I would like to compute the inverse Fourier transform of $\exp(4\pi^2i|\xi|^2t)$.

\begin{equation} f(x,t) = \int_{\mathbb{R}^n} e^{2\pi i x\cdot\xi} e^{4\pi^2i|\xi|^2t} \,\mathrm{d}\xi \end{equation}

Can I use the fact that the Fourier transform of $e^{-\pi|x|^2}$ is $e^{-\pi|\xi|^2}$, and then apply to the function above with the change of variable (dilation)? I wonder if it works for complex numbers.

Thank you.

  • $\begingroup$ What is the function (tempered distribution) defined on frequency space of which you want to take the inverse Fourier transform? $(\xi,x)\mapsto e^{4\pi^{2}i\left|\xi\right|^{2}x}$ is defined on $\mathbb{R}^{n}\times\mathbb{R}$. $\endgroup$ – Matt Rosenzweig May 6 '15 at 2:27
  • $\begingroup$ You can't transform a function in both $x$ and $\xi$, only one or the other. Unless you want $x$ to be a constant $\endgroup$ – Dylan May 6 '15 at 2:29
  • $\begingroup$ My mistake. There should be a "$t$". $\endgroup$ – dh16 May 6 '15 at 2:55
  • $\begingroup$ For real $t$ fixed, the function $g(\xi)=e^{4\pi^{2}i\left|\xi\right|^{2}t}$ is not integrable in the Lebesgue sense, so the expression you have above is not a Lebesgue integral. $g$ has an inverse Fourier transform in the sense of tempered distributions. $\endgroup$ – Matt Rosenzweig May 6 '15 at 3:02
  • $\begingroup$ @MattRosenzweig Could you please elaborate more? I am solving the Schrodinger equation $iu_t-\Delta u=0$ and $u(x,0)=g(x)$. $g$ is given in $L^2(\mathbb{R}^n)$. I am supposed to find the solution $u\in C^1_tC^2_x\cap L_t^\infty H^2_x$. I used Fourier transform and had a solution, but I do not know how to find the inverse of the kernel. $\endgroup$ – dh16 May 6 '15 at 3:08

Before editing the question you were asking the Inverse Fourier transform of $\xi\mapsto \exp(4\pi^2i|\xi|^2)$ \begin{eqnarray} f(x)&=&\int_{\mathbb{R}^n}e^{2\pi ix\cdot\xi}e^{4\pi^2i|\xi|^2}\,d\xi\stackrel{y=2\pi\xi}{=}\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}e^{i(x\cdot y+|y|^2)}\,dy=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}e^{i(|y+x/2|^2-|x|^2/4)}\,dy\\ &=&\frac{e^{-i\frac{|x|^2}{4}}}{(2\pi)^n}\int_{\mathbb{R}^n}e^{i|y+x/2|^2}\,dy\stackrel{z=y+x/2}{=}\frac{e^{-i\frac{|x|^2}{4}}}{(2\pi)^n}\int_{\mathbb{R}^n}e^{i|z|^2}\,dz=\frac{e^{-i\frac{|x|^2}{4}}}{(2\pi)^n}\left(\int_{\mathbb{R}}e^{it^2}\,dt\right)^n\\ &=&\frac{e^{-i\frac{|x|^2}{4}}}{(2\pi)^n}\left(\int_{\mathbb{R}}\cos(t^2)\,dt+i\int_{\mathbb{R}}\sin(t^2)\,dt\right)^n=\frac{e^{-i\frac{|x|^2}{4}}}{(2\pi)^n}\left[\sqrt{\frac\pi2}(1+i)\right]^n\\ &=&\left(\frac{1+i}{\sqrt{2}}\right)^n\cdot\left(\frac{\sqrt{\pi}}{2\pi}\right)^ne^{-i\frac{|x|^2}{4}}=\frac{e^{in\frac\pi4}}{(2\sqrt{\pi})^n}e^{-i\frac{|x|^2}{4}}=\frac{1}{(2\sqrt{\pi})^n}\exp\left(-i\frac{|x|^2-n}{4}\right). \end{eqnarray}

After editing the question you are asking the Inverse Fourier transform of $(\xi,t)\mapsto \exp(4\pi^2i|\xi|^2t)$ (w.r.t. $\xi$). For $t\ne 0$, let us denote by $s$ the sign of $t$. Then

\begin{eqnarray} f(x,t)&=&\int_{\mathbb{R}^n}e^{2\pi ix\cdot\xi}e^{4\pi^2i|\xi|^2t}\,d\xi\stackrel{y=2\pi\sqrt{|t|}\xi}{=}\frac{1}{(2\pi\sqrt{|t|})^n}\int_{\mathbb{R}^n}\exp\left(is(s|t|^{-1/2}x\cdot y+|y|^2)\right)\,dy\\ &=&\frac{1}{(2\pi\sqrt{|t|})^n}\int_{\mathbb{R}^n}\exp\left[is\left(\left|y+\frac{s|t|^{-1/2}}{2}x\right|^2-\frac{|x|^2}{4|t|}\right)\right]\,dy\\ &=&\frac{\exp\left(-is\frac{|x|^2}{4|t|}\right)}{(2\pi\sqrt{|t|})^n}\int_{\mathbb{R}^n}\exp\left(is\left|y+\frac{s|t|^{-1/2}}{2}x\right|^2\right)\,dy\\ &\stackrel{z=y+\frac{s|t|^{-1/2}}{2}x}{=}&\frac{e^{-it\frac{|x|^2}{4}}}{(2\pi\sqrt{|t|})^n}\int_{\mathbb{R}^n}e^{is|z|^2}\,dz=\frac{e^{-it\frac{|x|^2}{4}}}{(2\pi\sqrt{|t|})^n}\left(\int_{\mathbb{R}}e^{isu^2}\,du\right)^n\\ &=&\frac{e^{-it\frac{|x|^2}{4}}}{(2\pi\sqrt{|t|})^n}\left(\int_{\mathbb{R}}\cos(su^2)\,du+i\int_{\mathbb{R}}\sin(su^2)\,du\right)^n=\frac{e^{-it\frac{|x|^2}{4}}}{(2\pi\sqrt{|t|})^n}\left[\sqrt{\frac\pi2}(1+si)\right]^n\\ &=&\left(\frac{1+si}{\sqrt{2}}\right)^n\cdot\left(\frac{\sqrt{\pi}}{2\pi\sqrt{|t|}}\right)^ne^{-it\frac{|x|^2}{4}}=\frac{e^{isn\frac\pi4}}{(2\sqrt{\pi|t|})^n}e^{-it\frac{|x|^2}{4}}\\ &=&\frac{1}{(2\sqrt{\pi|t|})^n}\exp\left(-it\frac{|x|^2-sn}{4}\right)=\frac{1}{(2\sqrt{\pi|t|})^n}\exp\left(-i\frac{t|x|^2-|t|n}{4}\right). \end{eqnarray}

  • $\begingroup$ I am sorry for a change in the problem. For such a change in the exponent, I think the first substitution becomes $y=2\pi\xi t$; is that correct? $\endgroup$ – dh16 May 6 '15 at 2:57
  • $\begingroup$ I feel like the 2 should be under the root in the 1D case... In the spirit of cheating, you may try the following in Mathematica: Integrate[Exp[2*Pi*I*o*x]*Exp[4*Pi^2 I*o^2], {o, -Infinity, Infinity}] = ((1/2 + I/2) E^(-((I x^2)/4)))/Sqrt[2 \[Pi]] $\endgroup$ – Mikhail May 6 '15 at 3:02
  • $\begingroup$ @Mercy Is there $i$ when we subtract $|x|^2/4$? $\endgroup$ – dh16 May 6 '15 at 3:16
  • $\begingroup$ @Mercy Thank you for your help. I rated your reply as the answer before you edited it because I was able to do the problem using your method. Thanks again. $\endgroup$ – dh16 May 8 '15 at 1:52

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