Integral of delta distribution in spherical coordinates 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:


*

*Which one is the correct form for the delta in spherical coordinates?

*Is the integral $\int_0^\infty \delta(r) dr$ well defined? (See also Ruslan's answer).

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