Integral kernel of resolvent of Laplacian Consider the Laplacian $-\Delta: C^\infty (S^1) \to C^\infty(S^1)$ where $\mathbb R/\mathbb Z=S^1$ and for a periodic function $f:\mathbb R\to\mathbb R$ we have $-\Delta f=-f''$. 
For the orthonormal basis $\{\phi_n=e^{inx}\}_{n\in \mathbb Z}$ of $L^2(S^1)$ we have $-\Delta \phi_n= n^2 \phi_n $, so $\lambda_n=n^2$ are eigenvalues of $-\Delta$. 
One can define the resolvent $(1-\Delta)^{-1}f= \sum_{n\in \mathbb Z} \langle f, \phi_n \rangle (1+\lambda_n)^{-1} \phi_n $ and more generally for $\delta>0$ we have
\begin{align}
(1-\Delta)^{-\delta}f(x)&= \sum_{n\in \mathbb Z} \langle f, e^{in(\cdot)}  \rangle (1+n^2)^{-\delta} e^{inx} \\ &= \sum_{n\in \mathbb Z}\hat f(n) (1+n^2)^{-\delta} e^{inx}
\end{align}
which makes sense e.g. for $L^2$ functions, or for distributions.
I want to compute an integral kernel for $(1-\Delta)^{-\delta}$, which according to a paper I am reading has a singularity of order $|x-y|^{2\delta-1}$. What is a justification for that?
EDIT: the kernel should be $K(x,y)=\sum_{n\in \mathbb Z} (1+n^2)^{-\delta} e^{inx} e^{iny}$ but how can this be converted to an expression that includes $|x-y|^{2\delta-1}?$ 
I suspect that we should obtain an expression of the form
$$(1-\Delta)^{-\delta}f(x)= \int_0^{2\pi} \bigg(\textrm{linear expression of $f(x+y),f(x-y),f(x)...$} \bigg) \tilde K(x,y)dy $$
where $\tilde K$ has a singularity $|x-y|^{2\delta-1}$. I am also a bit worried about the periodicity of the kernel. Any ideas would be appreciated.
 A: Not sure if there is a closed-form solution for the kernel, but the
singularity might be related to the Poisson summation formula, 
\begin{align*}
\sum_{n\in\mathbb{Z}}\hat{f}\left(x+2\pi n\right) & =\sum_{n\in\mathbb{Z}}f\left(n\right)e^{ixn}.
\end{align*}
Let $f\left(x\right)=\left(1+x^{2}\right)^{-\delta}$, and it is known
that 
\begin{align*}
\hat{f}\left(t\right) & =C_{\delta}\left|t\right|^{\delta-\frac{1}{2}}K_{\frac{1}{2}-\delta}\left(\left|t\right|\right)
\end{align*}
where $C_{\delta}$ is a constant, and $K$ is the modified Bessel
function of the second kind. Since the kernel is $k\left(x,y\right)=\sum_{n\in\mathbb{Z}}\left(1+n^{2}\right)^{-\delta}e^{in\left(x-y\right)}$,
so a $2\pi$-periodic extension of $\hat{f}$ will give the type of
singularity as stated.
For singularity, only need the case when $\delta-\frac{1}{2}<0$.
From Abramowitz (pg. 375):
(1) $K_{\nu}\left(z\right)=K_{-\nu}\left(z\right)$, $\nu\in\mathbb{R}$; 
(2) for $\nu>0$, $K_{\nu}\left(z\right)\sim\frac{1}{2}\Gamma\left(\nu\right)\left(\frac{2}{z}\right)^{\nu}$,
$z\rightarrow0$. 
So 
\begin{align*}
\left|t\right|^{\delta-\frac{1}{2}}K_{\frac{1}{2}-\delta}\left(\left|t\right|\right) & \sim\left|t\right|^{\delta-\frac{1}{2}+\delta-\frac{1}{2}}=\left|t\right|^{2\delta-1},\;t\rightarrow0.
\end{align*}
