Asymptotic expansion of heat operator $e^{-\Delta{t}}$ and $e^{-\mathcal{D}t}$ of Dirac operator For a closed Riemannian manifold $M$ of $n$-dimension, we consider the Laplace-Beltrami operator $\Delta$.
It is known that we have an asymptotic expansion for the trace of heat operator $e^{-\Delta{t}}$ as follows
$$
\mathrm{tr}(e^{-\Delta{t}})=\sum_{\lambda}e^{-\lambda{t}}\overset{t\downarrow0}{\sim}t^{-\frac{n}{2}}\sum_{n} \alpha_{n}t^{n},
$$
where $\lambda$ runs over the set of spectrum of Laplacian $\Delta$.
My question is that

Denote by $\mathcal{D}$ the Dirac operator whose square coincides with the Laplacian, $i.e.$ $\mathcal{D}^{2}=\Delta$. 
  Then the sum of positive eigenvalues of an operator $e^{-\mathcal{D}t}$ $$\sum_{\lambda\in\mathrm{Sp}(\Delta)}e^{-\sqrt{\lambda}{t}}$$ has an asymptotic expansion around $t=0$?
  If it exists, then is it possible to induce a relation between coefficients?

I know that the proof for the case of heat operator follows from the construction of heat kernel. But I wonder that the same construction can be applied to the Dirac operator.
Thank you for your time and effort.
 A: We define the spectral zeta function of the Laplacian $\Delta$ by
$$
\zeta_{\Delta}(s) = \sum_{\lambda\in\mathrm{Sp}(\Delta)}\lambda^{-s}.
$$
It is known that $\zeta_{\Delta}(s)$ converges absolutely over $\mathrm{Re}(s)>\frac{n}{2}$, and has an analytic continuation to a meromorphic function over all complex plane.
It can be expressed by 
$$\zeta_{\Delta}(s) = \frac{1}{\Gamma(s)}\int_{0}^{\infty}t^{s-1}\sum_{\lambda\in\mathrm{Sp}(\Delta)}e^{-\lambda{t}}dt,$$
where $\Gamma(s):=\int_{0}^{\infty}t^{s-1}e^{-t}dt$ is the Gamma function. 
From the identities
$$
\Gamma(s)\lambda^{-s}=\int_{0}^{\infty}t^{s-1}e^{-\lambda{t}}dt
$$
and
$$
\Gamma(2s)\lambda^{-s}=\int_{0}^{\infty}t^{2s-1}e^{-\sqrt{\lambda}t}dt,
$$
we deduce the following relation
$$
\zeta_{\Delta}(s) = \frac{1}{\Gamma(s)}\int_{0}^{\infty}t^{s-1}\sum_{\lambda\in\mathrm{Sp}(\Delta)}e^{-\lambda{t}}dt \\
=\frac{1}{\Gamma(2s)}\int_{0}^{\infty}t^{2s-1}\sum_{\lambda\in\mathrm{Sp}(\Delta)}e^{-\sqrt{\lambda}t}dt.
$$
Since the spectral zeta function $\zeta_{\Delta}(s)$ is meromorphic over a complex plane, $i.e.$ is analytic except for discrete singularities, the series 
$$
\sum_{\lambda\in\mathrm{Sp}(\Delta)}e^{-\sqrt{\lambda}t}
$$
has an asymptotic expansion.
