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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 $\int_0^\infty \delta(x) dx$?

Laplacian
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    Isn't $\delta$ defined as the weak* limit of $\delta_n$ where $\delta_n(x) = n\delta_1(nx)$ and $\delta_1$ is any function satisfying a whole list of properties (non-negative, smooth, compactly supported, symmetric about $0$, $\int_\mathbb{R}\delta_1 = 1$)? With that, we'd have $\int_{[0,\infty)} \delta_n = 1/2$ for all $n$, the limit of which is still $1/2$. – Brian Moehring May 10 '22 at 03:17
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    @BrianMoehring, As for my understanding, $\delta$ is defined as the linear functional on the space of test functions satisfying $\langle\delta,\varphi\rangle=\varphi(0)$. So, it can be realized in various ways: $$\delta=\lim_{n\to\infty} n\mathbf{1}{[-1/2n,1/2n]}=\lim{n\to\infty} n\mathbf{1}_{[0,1/n]}$$ in the sense of weak-* limit. – Sangchul Lee May 10 '22 at 03:24
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    When $\delta$ is applied to the theory of circuit, $\delta(t)$ is often interpreted as the 'impulse immediately after time $0$'. So, one actually chooses the convention $$\int_{0}^{\infty}\varphi(t)\delta(t),\mathrm{d}t=f(0).$$ For example, the Laplace transform of $\delta$ is considered as $1$ instead of $\frac{1}{2}$. That said, how $\delta$ should be defined over a broader class of test functions seems purely at our discretion, depending on how $\delta$ idealizes the ambient physical process. – Sangchul Lee May 10 '22 at 03:28
  • This uses a symmetric function to yield behavior of the dirac delta function. I'm thinking integrating only over positive reals might produce the desired result: https://math.stackexchange.com/questions/2833912/reference-for-dirac-delta-function-as-gaussian – TurlocTheRed May 10 '22 at 03:28
  • The definition $\int_S\delta(x)f(x)dx:=\lim_{\epsilon\to0^+}\int_S\tfrac{1}{\epsilon}\eta(\tfrac{x}{\epsilon})f(x)dx$ with $\eta$ a nascent delta is dependent on the choice of $\eta$ if $S=[0,,\infty)$. If you know of a context where physicists salvage this, it will be by restricting the choice of $\eta$ (or doing something equivalent, as you don't have to define Dirac deltas this way). – J.G. May 10 '22 at 16:16
  • The best is to not use the Dirac delta for such cases, and rather keep the notation indicating that a limit has to be taken. Or following the advice given here https://math.stackexchange.com/questions/4427947/what-is-the-value-of-the-integral-int-inftya-deltax-a-dx-and-relate/4427952#4427952, you could write $$ \frac{1}{2} \left(\int_{[0,\infty)} \delta + \int_{(0,\infty)} \delta\right) $$ – LL 3.14 Jan 26 '23 at 21:20

2 Answers2

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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.

reuns
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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 $\delta (x)\,d x:=\frac1{2}\delta (dx)$.

Masacroso
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  • Obviously the OP wants the integration from -1 to 0 to be 1/2 and the integration from 0 to 1 to be also 1/2 – High GPA May 13 '22 at 02:19