-1

What is the value of integral of $\int_{-\infty}^0 δ(t)dt$ (dirac function)? Wouldn't it make sense for it to be 0.5 since it is an even "function" ? Show proof of whatever the answer is if you know it please.

Spyrakoss
  • 3
  • 1
    $\int_a^b\bullet,dt$ is a notation which is used inherently for situations where considering or removing the extremal points of the interval $[a,b]$ is inconsequential. Therefore, it is not not suited for integrating discrete distributions. If you want precision in that regard, you should either use $\int_{(-\infty,0]}$ and $\int_{(-\infty, 0)}$, or look for specific reasons in your problem at hand for choosing one value of $\int_{-\infty}^0$. –  Nov 23 '20 at 16:59
  • A good number of authors do prefer to assign $H(0)=\frac12$ for the Heaviside step function, though. –  Nov 23 '20 at 17:02
  • The Heaviside function $H$ is not a suitable test function and hence the distribution $\langle \delta, H\rangle$ is meaningless. – Mark Viola Nov 23 '20 at 17:59
  • Does this answer your question? integral of delta function from zero to infinity – LL 3.14 Nov 24 '20 at 08:51

2 Answers2

0

Actually, $\int_{-\infty}^0\delta(t)dt$ is undefined, because $\lim_{a\to0^+}\int_{-\infty}^0\tfrac1a\delta_0(\tfrac{t}{a})dt$ is not the same for all choices of PDFs $\delta_0$ we can use as a nascent delta. For example,$$\begin{align}\delta_0(u)&:=e^{-u}1_{[0,\,\infty)}(u)\\\implies\lim_{a\to0^+}\int_{-\infty}^0\tfrac1a\delta_0(\tfrac{t}{a})dt&=0,\,\\\delta_0(u)&:=e^u1_{(-\infty,\,0]}(u)\\\implies\lim_{a\to0^+}\int_{-\infty}^0\tfrac1a\delta_0(\tfrac{t}{a})dt&=1.\end{align}$$

J.G.
  • 115,835
  • The Heaviside function $H$ is not a suitable test function (i.e., $H\notin C_C^\infty$, $H\notin \mathbb{S}$) and hence the distribution $\langle \delta, H\rangle$ is meaningless. As pointed out herein, evaluating $\langle \delta,H\rangle$ using a regularization of $\delta$ is a non-unique result. – Mark Viola Nov 23 '20 at 18:04
  • Actually this is because $H$ is not continuous, not because it is not smooth (the Dirac delta is a measure, so it can be multiplied by continuous functions) – LL 3.14 Nov 24 '20 at 08:56
0

For the integration $\int_{-\infty}^0$, 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 ${-\infty}~ \to~ 0$ 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)=1;~~~~~~~~\int_0^{ \eta_2}dx~\delta (x)=0 \end{equation*}

Then,

\begin{equation*} \int_{-\infty}^0~dx~\delta (x)= \int_{-\infty}^{ -\eta_1}dx~\delta (x)+ \int_{ -\eta_1}^0~dx~\delta (x)=0+1=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’

user151522
  • 467