Fourier transform of $e^{it|\xi|^{\alpha}}$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T10:56:31Z http://mathoverflow.net/feeds/question/109443 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109443/fourier-transform-of-eit-xi-alpha Fourier transform of $e^{it|\xi|^{\alpha}}$ Shanlin Huang 2012-10-12T08:48:23Z 2012-10-13T08:53:39Z <p>Consider the fourier transform of $e^{it|\xi|^{2\alpha}}$ ($\alpha>0$)in $\mathbb{R}^n$,let $K_{\alpha}=\mathcal{F}(e^{it|\xi|^{\alpha}})$,so $K$ is a tempered distribution.Now I want to know if there is a explicit expression of $K$,for the simpliest case,namely $\alpha=1$,it's well known that $$K_1=(4\pi it)^{-\frac{n}{2}}e^{-\frac{|x|^2}{4it}}$$ Another special case is $\alpha=\frac{1}{2}$,since we know that $\mathcal{F}e^{-t|\xi|}=C_{n}\frac{t}{(t^{2}+|\xi|^2)^{\frac{n+1}{2}}}$,where $t>0$,let $t=-it$,so at least formally, $$K_{\frac{1}{2}}=C_{n}\frac{-it}{(|\xi|^2-t^{2})^{\frac{n+1}{2}} }$$</p> <p>My question is how about general $\alpha$ ?,so far I have known that when $0&lt;\alpha&lt;\frac{1}{2}$,$\alpha=\frac{1}{2}$,$\alpha>\frac{1}{2}$,the singularity of $K$ lies at $0$,$t=|x|$,$\infty$ respectively.</p> http://mathoverflow.net/questions/109443/fourier-transform-of-eit-xi-alpha/109457#109457 Answer by Abdelmalek Abdesselam for Fourier transform of $e^{it|\xi|^{\alpha}}$ Abdelmalek Abdesselam 2012-10-12T12:26:26Z 2012-10-12T13:16:14Z <p>For $t=i$ there is a formula involving an integral of a Bessel function, so I doubt there is a simple closed formula for $K_{\alpha}$ in general. You can find the formula I mentioned in the first page of the article <a href="http://www.ams.org/journals/tran/1960-095-02/S0002-9947-1960-0119247-6/" rel="nofollow">"Some theorems on stable processes"</a> by Blumenthal and Getoor. Also see <a href="http://mathoverflow.net/questions/91263/asking-for-a-fourier-inverse-transform-which-is-related-to-stable-laws/91401#91401" rel="nofollow">this related MO question</a>.</p> http://mathoverflow.net/questions/109443/fourier-transform-of-eit-xi-alpha/109520#109520 Answer by Shanlin Huang for Fourier transform of $e^{it|\xi|^{\alpha}}$ Shanlin Huang 2012-10-13T08:40:57Z 2012-10-13T08:53:39Z <p>We consider $t=1$ for simplicity and wright $K_{\alpha}=\mathcal{F}(\eta(|\xi|) e^{i|\xi|^\alpha})+\mathcal{F}((1-\eta )e^{i|\xi|^\alpha})$,where $\eta\in C^{\infty}(\mathbb{R})$,and $\eta=0$,near 0,$\eta(t)=1$,when $t\ge 1$,the second term in the RHS is smooth and has good behaviour at $\infty$,so we look at the first term,in A.Miyachi's paper "On some singular fourier multipliers"see <a href="http://repository.dl.itc.u-tokyo.ac.jp/dspace/bitstream/2261/6297/1/jfs280206.pdf" rel="nofollow">http://repository.dl.itc.u-tokyo.ac.jp/dspace/bitstream/2261/6297/1/jfs280206.pdf</a> it has a thoroughly analysis on it.When $0&lt;\alpha&lt;\frac{1}{2}$,we have $K\in C^{\infty}(\mathbb{R}^{n}\backslash{0})$ and $$K_{\alpha}=C|x|^{\frac{n(\alpha-1)}{1-2\alpha}}e^{iB|x|^{-\frac{2\alpha}{1-2\alpha}}}+o(|x|^{\frac{n(\alpha-1)}{1-2\alpha}})\quad \text{as}|x|\to 0$$</p> <p>When $\alpha>\frac{1}{2}$,$K$ is smooth throughout $\mathbb{R}^{n}$,and $$K_{\alpha}=C|x|^{\frac{n(\alpha-1)}{1-2\alpha}}e^{iB|x|^{-\frac{2\alpha}{1-2\alpha}}}+o(|x|^{\frac{n(\alpha-1)}{1-2\alpha}})\quad \text{as}|x|\to\infty$$ In this case we can see that unlike $\alpha=1$, for $\alpha>1$,$K_{\alpha}$ has decay of $|x|^{-\frac{n(\alpha-1)}{2\alpha-1}}$ at $\infty$.</p>