What's the Fourier transform of $\delta_{x-y^2}$

Solution 1:

Are you sure $\langle T,\varphi\rangle = \int_\Bbb{R} \varphi(y^2,y) \,dy$ is the distribution you are interested in, not the arc-length parametrization? If so then let

$$\langle T_h,\varphi\rangle = \int_\Bbb{R} e^{-hy^2}\varphi(y^2,y) \,dy$$ So that $$\widehat{T}(u,v)=\lim_{h\to0}\widehat{T}_h(u,v)=\lim_{h\to0}\frac1\pi\int_{-\infty}^\infty e^{-h y^2} e^{-i (uy^2+vy)} dy$$ $$=\lim_{h\to0}\frac1\pi e^{-\frac{v^2}{4(h+i u)}} \int_{-\infty}^\infty e^{-(h+i u)(y+\frac{i v}{2(h+i u)})^2} dy=\lim_{h\to0}\frac1\pi e^{-\frac{v^2}{4(h+i u)}} \int_{-\infty}^\infty e^{-(h+i u)y^2} dy$$ (Cauchy integral formula) $$=\lim_{h\to0}\frac{e^{-\frac{ v^2}{4(h+i u)}}}{\pi (h+i u)^{1/2}} \int_{-\infty}^\infty e^{-y^2} dy=\frac{e^{-\frac{ v^2}{4i u}}}{\pi (i u)^{1/2}} \sqrt{\pi}$$ where around $u=0$ the last step is convergence in $L^1_{loc}$ sense.