Retrieving Euler characteristic from spectral zeta function I have a question about the relation between the spectral zeta function of the Laplacian on a given space and the associated heat kernel.
From what I understand, one can retrieve the heat kernel from the spectral zeta function via an inverse Mellin transform:
$$Z(t)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}ds\zeta(s)\Gamma(s)t^{-s}$$
You can then use the residue theorem in order to express the heat kernel as an asymptotic series depending on t. 
Say I'm interested in the constant term of this asymptotic series (corresponding to the power $t^0$). Then assuming my zeta function does not have a pole at $s=0$, this term only emerges as a result of the pole of the Gamma function at $s=0$, and it will be equal to the value of the spectral zeta function at $s=0$. Am I correct?
Since in the case of compact manifolds, the constant term in the heat kernel expansion is related to the Euler characteristic, this means that I can hypothetically find the Euler characteristic using only the value of the spectral zeta function as $0$, right? 
Thanks in advance!
 A: Indeed, $\zeta(0)$ is the coefficient on $t^0$ in the $t \to 0$ asymptotic expansion of $Z(t)$.
Your idea about the Euler characteristic is an interesting one, but in the compact manifold case, there's an issue with constant terms. I usually think about this from the other direction: Given the heat trace $\operatorname{Tr} e^{-t\Delta}$, one defines the zeta function via the Mellin transform. But actually one needs to subtract the trace on harmonic forms, i.e., to take the Mellin transform of $\operatorname{Tr} e^{-t\Delta} \Pi_{(\ker \Delta)^\perp} = \operatorname{Tr} e^{-t\Delta} - \dim \ker \Delta$, which decays exponentially as $t \to \infty$, which ensures integrability.
If $\Delta_k$ denotes the Laplacian on $k$-forms, then $\chi(M) = \sum_{k} (-1)^k \dim \ker \Delta_k$ by the Hodge theorem, which also equals $\sum_k (-1)^k \operatorname{Tr} e^{-t \Delta_k}$, which is constant in $t$ (this is the McKean-Singer formula). But then $\sum_k (-1)^k \operatorname{Tr} e^{-t \Delta_k} \Pi_{(\ker \Delta)^\perp} = \sum_k (-1)^k \operatorname{Tr} e^{-t \Delta_k} - \sum_k (-1)^k \dim \ker \Delta$ is identically zero. So one may be tempted to say that $\sum_k (-1)^k \zeta_{\Delta_k}(0)$ is the Euler characteristic, but in fact it's zero. Essentially, we had to subtract the Euler characteristic to get the integral in the Mellin transform to converge.
