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In $\mathbb{R}^d$, let $f(x)=|x|^{-t}$, its Fourier tranform $F(f)(ξ):=(2\pi)^{-\frac{d}{2}}∫_{\mathbb{R}^d} e^{ix\cdot ξ}f(x)dx$, is there any fast way to see that this integral converge at $ξ \neq0$ for all $t\in (0,d)$?

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switch to polar coordinates, to get radial dependence. The Jacobian in $d$ dimensions should take care of the singularity. –  Alex R. Feb 5 '13 at 3:21
    
@Alex Near $0$, yes. Near $\infty$, quite the opposite. –  user53153 Feb 8 '13 at 3:42
    
This must be done via Cauchy Principle Value –  Alex R. Feb 8 '13 at 17:21
    
Is this means only for $t\in (d-1,d)$, its Fourier transform is a regular function(with out the point $\xi=1$); for $t\in (0,d-1]$, its Fourier transform is a distribution? –  Alron Feb 8 '13 at 22:13
    
@Alron Yes, and by the way, this question came up before: math.stackexchange.com/a/48446 –  user53153 Feb 8 '13 at 22:23

2 Answers 2

Thanks for the reply. Maybe I've made a stupid mistake, but here is what I get: $\int_{\mathbb{R}^d}\frac{e^{ix\cdot \xi}}{|x|^t}dx=\int_{\mathbb{R}^d}\frac{\cos(x\cdot \xi)}{|x|^t}dx=\frac{1}{|\xi|^{d-t}}\int_{\mathbb{R}^d}\frac{\cos(y_1)}{|y|^t}dy$, after the change of variable $\frac{y}{|\xi|} \to x$ and a rotation on the coordinate system. Then I switch to the spherical, $y_1=r\cos\theta$, $y_2=r\sin\theta \cos\beta_1, \cdots$, end up with this integral: $\int_0^{\pi}d\theta \int_0^{\pi} d\beta_1 \cdots \int_0^{\pi}d\beta_{d-3}\int_0^{2\pi}d\beta_{d-2}\int_0^{\infty}\frac{\cos(r\cos\theta)}{r^t}r^{d-1}\sin^{d-2}\theta\sin^{d-3}\beta_1\cdots \sin\beta_{d-3}dr.$ Let $C(d)= \int_0^{\pi} d\beta_1 \cdots \int_0^{\pi}d\beta_{d-3}\int_0^{2\pi}d\beta_{d-2}\sin^{d-3}\beta_1\sin^{d-3}\beta_2 \cdots \sin\beta_{d-3}$ be the constant obtained by integration on the $\beta_i$ angles. Then the sigular part is $\int_0^{\pi}\int_0^{\infty}\frac{\cos(r\cos\theta)}{r^t}r^{d-1}\sin^{d-2}\theta drd\theta$ . I change variable using $r\cos\theta \to z$, end up with this $(\int_0^{\pi}\sin^{d-2}\theta \cos^{d-1-t}\theta d\theta)(\int_0^{\infty}\frac{\cos z}{z^{t+1-d}}dz)$, the first "()" is fine since $d-1-t>-1$ for all $t\in (0,d)$, it seems to me that the second "()"is finite only when $t\in (d-1, d)$(given $t$ is positive), missing the $(0,d-1]$ part.

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First, consider the meaning of "converge". Since the absolute value of the integrand is $|x|^{-t}$, the integral does not converge absolutely for any $t\in (0,d)$. Conditional convergence is the most we can hope for; that is, convergence under a certain way of exhausting $\mathbb R^d$ by sets of finite measure. We can choose to exhaust by balls $\|x\|\le R$, or by cubes $\max |x_i|\le R$, etc. For some exhaustions the integral will diverge, for others it will converge (to a value that depends on the exhaustion method). The symmetry of $f$ makes the spherical approach more appealing than others, but does not make it more "correct" than others.

I don't quite understand your integral $C(d)\int_0^{\pi}\int_0^{\infty}\frac{\cos(r\cos\theta)}{r^t}r^{d-1}drd\theta$ unless, of course, you meant to put $d=2$ here. In higher dimensions the integral over sphere is not as simple.

Also, it would be more natural to integrate over $\theta$ first. This wins you a bit of cancellation: the integral $\int_0^\pi \cos(r\cos\theta)\,d\theta$ tends to zero as $r\to\infty$, and also oscillates. Unfortunately, it is not very small: about $\frac{\cos r}{\log r}$. Multiplying this by any positive power of $r$ definitely creates a divergent integral at $\infty$. So, at least in two dimensions and with polar coordinate approach, the integral diverges when $0<t<1$.

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sorry for the confusion. I've updated it, there should be a "$\sin^{d-2}\theta$" term missing, yes, only when $d=2$, it's gone. I have the same conclusion as you. –  Alron Feb 8 '13 at 22:15

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