# Delta function integrated from zero

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?..

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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.

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Let $f_n$ be a sequence of smooth functions such that $f_n\to\delta(x)$ in $\cal D'(\mathbb R)$. Then for any test function $\varphi\in \cal D(\mathbb R)$ we have $\lim_{n\to\infty}\int_0^\infty f_n(x)\varphi(x)dx=\varphi(0)/2\;\;$. The same goes for integrating the half-line $(-\infty,0)$. The sum of two limits is $\varphi(0)$ as it should. So to say the action of $\delta$ is symmetrically divided between the left and right sides of the origin.

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