Integration of Dirac delta

Question

According to Mathematica, $$\int_{0}^{b}f(x)\delta(x-a)\,dx\,=\,f(a)\,(2\Theta(b)-1)\,\Theta(a-b\Theta(-b))\,\Theta(-a+b\Theta(b)).$$ for $a,b\in\mathbb{R}$. Is there a quick way to see how this relation follows directly from the original definition of the delta function?

Answer 1

Let's analyze the cases carefully. First, $b$'s sign will determine the sign of the integral. Second, for the integral to assume a value other than $0$, $a$ must lie between $0$ and $b$. \begin{equation} \int_{0}^{b}f(x)\delta(x-a)\,dx= \begin{cases}\color{green}{b>0} \begin{cases} \color{red}{0<a<b}\Rightarrow f(a) \\ others\Rightarrow 0 \end{cases} \\\color{green}{b<0} \begin{cases} \color{red}{b<a<0}\Rightarrow -f(a) \\ others\Rightarrow 0 \end{cases} \end{cases} \end{equation} Then we take a closer look at the expression $$f(a)\,(2\Theta(b)-1)\,\Theta(a-b\Theta(-b))\,\Theta(-a+b\Theta(b))$$ $(2\Theta(b)-1)$ corresponds to determining whether $\color{green}{b>0\: or\:b<0}$. If $b>0$ it gives a plus sign, if $b<0$ it gives the minus sign in front of $f(a)$.

The product $$\Theta(a-b\Theta(-b))\,\Theta(-a+b\Theta(b))$$ together says that for the integral to assume a value other than $0$, $\color{red}{either\; 0<a<b\;or\;b<a<0}$. $$\Theta(a-b\Theta(-b))=1\Leftrightarrow (b<0,a>b)\vee (b>0,a>0)$$ $$\Theta(-a+b\Theta(b))=1\Leftrightarrow (b>0,b>a)\vee (b<0,a<0)$$ Therefore, $$\Theta(a-b\Theta(-b))\,\Theta(-a+b\Theta(b))=1\neq 0$$ if and only if $$[(b<0,a>b)\vee (b>0,a>0)]\quad \wedge\quad [(b>0,b>a)\vee (b<0,a<0)]$$ Let $A=(b<0,a>b),\;B=(b>0,a>0),\;C=(b>0,b>a),\;(D=b<0,a<0)$, we have \begin{align} &(A\vee B)\wedge (C\vee D)\\=&(A\wedge C)\vee(A\wedge D)\vee(B\wedge C)\vee(B\wedge D)\\=&(A\wedge D)\vee (B\wedge C)\\=&\;\color{red}{b<a<0\quad or\quad 0<a<b} \end{align}

Answer 2

$$ \begin{align}I~:=~& \int_0^b \!dx~f(x)\delta(x\!-\!a)\cr ~=~& \left\{\theta(b) \int_0^b\!dx~-~\theta(-b) \int_b^0\!dx\right\} f(x)\delta(x\!-\!a) \cr ~=~& \int_{\mathbb{R}}\!dx \left\{\theta(b)\theta(b\!-\!x)\theta(x) ~-~\theta(-b) \theta(x\!-\!b)\theta(-x)\right\} f(x)\delta(x\!-\!a)\cr ~=~& f(a)\left\{\theta(b)\theta(b\!-\!a)\theta(a) ~-~\theta(-b) \theta(a\!-\!b)\theta(-a)\right\} \cr ~=~& f(a){\rm sgn}(b)\left\{\theta(a)\theta(b\!-\!a) ~+~\theta(a\!-\!b)\theta(-a)\right\}\cr ~=~& f(a){\rm sgn}(b)~\theta(a\!-\!\theta(-b)b)~\theta(-a\!+\!\theta(b)b) .\end{align}$$