How to make $\int_0^\infty \delta(x) dx = \frac12 $ rigorous using generalized functions? There are many versions of the theory on generalized functions. The most famous one is the distribution theory of Schwartz, where test functions are smooth and compactly supported. In the Schwartz's theory, the expression
$$\int_0^\infty \delta(x) dx = \langle\delta, \chi_{[0,\infty)}\rangle,$$
where $\chi$ is the indicator function, is undefined since $\chi_{[0,\infty)}$ is not smooth. However, the prescription(?) of setting  $\int_0^\infty \delta(x) dx = \frac12 $ is very useful in physics, and I want to make this expression rigorous.
Another famous theory of generalized function is the hyperfunction theory of Sato. Hyperfunctions can be regarded as differences of boundary values of two holomorphic functions on upper and lower half plane. As far as I know, the integral
$$\int_a^b f(x) dx$$
of a hyperfunction $f$ is defined only if $f$ is real analytic on two endpoints $a$ and $b$. However, the hyperfunction $\delta(x)$ is not real analytic at 0, and the integral $\int_0^\infty \delta(x) dx$ is not defined.
Is there any way to make rigorous meaning to $\int_0^\infty \delta(x) dx$?
 A: No problem to take $\phi(x) = e^{-1/(1-x^2)} 1_{|x| <1}$, $C=\int_{-1}^1 \phi(x)dx,\phi_n(x)=\phi(x/n),\varphi_n(x)=\frac{n\phi(nx)}{C}$ and to set $$( T,f) = \lim_{n\to \infty} \langle T,\phi_n \ (f\ast \varphi_n)\rangle$$
for any distributions $f,T$.
When $f\in C^\infty_c$ it is the standard pairing.
Otherwise the limit will often diverge but sometimes it will converge.
The point is that there is no nice topological interpretation as for the pairing of distributions with $C^\infty_c$ functions, and that the pairing will be different when replacing $\phi$ by some other (non-even) $C^\infty_c$ function.
A: In the distribution theory the $\delta $ is the Dirac measure at zero, that is
$$
\delta (A)=\int_{\mathbb{R}}\mathbf{1}_{A}(x)\delta (d x)=\begin{cases}
1,&0\in A\\
0,&\text{ otherwise }
\end{cases}
$$
And in general
$$
\int_{\mathbb{R}}f(x) \delta (d x)=f(0)
$$
for any measurable function $f$, smooth or not. By abuse of notation usually $\delta $ is written as a "function" in the integrand instead of a measure, writing $\delta (x)\,d x$ instead of $\delta (dx)$.
If you want that $\int_{\mathbb{R}}\mathbf{1}_{[0,\infty )}(x)\delta (x)dx=\frac1{2}$ then you can set $\delta (x)\,d x:=\frac1{2}\delta (dx)$.
