Can anyone prove this equality? $$\int_{\mathbb R^2}f(x)\int_{|y|<1}\frac{e^{ix\cdot y}\ -1}{|y|^2}dydx=\int_{\mathbb R^2}f(x)\int_0^1 \frac{1}{r}\int_0^{2\pi}(e^{i|x|r\cos\theta}\ -1)d\theta dr dx,$$ where $f:\mathbb R^2\to \mathbb C$ is in Schwartz class.
This equality is in my text book and this says that this can be shown by changing variables : $y_1=r\cos\theta, y_2=r\sin\theta.$
Using the method, I have \begin{align} LHS &=\int_{\mathbb R^2}f(x)\int_0^1 \int_0^{2\pi}\frac{e^{i(x_1\ r\cos\theta+x_2 \ r\sin\theta)}\quad \ -1}{r^2}rd\theta drdx\\ &=\int_{\mathbb R^2}f(x)\int_0^1\frac{1}{r}\int_0^{2\pi}(e^{i(x_1\ r\cos\theta+x_2 \ r\sin\theta)}\quad \ -1) d\theta dr dx. \end{align}
Is this equal to $\int_{\mathbb R^2}f(x)\int_0^1 \frac{1}{r}\int_0^{2\pi}(e^{i|x|r\cos\theta}\ -1)d\theta dr dx$ ?
I cannot see.
(The textbook is written in Japanese so most of you cannot read it.)