I have to find the asymptotic behavior of the Fourier transform of a function that is non analytical, but has a cusp. Given $$f(p)=\frac{1}{|p|^\sigma+1}$$ with $\sigma \in \mathbb{R}$, $\sigma >0$, what is the behavior of $$\hat{f}(x)=\int_{-\infty}^\infty dp \frac{e^{i p x}}{|p|^\sigma+1}$$ for $x \gg 1$?
I have computed this numerically and I know that $\hat{f}(x) \sim \frac{1}{x^{\sigma+1}}$ (I don't care about any factor, I just need the power law exponent). I would like to obtain this result analytically (and rigorously).
Since the power law behavior is given by the cusp of $f(p)$ at $p=0$ (if not for this cusp, the function would be analytical and $\hat{f}(x)$ would decay exponentially), I thought about integrating over a small region around zero, and that should tell me something about the asymptotics of $\hat{f}(x)$. However, considering the integral over a finite region would mean that I'm inserting some other non-analyticity in the problem (I would effectively be multiplying my function $f(x)$ by some step function), so I would get other powers of $x$.
Any ideas?
EDIT: saddle point approximation doesn't lead anywhere. Write the integral as $\hat{f}(x)=2\int_0^\infty \frac{\cos(px)}{p^\sigma+1}=\int_0^\infty \frac{e^{i p x}}{p^\sigma+1}+\int_0^\infty \frac{e^{-i p x}}{p^\sigma+1}$. Now write $\frac{e^{\pm i p x}}{p^\sigma+1}=e^{\pm i p x - \log (p^\sigma+1)}$. To find the where the saddle point is located I need to solve the equation $\pm i x -\sigma \frac{p^{\sigma-1}}{(p^\sigma +1)}=0$, which I cannot solve analytically.