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The question come from here and is about the answer of Chappers with. I suppose $2s<1$. The following integral is of course not absolutely convergent $$\int_{\mathbb R}\frac{e^{-2i\pi x\xi}}{|\xi|^{2s}}d\xi,$$ but does it exist or not ? I mean, is $$\xi\longmapsto \frac{e^{-2i\pi x\xi}}{|\xi|^{2s}},$$ integrable (in Riemann sense since it's not in Lebesgue sense).

In the answer of Chappers he use the fact that $$\frac{1}{|\xi|^{2s}}=\frac{2\pi^s}{\Gamma(s)}\int_0^\infty t^{2s-1}e^{-\pi t^2|\xi|^2}dt,$$ and used Fubini to get : $$\int_{\mathbb R}\frac{e^{-2i\pi x\xi}}{|\xi|^{2s}}d\xi \underset{(*)}{=}C\int_0^\infty \int_{\mathbb R} t^{2s-1}e^{-\pi t^2|\xi|^2}e^{-2i\pi x\xi}d\xi dt=C|x|^{-(1-2s)}.$$

First of all, is this argument valid in the sense that $\xi\longmapsto \frac{e^{-2i\pi x\xi}}{|\xi|^{2s}}$ is not integrable a priori and thus I'm very sceptic with equality $(*)$. Secondly, is Fubini work for improper integral that are convergent but not Lebesgue integrable ? Because in this wikipédia, they want them to be Lebesgue integrable and in the Riemann integrable case, I only saw it for Rieman integrable function on compact set. So even if $\xi\longmapsto \frac{e^{-2i\pi x\xi}}{|\xi|^{2s}}$ would be a convergent integral, the argument, doesn't work, does it ?

idm
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1 Answers1

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For the interval to converge, one must have $s\geq 0$. Since $2s<1$, the integral converges absolutely in a neighborhood of $0$. Thus it is sufficient to prove convergence of $\int_{1}^\infty\frac{e^{-2i\pi x\xi}}{|\xi|^{2s}}d\xi$.

Integrating by parts, $$\int_{1}^M\frac{e^{-2i\pi x\xi}}{\xi^{2s}}d\xi = \left[ \frac{e^{-2i\pi x\xi}}{-2i\pi x}\frac{1}{\xi^{2s}}\right]_1^M - \frac{2s}{2i\pi x}\int_1^M \frac{e^{-2i\pi x\xi}}{\xi^{2s+1}}d\xi$$

Since $2s+1>1$, the second summand has a finite limit, hence convergence of the integral.

Gabriel Romon
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