integral of delta function from zero to infinity

Question

I would appreciate it if someone could help me with the following problem. I can not understand how a delta function $\delta(x)$ is integrated from zero to infinity. Because the integration interval should contain zero.

Answer 1

So first the Dirac delta is not a function, it is a distribution of order $0$, and so can be also interpreted as a bounded measure. As a measure, it is defined as acting on a set $A$ by $$ \delta_0(A) = \left\{ \begin{array}{} 1 &\text{ if } 0∈ A \\ 0 &\text{ if } 0\notin A. \end{array}\right. $$ This can be written $\delta_0(A) = \mathbb{1}_A(0)$ which is why it can be identified as the linear form over continuous functions defined for every continuous function $\varphi∈C^0$ by $$ \langle \delta_0,\varphi \rangle = \varphi(0) = ∫_{\mathbb{R}} \varphi(x)\, \delta_0(\mathrm{d}x). $$ In the same way that integrable functions $f∈L^1$ can be identified as linear forms over bounded functions defined for every $\varphi∈L^\infty$ by $$ \langle f,\varphi \rangle = ∫_{\mathbb{R}} \varphi(x)\, f(x)\,\mathrm{d}x. $$ From this point of view, $f$ can be identified with the measure $\mu_f$ such that $\mu_f(\mathrm{d}x) = f(x)\,\mathrm{d}x$, and so $$ ∫_0^\infty f(x)\,\mathrm{d}x = ∫_0^\infty \mu_f(\mathrm{d}x) = ∫_{\mathbb{R}} \mathbf{1}_{(0,\infty)}(x)\,\mu_f(\mathrm{d}x) = \mu_f((0,\infty)) = \mu_f([0,\infty]). $$ Remark however that $\mu_f((0,\infty)) = \mu_f([0,\infty])$ is a special feature of locally integrable functions (coming from the fact that the integral is the same if we remove a set of measure $0$). This is not the case of the dirac measure. Therefore we have to specify if we are looking at the integral of the dirac over $[0,\infty)$ or $(0,\infty)$. And then we can define $$ \begin{align*} \int_{[0,\infty)} \delta_0(\mathrm{d}x) := \delta_0([0,\infty)) = 1 \\ \int_{(0,\infty)} \delta_0(\mathrm{d}x) := \delta_0((0,\infty)) = 0. \end{align*} $$

Answer 2

For the integration $\int_0^\infty$, you do not use the usual definition of the delta function, which Barton, Reference 1, pg 12 , refers to as the ‘weak definition’ of $\delta(x)$, and which he gives in (1.1.1) on pg 10 as ( I quote )

\begin{equation*} \delta(x)=0~~~~~~if ~~x\neq0;~~~~~~\int_{- \eta_1}^{ \eta_2}dx~\delta (x)=1~~~~~~~~~~~(1.1.1) \end{equation*}

Note that ( $\eta_1~>~0,~~ \eta_2~>~0$ ).

Also, (1.1.1) says nothing about \begin{equation*} ~\int_{- \eta_1}^0dx~\delta (x),~~~~~~~\text{or}~~~~~~~\int_0^{ \eta_2}dx~\delta (x) \end{equation*}

Barton, puts in his (1.1.5), I quote \begin{equation*} \int_L \equiv~\int_{- \eta_1}^0dx~\delta (x),~~~~~~~\int_R \equiv \int_0^{ \eta_2}dx~\delta (x)~~~~~~~~~~(1.1.5) \end{equation*}

subject only to $\int_L + \int_R=1.$

A definition of $\delta(x)$ that assigns values to $\int_L$ and $\int_R$ is called a ‘strong definition’ of $\delta(x)$ by Barton.

So, here, to perform our $0 \to ~\infty$ integral, we define $\delta(x)$ as, c.f (1.1.1) above

\begin{equation*} \delta(x)=0~~~~~~if ~~x\neq0;~~~~~~\int_{- \eta_1}^0dx~\delta (x)=0;~~~~~~~~\int_0^{ \eta_2}dx~\delta (x)=1 \end{equation*}

Then,

\begin{equation*} \int_0^\infty~dx~\delta (x)= \int_0^{ \eta_2}dx~\delta (x)+ \int_{ \eta_2}^\infty~dx~\delta (x)=1+0=1 \end{equation*}

Reference:

G.Barton, Elements of Greens Functions and Propagation. Potentials, Diffusion, and Waves, Clarendon Press Oxford, 1989.

NB: Pgs7-40 , contain material on the ‘Dirac Delta Function’

Other Information

Note, when using a “radial variable”, negative values of that variable are meaningless.

Barton, pgs 32, 33 calls the following definition of $\delta(r)$, both a strong definition, and a ‘one-sided’ definition. I quote

the one-sided definition

\begin{equation*} \delta(r)=0~~~~~~if ~~r\neq0;~~~~~~\int_0^{ \eta_1}dr~\delta (r)=1~~~~~~~~~~(1.4.12) \end{equation*}

He also says, pg 9

Strong definition of radial $\delta(r)$:

\begin{equation*} \int_0^{ \infty}dr~f(r)~\delta (r)=f(0) \end{equation*}