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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.

bnado
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  • this is the same then asking for the asymptotics of $2\int_0^{\infty}\frac{\cos(kx)}{k^{\rho}+1}$,right? – tired Jul 27 '16 at 14:54
  • yes, it is the same – bnado Jul 27 '16 at 14:58
  • @user1952009 I cannot use the Laplace method for the integral at the beginning, since there is no saddle point (you get an equation that is zero for a complex $|p|$). If I consider only the real positive axis, like in tired's comment, to find the saddle point, i then have to solve an equation that I cannot solve analytically. – bnado Jul 27 '16 at 15:02
  • First of all i think you should replace $p x=y$ and then restict the integration range to a region around the origin where you can taylor expand – tired Jul 28 '16 at 08:38
  • @tired say I do this and I integrate in a region between $-1/x$ and $1/x$, then I will be doing the integral $\int_{-1/x}^{1/x} f(p)e^{i p x}dp$ (forget about the change of variable, it doesn't change this). Then I will also get terms that go like $1/|x|$. This happens because effectively I'm doing the Fourier transform of the function $f(p)\theta(p+1/x)\theta(1/x-p)$. This function has other non analyticy points (the points $\pm 1/x$) and therefore I get different asymptotics – bnado Jul 28 '16 at 08:44
  • i would stick to the rep with cosine i posted above – tired Jul 28 '16 at 09:06
  • @tired I can write it as cosine but it doesn't change what I just said – bnado Jul 28 '16 at 09:08
  • ok, i will try to think more deeply about your question later on... – tired Jul 28 '16 at 09:13

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