Consider the integral depending on 2 parameters
$$f(\tau,x):=\int\_{-\infty}^{+\infty}\frac{dp}{\sqrt{p^2+1}}e^{-\sqrt{p^2+1}\tau+ipx},$$
where $\tau >0,x\in \mathbb{R}$. This integral absolutely converges in this region.


**Question. Is it true that $f(\tau,x)$ is $SO(2)$ invariant in the above region, i.e. depends only on $\tau^2+x^2$?**


Remark. My motivation to ask this question comes from the well known fact that after the so called Wick rotation $\tau=it$ one gets an integral invariant under the Lorentz transformations in the plane $(t,x)$. 



Rotation invariance of an integral
Yes. Denote $p=\sinh s$, then $s\in (-\infty,\infty)$, $\sqrt{p^2+1}=\cosh s$, $dp=\cosh s\, ds$. Next, denote $r=\sqrt{\tau^2+x^2}$, $\tau=r\cos \varphi$, $x=r\sin \varphi$, where $\varphi\in (-\pi/2,\pi/2)$. So our integral rewrites as $$\int\_{-\infty}^\infty ds \exp(-r\cosh(s-i\varphi)).$$
It does not depend on $\varphi$ as may be seen from the rectangular contour (integrals over vertical sides tend to 0). 


