I am trying to understand the motivation behind the following identity stated in Bracewell's book on Fourier transforms: $\delta^{(2)}(x,y)=\frac{\delta(r)}{\pi r},$ where $\delta^{(2)}$ is a 2-dimensional delta function. Starting with something we know to be true, we can do $1 = \iint \delta^{(2)}(x,y) dx\,dy = \int_0^\infty \int_0^{2\pi} \frac{\delta(r)}{\pi r} r\,dr\,d\theta = 2 \int_0^\infty \delta(r).$
This suggests that the integral of delta function from 0 to infinity is 1/2. In fact, this seems to make sense if we treat the delta function as a limiting case of an even function peaked at zero (Gaussian, sinc, etc.) However, Wikipedia, citing Bracewell, claims the following to be true:
$\int_0^\infty \delta(r-a) e^{-s r} dr = e^{-s a},$ and plugging in 0 for a and s we get $\int_0^\infty \delta(r) dr = 1.$
What is going on here?.. Where's the screw-up?.. If the integral from 0 to infinity is not 1/2, then how do we justify the above polar-coordinate expression for a 2D delta function?..