$\sum_{k \geq 1} e^{it \sqrt{\lambda_k}}$ - Theory of distribution An exercise asks to find the wave trace $w(t)=\operatorname{tr} \left(e^{it \sqrt\Delta}\right)=\sum_{k \geq 1} e^{it \sqrt{\lambda_k}}$   as a distribution (or generalized function) of the Laplacian on the $1$-D circle. (Hint : spectrum ($\mathbb{S}^1)=\{k^2 : k \in \mathbb{N}^*\}$) 
I thought I should compute $\langle w,\varphi\rangle =\int_{\mathbb{S}^1} \left(\sum_{k \geq 1} e^{itk}\right) \varphi(t) \, dt$, where $\varphi \in D(\mathbb{S}^1)$, However, this is a wrong answer, but how do I have to interpret the Wave trace as a distribution in considering spect ($\mathbb{S}^1$)? Is there anyone could give me a track of reflexion? Do I have to consider the transformation $t \to |t|$ such that $t=(x,y) \in \mathbb{S}^1$?
So far I think we should obtain $$\langle w,\varphi\rangle =\int_{\mathbb{S}^1} \left(\sum_{k \geq 1} e^{i|t|k}\right) \varphi(t) \, dt =  \sum_{k \geq 1} \int_{\mathbb{S}^1} \left(e^{i|t|k} \varphi(t) \right) \, dt \stackrel{??}{=} \sum_{k \geq 1} e^{ik} \int_0^{2 \pi} \int_0^1  \varphi(r, \theta) r \, dr d\theta. $$
From this time, we could compute it with the Poisson Summation Formula.
From the bottom of the page $27$ of the pdf, we could find an example on the real line instead of $\mathbb{S}^1$
 A: the Laplacian is diagonalized in the orthonormal basis of eigenfunctions $(u_k)$ and $$\Delta \varphi = \sum_k u_k \langle u_k, \varphi \rangle \lambda_k$$ this operator is normal hence $\Delta^a$ means : $$\Delta^a \varphi = \sum_k u_k \langle u_k, \varphi \rangle (\lambda_k)^a$$ and for any complex number $t$ : $$e^{i t \Delta^{1/2}} \varphi = \sum_{m=0}^\infty (it)^m \frac{\Delta^{m/2}}{m!} \varphi =  \sum_{m=0}^\infty \frac{(it)^m}{m!}  \sum_k u_k \langle u_k, \varphi \rangle (\lambda_k)^{m/2} =  \sum_k u_k\langle u_k, \varphi \rangle\sum_{m=0}^\infty \frac{(it)^m}{m!}  (\lambda_k)^{m/2} = \sum_k u_k\langle u_k, \varphi \rangle e^{it (\lambda_k)^{1/2}}$$
from this the trace of the operator is : $$tr(e^{i t \Delta^{1/2}})  = \sum_k e^{it (\lambda_k)^{1/2}}$$
given that in $\mathbb{S}^1$ the eigenvalues are $\lambda_k = k^2, k \ge 1$, if $Re(it) < 0$ (so that $|e^{it}| < 1$) you get : $$tr(e^{i t \Delta^{1/2}})  = \sum_{k =1}^\infty  e^{it k} = \frac{e^{it}}{1-e^{it}}$$
and this is indeed the answer. in terms of distributions on $\mathbb{R}$ (see page 26 of your pdf ), 
with $w(t) = tr(e^{i t \Delta^{1/2}})$ : $$\langle w, \phi \rangle = \lim_{\epsilon \to 0^+} \int_{-\infty}^\infty w(t+i \epsilon) \phi(t) dt = -i\int_{-\infty}^\infty \phi'(t) \ln(1-e^{it})dt\qquad \qquad \phi \in C^\infty_c(\mathbb{R})  $$ 

Note that with $t \in \mathbb{R}$, $\sum_{k=1}^\infty e^{itk}$ is not well-defined as a function, but it is as a distribution. By the properties of test functions $\phi \in C^\infty_c(\mathbb{R})$, we know that the Fourier transform $\hat{\phi}(\omega) = \int_{-\infty}^\infty e^{i \omega t} \phi(t) dt$ is a Schwartz function (see Schwartz space), hence $T(\phi)   =\sum_{k= 1}^\infty \langle e^{i t k},\phi \rangle = \sum_{k= 1}^\infty \hat{\phi}(k)$
is an absolutely convergent series. by integration by parts : $$T(\phi) = \sum_{k= 1}^\infty \frac{i}{k}\langle e^{i t k},\phi' \rangle = -i\int_{-\infty}^\infty \ln(1-e^{it}) \phi'(t) dt$$
(where $\ln(1-e^{it})$ is locally integrable ($L^1_{loc}(\mathbb{R})$) hence $\phi \mapsto -i\int_{-\infty}^\infty \ln(1-e^{it}) \phi'(t)$ is automatically a well-defined distribution).
Hence, in the sense of distributions, with $\Delta$ the Laplacian on the circle : $$\langle tr(e^{it \Delta^{1/2}}), \phi \rangle = \sum_{k= 1}^\infty \langle e^{i t k},\phi \rangle  = -i\int_{-\infty}^\infty \ln(1-e^{it}) \phi'(t) dt$$
