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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?..

2 Answers 2

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The Wikipedia formula is only valid for $a>0$, but not for $a<0$ or $a=0$.

The left hand side of their formula makes sense, however, and equals zero when $a<0$ and equals one-half (as you expect) when $a=0$.

It may be easier to understand by rewriting the integral as $ \int_{-\infty}^\infty \delta(r-a)\ e^{-s r}\ \mathbb{I}_{[0,\infty)}(r) \ dr$ where $\mathbb{I}$ is the indicator function of the positive half of the line. If you treat the delta function above as a limiting case of even functions peaked at zero, you will get the result for any value of $a$.

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The equation you quote from Wikipedia (from this section) specifies a Laplace transform, and the article on the Laplace transform (in this section) states that the intended meaning of that integral is the limit as $0$ is approached from the left.