I am trying to verify that the heat's kernel is positive using the inverse Fourier transform. For this, I calculate the heat's kernel by means of a contour integral. After verifying this fact, my question arises.
Let \begin{align} p_t(x):=\int_{\mathbb{R}^n} \mathrm{e}^{ix\cdot \xi} \mathrm{e}^{-t|\xi|^2}\,d\xi=:I \end{align} heat's kernel.
Since \begin{align}-t|\xi|^2+ix\cdot \xi=-t(\xi_1^2+\cdots+\xi_n^2)+i(x_1\xi_1+\cdots+x_n\xi_n) \end{align}then
\begin{align} I=I_1\cdots I_n \end{align}
where $$I_i=\int_{-\infty}^{\infty} \mathrm{e}^{-t\xi_i^2+i x_i\xi_i}\,d\xi_i$$
Now, let $$I_s:=\int_{-\infty}^{\infty} \mathrm{e}^{-ts^2+i\lambda s}\,ds$$
Let us note that $$-ts^2+i\lambda s=-(\sqrt{t}s-i\frac{1}{2}\lambda \frac{1}{\sqrt{t}})^2-\frac{1}{4}\lambda^2\frac{1}{t}$$
then \begin{align} I_s=\mathrm{e}^{-\frac{1}{4}\lambda^2\frac{1}{t}}\int_{-\infty}^{\infty}\mathrm{e}^{-(\sqrt{t}s-i\frac{1}{2}\lambda \frac{1}{\sqrt{t}})^2}\,ds \end{align}
and with $u=\sqrt{t}$ and $\int_{-\infty}^{\infty} \mathrm{e}^{-(u-i\frac{1}{2}\lambda\frac{1}{\sqrt{t}})^2}\,du=\int_{-\infty}^{\infty}\mathrm{e}^{-u^2}\,du=\sqrt{\pi}$
(Gaussian integral $\int_{-\infty}^\infty \exp(-(x+\mathrm iY)^2)\,\mathrm d x$ along $[-R,R]+\mathrm i[0,Y]$)
\begin{align} I_s=\frac{\sqrt{\pi}}{\sqrt{t}}\mathrm{e}^{-\frac{1}{4}\lambda^2\frac{1}{t}} \end{align}
Therefore, $I=(I_s)^n>0$
If $f(\xi)=|\xi|^2$, we see that $f(\xi)$ is positive (and elliptic).
Question 1. In general, if $f(\xi)$ is positive and elliptic, then is the kernel $p_t(x):=\int_{\mathbb{R}^n}\mathrm{e}^{ix\cdot\xi}\mathrm{e}^{-tf(\xi)}\,d\xi$ is positive? Thank you.