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We know that, if $\mathcal D$ is a domain containing the origin $(0,0,0)$, then

$$\int_{\mathcal D} \delta(\vec r) d \vec r= \int_{\mathcal D} \delta(x) \delta(y) \delta(z) dx dy dz=1$$

However, we also know that the delta distribution can be expressed in spherical coordinates as

$$\delta(r,\theta,\phi)=\frac{\delta(r)}{2 \pi r^2}$$

If we take $\mathcal D = \mathbb R^3$, we would then have

$$\int_{\mathbb R^3} \delta(\vec r) d\vec r =\int_0^\infty 4 \pi r^2 \frac{\delta(r)}{2 \pi r^2} dr= \int_0^\infty 2 \delta(r) dr = 1$$

That is to say,

$$ \int_0^\infty \delta(r) dr = \frac 1 2$$

Now, this seems very odd, but maybe it can have some sense. Indeed, we know that for every $\epsilon >0$

$$\int_{-\epsilon}^{\epsilon} \delta(x) dx = 1$$

So it is possible that we can say (maybe in a not very rigorous way...), being $\delta(x)$ even, that

$$\int_{0}^{\epsilon} \delta(x) dx = \int_{-\epsilon}^{0} \delta(x) dx = \frac 1 2$$

Does this have sense? If so, can we make it rigorous, i.e. showing that every succession of function converging to $\delta$ has this property?

And if not, can we nevertheless give some sense to

$$ \int_0^\infty \delta(r) dr = \frac 1 2 \ \ ?$$

Update

Here, I found another formula for the delta distribution in spherical coordinates, that is to say:

$$\delta(\vec r ) = \frac{\delta (r)}{4 \pi r^2}$$

This seems to make much more sense, because we would have

$$\int_0^\infty \delta(r) dr = 1$$

However, there are two issues at this point:

  1. Which one is the correct form for the delta in spherical coordinates?
  2. Is the integral $\int_0^\infty \delta(r) dr$ well defined? (See also Ruslan's answer).
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2 Answers 2

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First, your integral identity doesn't make any sense. An example of an asymmetric nascent delta function, which will have another result is

$$\delta_s(x)=\frac1{s\sqrt\pi}\exp\left(-\frac{(x-s)^2}{s^2}\right).$$

For it we'll have

$$\int\limits_0^\infty \delta_s(x)\,dx=\frac12+\frac{\operatorname{erf}(1)}2.$$

Now to your transformation of variables. It'd be better to avoid using ill-defined integrals when working with $\delta$ distribution. To get well-defined integrals we can use spherical coordinates with another ranges of variables. For example, we can take $r\in(-\infty,\infty)$, $\phi\in[0,\pi)$, $\theta\in[0,\pi)$. Then your final integral will be

$$\int\limits_{\mathbb R^3} \delta(\vec r) d\vec r = \int\limits_{-\infty}^\infty 2 \pi r^2 \frac{\delta(r)}{2 \pi r^2} dr= \int\limits_{-\infty}^\infty \delta(r) dr = 1.$$

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  • $\begingroup$ Ok, I understand that it is not a general property. But shouldn't the result of the integral be independent from the particular change of variables we choose? What is wrong in my calculation (the one which holds the result 1/2)? $\endgroup$
    – valerio
    Mar 9, 2017 at 16:34
  • $\begingroup$ @valerio92 your calculation is wrong merely because you end up with an ill-defined integral. You just can't get past it without some additional assumptions about the $\delta$. Thus it's easier to just go the well-defined way. $\endgroup$
    – Ruslan
    Mar 9, 2017 at 16:36
  • $\begingroup$ Why is it ill-defined? Is it because I am including the origin in it? $\endgroup$
    – valerio
    Mar 9, 2017 at 17:11
  • $\begingroup$ @valerio92 it's ill-defined because it's equivalent to $\int\limits_{-\infty}^\infty \theta(x)\delta(x)dx$, where $\theta$ is Heaviside step function. A distribution is only well-defined when acting on a smooth function — or at least when the distribution's singular support and argument function's set of singularities don't intersect. In your case $\theta$ is not smooth at $x=0$, neither is $\delta$. Thus the action of $\delta$ on Heaviside step function is ill-defined. $\endgroup$
    – Ruslan
    Mar 9, 2017 at 18:09
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I would be inclined to say that your intuition is wrong, mostly because $\delta$ is not a function.

I'll explain. Take the function $\delta_n$ defined by $\delta_n(x) = n$ if $x \in (-1/n,0)$ and $\delta_n(x) =0$ otherwise. Clearly $\delta_n(x) = 0$ for $x \ge 0$, and so $$ \int_0^\infty \delta_n(x) = 0 $$ for all $n$.

Nevertheless, $\delta_n \to \delta$ in the sense of distributions. In fact, take a test function $\varphi$. We have

$$ \int_{-\infty}^\infty \varphi(x) \delta_n(x)dx = n \int_{-1/n}^0\varphi(x)dx = \varphi(\xi_n),$$

where $\xi_n$ is between $-1/n$ and zero. Therefore by the continuity of $\varphi$,

$$\int_{-\infty}^\infty \varphi(x) \delta_n(x)dx \to \varphi(0), $$

and so $\delta_n \to \delta$.

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