Integral of Dirac[t] delta function

Question

When calculating $$I=\int_{-\infty}^{\infty}f(x)\,\delta\big(g(x)\big)\, \mathrm dx$$ one can use the fact that $$\delta\big( g(x) \big) = \sum_i\dfrac {\delta(x-x_i)}{\left|g'(x_i)\right|}$$ where $g(x_i)=0$ to determine $I$.

However, what happens in the case where we're not integrating over the whole real line, but instead one (or multiple) such $x_i$s happen to be integration bounds -- e.g. $$\int_0^1 \delta(x^2-x)\,\mathrm dx$$

Or, more broadly, what is $$\int_0^\infty \delta(x)\, \mathrm dx$$?

Answer 1

Firstly, the integrals $\newcommand\dif{\mathop{}\!\mathrm{d}}$ \begin{equation} \int_0^\infty \delta(x)\dif x \end{equation} and \begin{equation} \int_0^\infty \delta(x^2-x)\dif x \end{equation} are not defined in the context of distribution theory. Here is one way to justify this: Let $\delta_\epsilon$ be a nascent dirac delta, then the limit \begin{equation} \lim_{\epsilon\to 0}\int_0^\infty \delta_\epsilon(x)\dif x \end{equation} is not independent of the chosen sequence and may even not exist (here is the proof).

That being said, one may still assign a value to these integrals: As explained in Mathematics for Physicists we can define such integrals by restricting ourselves to "appropriate" representations of the $\delta$-function, e.g. the Gaussian representation \begin{equation} \delta_\epsilon(x)=\frac{\exp(-x^2/\epsilon^2)}{\epsilon\sqrt\pi} \end{equation} or the Lorentzian representation: \begin{equation} \delta_\epsilon(x)=\frac{1}{\pi}\frac{\epsilon}{\epsilon^2+x^2} \end{equation} Then we obtain the following results: \begin{equation} \lim_{\epsilon\to 0}\int_0^\infty \delta_\epsilon(x)\dif x=\frac{1}{2} \end{equation} and \begin{equation} \lim_{\epsilon\to 0}\int_0^\infty \delta_\epsilon(x^2-x)\dif x=1 \end{equation}

Sketch of the proof

More generally, we can consider integrals of the form \begin{equation} \int_A\delta(g(x))\dif x:=\lim_{\epsilon\to 0}\int_A\delta_\epsilon(g(x))\dif x \end{equation} To be precise, we consider the following setting: $g$ is defined on an open set $D\subset\mathbb R$ and $A\subset D$ such that $g$ has finitely many zeros $$x_1<\ldots<x_n$$ in $A$. $A$ is not necessarely open. Furthermore, we assume that $g$ is a diffeomorphism on a neighbourhood of each zero. Now let $U_1,\ldots,U_n$ be a list of disjoint subsets in $A$ which are open w.r.t. to the relative topology and with $x_i\in U_i$. Then: \begin{align} \forall \epsilon:\lim_{\epsilon\to 0}\int_A\delta_\epsilon(g(x))\dif x=\lim_{\epsilon\to 0}\int_{\bigcup_i U_i}\delta_\epsilon(g(x))\dif x+\lim_{\epsilon\to 0}\int_{A\setminus\bigcup_i U_i}\delta_\epsilon(g(x))\dif x\\=\sum_i\lim_{\epsilon\to 0}\int_{U_i}\delta_\epsilon(g(x))\dif x+\underbrace{\int_{A\setminus\bigcup_i U_i}\underbrace{\lim_{\epsilon\to 0}\delta_\epsilon(g(x))}_{=0}\dif x}_{=0}\\=\sum_i\lim_{\epsilon\to 0}\int_{g(U_i)}\frac{\delta(y)}{|(g'\circ g^{-1})(y)|}\dif y=\sum_i\frac{c_i}{|g'(x_i)|} \end{align} with $c_i=1$ if $g(x_i)$ is an inner point of $g(U_i)$ which is equivalent to $x_i$ being an inner point of $D$ since $g|_{U_i}$ is a diffeo (I am not 100% sure regarding the equivalence) and $c_i=1/2$ otherwise (i.e. if $x_i$ is a boundary point of $U_i$).

Not let us apply the result above to your example: In our case, we choose some $0<\epsilon<1/2$ such that $g(x)=x^2-x$ is a diffeo on $[0,\epsilon]$ and on $[1-\epsilon,1]$. Note that $g(\epsilon)=\epsilon^2-\epsilon<0$ and $g(1-\epsilon)=-\epsilon(1-\epsilon)<0$ and we obtain: \begin{align} \forall\epsilon:\int_0^1 \delta_\epsilon(x^2-x)\dif x=\int_0^\epsilon \delta_\epsilon(g(x))\dif x+\int_{1-\epsilon}^1 \delta_\epsilon(g(x))\dif x\\=\int_{\epsilon^2-\epsilon}^0 \frac{\delta_\epsilon(y)}{|(g'\circ g^{-1})(y)|}\dif y+\int_{-\epsilon(1-\epsilon)}^0 \frac{\delta_\epsilon(y)}{|(g'\circ g^{-1})(y)|}\dif y \end{align} and thus: \begin{equation} \lim_{\epsilon\to 0}\int_0^1 \delta_\epsilon(x^2-x)\dif x=\frac{1}{2}\frac{1}{|g'(0)|}+\frac{1}{2}\frac{1}{|g'(1)|}=1 \end{equation}

Answer 2

An integral of the type $\int _0 ^1 \delta(x^2-x)dx$ may appear confusing at first, but can be understood by choosing a suitable representation of the Dirac delta function. The block-form is the simplest (and often the best):

$\delta_\epsilon (y) = 1/(2\epsilon) \qquad$ if $\qquad -\epsilon < y < \epsilon \qquad and \qquad =0 \qquad $ otherwise

Application of this representation to the integral above means finding the values of $x$ where the argument is larger than $-\epsilon$. Up to linear order in $\epsilon$ this results in the two regions $0 < x < \epsilon$ and $1-\epsilon < x < 1$. The integral therefore simplifies to

$$\int _0 ^\epsilon 1/(2\epsilon) dx + \int _{1-\epsilon} ^1 1/(2\epsilon)dx = 1/2 + 1/2 = 1$$

EDIT: Here is a more formal way of evaluating the intgral. First we split the intgration interval into two regions: $(0,1/2)$ and $(1/2, 1)$. Now introduce the new variable $y = x^2-x$. This is equivalent to $x = 1/2 + \sqrt{y + 1/4}$ or $x = 1/2 - \sqrt{y + 1/4}$. The minus sign corresponds to the first interval, the plus to the second. Differentiation with respect to $y$ yields the relation between $dx$ and $dy$. Substitution into the original expression gives:

$$\int_{0}^{-1/4}(-1/2)/\sqrt{y+1/4} \ \delta(y)dy + \int_{-1/4}^{0}(1/2)/\sqrt{y+1/4} \ \delta(y)dy$$

This is a trivial integral that reduces to $1/2 + 1/2 = 1$. So the "loose" method and the "formal" method lead to the same result. Note that if the original intgral contained a smooth test function $f(x)$, the result with both methods would be: $(f(0)+f(1))/2$.