heat terms gives, near $s=0$,where $H$ is holomorphic for $\mathrm{Re}s>-1$. As $1/\Gamma (s)=s+O(s^2)$, only the second term survives at zero. Thus the zero-mode correction to the spectral zeta value isThe Gauss-Bonnet theorem, with $R$ twice the Gaussian curvature, makes this $\boxed{{\zeta }_L(0)=\chi (M)/6-n_0}$. For a connected surface $n_0=1$, so the answer is $\chi (M)/6-1$; a sphere gives $-2/3$, and a flat torus gives $-1$.

The purely local heat coefficient is $A_1=\int RdV/(24\pi )$. Sometimes a formal formula for $\zeta (0)$ quotes this coefficient with zero modes left implicit. For the actual positive-spectrum spectral zeta function defined above, the compact scalar problem necessarily has the additional subtraction $-n_0$. A strictly positive shifted operator is a different problem and has its own heat coefficients.

For a Laplace-type operator on a closed manifold, let $a_{t^0}$ be the constant coefficient of its full heat trace expansion. In the Mellin transform defining the spectral zeta function, the reduced trace has constant coefficient $a_{t^0}-\dim \ker L$. Its contribution is $(a_{t^0}-\dim \ker L)/s$, and $1/\Gamma (s)=s+O(s^2)$ proves the identity. For the scalar positive Laplace-Beltrami operator on a closed surface, $a_{t^0}=\int RdV/(24\pi )$ and $\dim \ker L$ is the number of connected components. In particular, a connected sphere has value $-2/3$ and a connected flat torus has value $-1$.