I am trying to understand Chapter 1 of "Advances in IMaging and Electron Pysics and came across the following derivations:
The 2D Fourier transform of a radially symmetric function ($f(r,\theta)=f(r))$ is given by$$F(\rho, \psi) = \int_{0}^{\infty} rf(r) dr \int_{-\pi}^{\pi}e^{-ir\rho\cos(\psi-\theta)} d\theta$$ Using the integral definition of the zeroth-order Bessel function, $$J_0(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-ixcos(\psi-\theta)}d\theta = \frac{1}{2\pi} \int_{-\pi}^{\pi}e^{-ix\cos{\alpha}} d\alpha$$ Therefore, the first equation can be written as $$F(\rho) = 2\pi \int_{0}^{\pi} f(r) J_0(\rho r) rdr$$
In the second equation, it seems like we used the change of variables ($\alpha = \psi-\theta$) but why don't we have to change the integral range? Doesn't it have to be $\alpha: \psi-\pi \rightarrow \psi+\pi$ ?