enter image description here

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

  • 1
    $\begingroup$ $\theta$ 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 $dx dy$ $\endgroup$
    – user619894
    Apr 30, 2020 at 13:57

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.