Fourier transform of $|x|^{-t}$

Question

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 $t\in (0,d)$?

Answer 1

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 $$\int_0^{\pi}\int_0^{\infty}\frac{\cos(r\cos\theta)}{r^t}r^{d-1}\sin^{d-2}\theta drd\theta.$$ Then 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.

Answer 2

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 $0<t<1$.