# Fourier transform difficulty 2D

I have solved easily the (1) but I am really struggling with (2) and (3).

Do i need to use change of order of integration, because of so many parameters... i am confused a little bit.

My work:

$$\hat{Xf}(\xi,\theta)=\int\limits_{-\infty}^{\infty}\left(\int\limits_{-\infty}^{\infty}f(t\cos\theta+s\sin\theta,t\sin\theta - s\cos\theta) ds\right)e^{-2\pi i \xi t} dt = ??$$

How do I achieve $$d\theta$$, any hints please?

I got Jacobian as $$t$$ So, $$dxdy=t dt d\theta$$

I guess I am very close, I have also obtained $$\hat{f}(\xi\cos\theta,\xi\sin\theta)=\int\limits_{-\infty}^{\infty}\int\limits_{-\pi}^{\pi}f(t\cos\theta+s\sin\theta,t\sin\theta - s\cos\theta)e^{-2\pi i \xi t} t d\theta dt$$

• $\theta$ is not to be integrated over, you should not have a $d\theta$. Hint: try writing the integral over $dt ds$ in terms of $x=t\cos\theta+s\sin\theta,y=t\sin\theta - s\cos\theta$ and go to $dx dy$ Apr 30 '20 at 13:57

For point (2): you correctly wrote the transform $$\widehat{Xf}(\xi,\theta)$$:

$$\widehat{Xf}(\xi,\theta)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(t\cos\theta+s\sin\theta,t\sin\theta - s\cos\theta)e^{-2\pi i \xi t} ds dt$$

Following the hint in the text, let $$\gamma := (\cos(\theta),\sin(\theta))$$ and $$\gamma_\bot := (\sin(\theta),-\cos(\theta))$$. It is easy to check that

$$\xi t = \xi \gamma \cdot( t\gamma + s\gamma_\bot)$$ so replace this in the integral: $$\widehat{Xf}(\xi,\theta)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(t\gamma + s\gamma_\bot)e^{-2\pi i (\xi \gamma \cdot( t\gamma + s\gamma_\bot)) } ds dt$$

Now notice that $$\gamma$$ and $$\gamma_\bot$$ are a orthonormal basis for $$\mathbb{R}^2$$. Intuitively, this tells you that integrating over $$ds dt$$ is "the same" as integrating over $$\mathbb{R}^2$$ in standard coordinates. Formally notice that we can write

$$\begin{bmatrix} x \\ y \end{bmatrix} = \underbrace{\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}}_{A} \begin{bmatrix} t \\ s \end{bmatrix}$$

So using the formula for the multidimensional change of variables and knowing that $$|\det(A)|=1$$ we have

$$\widehat{Xf}(\xi,\theta)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)e^{-2\pi i \xi \langle \gamma, [x,y]\rangle } dx dy = \widehat{f}( \xi \gamma ) = \widehat{f}(\xi \cos(\theta), \xi \sin(\theta) )$$

For point (3): knowing that $$f\in \mathcal{S}$$ we can write

$$f(x,y) = \int_{\mathbb{R}^2} \hat f(a,b) e^{i 2\pi (ax+by)} da db$$

Writing the integral in polar coordinates and using (2):

\begin{align*} f(x,y) & = \int_{-\pi}^{\pi} \int_{-\infty}^\infty \rho \hat f(\rho \cos(\theta), \rho \sin(\theta)) e^{i2\pi \rho (x\cos(\theta)+ y\sin(\theta))} d\rho d\theta = \\ & = \int_{-\pi}^{\pi} \int_{-\infty}^\infty \rho \widehat{Xf}(\rho,\theta) e^{i2\pi \rho (x\cos(\theta)+ y\sin(\theta))} d\rho d\theta =0 \end{align*}

where the last step follows from the fact that the Fourier transform of the constant function $$0$$ is identically null.