Let $T=\sum_{|\alpha|=q}a_{\alpha}\frac{\partial^{|\alpha|}}{\partial x^{\alpha}}$ be an order $q$ constant coefficient differential operator on $\mathbb{R}^n$ and $\Delta=\sum_{|\alpha|=2}b_{\alpha}\frac{\partial^{|\alpha|}}{\partial x^{\alpha}}$ be a positive order two differential operator. Using Cauchy integrals one can define $\Delta^{z}$ for complex values of $z$. If $s$ has sufficiently big real part, then one can show that the operator $T\Delta^{-s}$ will be trace class and in particular, Hilbert Schmidt. Hilbert Schmidt operators have kernels and one can show that if the operator is furthermore trace class, then its trace is an integral of the kernel evaluated at the diagonal. Let us denote the kernel of $T\Delta^{-s}$ by $k_s$.

How to prove that $k_s(x,x)=\int_{\mathbb{R}^n}tr(\sigma(T)\sigma(\Delta)^{-s})d\xi$ where $\sigma(T)$ is the symbol of $T$ meaning the function $\xi \mapsto \sum_{|\alpha|=q}a_{\alpha} (i\xi)^{\alpha}$ where $\xi^{\alpha}:=\xi_1^{\alpha_1} \cdot ... \cdot \xi_n^{\alpha_n}$ (we deal with constant coefficient operators).

One can also show that the mapping $z \mapsto k_z(x,x)$ is $C^{\infty}$-valued meromorphic function and also that the function $Tr(T\Delta^{-s})$ extends to the meromorphic function with at most simple poles at $(n+q)/2,(n+q-1)/2,...$ we already know that for sufficiently large $s$ we have $Tr(T\Delta^{-s})=\int_{\mathbb{R}^n}k_s(x,x)dx$. It turns out that for the poles we have the following formula: $Res_{s_0}Tr(T\Delta^{-s})=\int_{\mathbb{R}^n}Res_{s_0}k_s(x,x)dx$.

Why we have the following formula: $$Res_{(n+q)/2}k_s(x,x)=c_n \int_{S^n}tr(\sigma(T)\sigma(\Delta)^{-\frac{n+q}{2}})d\xi$$ where $c_n$ is some constant and $S_n$ is the $n$-dimensional sphere?

This question is inspired by the index theory (Atiyah-Bott approach to index as a value of zeta function).

EDIT: some typos corrected.