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.
Dirac function integrated from -infinity to 0
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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’
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,$$\begin{align}\delta_0(u)&:=e^{-u}1_{[0,\,\infty)}(u)\\\implies\lim_{a\to0^+}\int_{-\infty}^0\tfrac1a\delta_0(\tfrac{t}{a})dt&=0,\,\\\delta_0(u)&:=e^u1_{(-\infty,\,0]}(u)\\\implies\lim_{a\to0^+}\int_{-\infty}^0\tfrac1a\delta_0(\tfrac{t}{a})dt&=1.\end{align}$$