I would like to provide a proof for the Fourier rotation theorem:
$\begin{equation} \mathcal{F}\left\lbrace I\left(R\vec{x}\right) \right\rbrace(\vec{k}) =\mathcal{F}\left\lbrace I\left(\vec{x}\right) \right\rbrace(R\vec{k}) \end{equation}$
where $\mathcal{F}$ is the Fourier transform, $R$ is a rotation in 3D, $I\left(\vec{x}\right)$ is a complex function in space, $\vec{k}$ is the position in frequency space (related to $\vec{x}$ via $\mathcal{F}$).
Using $\det(R^{-1})=1$, I started as follows:
$\begin{alignat}{10}\mathcal{F}\left\lbrace I\left(R\vec{x}\right) \right\rbrace(\vec{k}) &= \dfrac{1}{\sqrt{2\pi}} \int_{\infty} I (R \vec{x}) e^{-i 2 \pi ~\vec{k} \cdot \vec{x}} d\vec{x}\\ &\mathrm{~~~~~~\qquad Subs.~~~~} \vec{x}=R^{-1}\vec{y} ~~~\Rightarrow ~~~ d\vec{x}=\det \left(R^{-1}\right) d\vec{y}\\ &= \dfrac{1}{\sqrt{2\pi}} \int_\infty I (\vec{y}) e^{-i 2 \pi ~\vec{k} \cdot \left(R^{-1}\vec{y}\right)}~\det\left(R^{-1}\right) d\vec{y} \\ &= \dfrac{1}{\sqrt{2\pi}} \int_\infty I (\vec{y}) e^{-i 2 \pi ~\left(R \vec{k}\right) \cdot\vec{y}} ~d\vec{y} \\ &=\mathcal{F}\left\lbrace I\left(\vec{y}\right) \right\rbrace(R\vec{k})\end{alignat}$
This is almost what I want, but now I have $\vec{y}$ on the right side instead of $\vec{x}$ as shown above. Where is my mistake? I can't just replace $\vec{y}$ with $\vec{x}$, can I? I suppose I'm missing something quite obvious.