Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $t\in (0,d)$?

share|cite|improve this question
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: – user53153 Feb 8 '13 at 22:23

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.

share|cite|improve this answer

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 $0<t<1$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.