Dirac Delta function

Question

I know that $\int_{-\epsilon}^\infty f(x)\delta(x)dx=f(0)$ but what about $\int_0^\infty f(x)\delta(x)dx$? I suppose we have to do this by definition since the lower limit is bang on $0$?

Answer 1

The dirac delta function is a strange beast. The entire notation "$\delta(x)dx$" is a little bit of a lie, because there is not actually a function that has the properties that the $\delta$ function is axiomatized to have, which are:

• $\delta(x) = 0$ when $x \not = 0$
• $\int_{-\infty}^\infty \delta(x)\, dx= 1$

Using these, and the usual properties of the integral, it is not hard to prove that $\int_0^\infty f(x)\delta(x)\,dx = f(0)$ for every function $f$. The main point is that

$$\int_{-\infty}^\infty f(x)\delta(x)\,dx = \int_{(-\infty,0)} f(x)\delta(x)\,dx + \int_{[0,\infty)}f(x)\delta(x)\,dx = 0 + \int_0^\infty f(x)\delta(x)\,dx$$ and $$\int_{-\infty}^\infty f(x)\delta(x)\,dx = \int_{-\infty}^\infty f(0)\delta(x)\,dx = f(0)\int_{-\infty}^\infty \delta(x)\,dx = f(0)$$ Of course you can show by similar methods that $\int_0^0 f(x)\delta(x)\,dx= f(0)$, although this integral should always be 0. This is one way to see that no $\delta$ with the two properties listed above really exists as a function.

A way to make sense of this is to use measure-theoretic techniques and Lebesgue integration. We can treat the $\delta$ function as a measure; it puts 1 unit of "mass" at the origin and no mass anywhere else. If we call that measure $\phi$ and we integrate a function $f$ with respect to that measure, we really do get $\int_{-\infty}^\infty f(x) \,d\phi = \int_0^0 f(x)\,d\phi = f(0)$. The problem is that this measure does not have a density function (a Radon-Nikodym derivative) with respect to Lebesgue measure, and so we cannot replace the $d\phi$ with anything of the form $\delta(x)dx$ where $dx$ indicates Lebesgue measure.

In practice, people use the notation $\delta(x)dx$ as a purely formal device for calculations. That works out OK as long as only sound methods used for manipulating integrals involving $\delta(x)$. In introductory books, the entire issue is usually glossed over, with just a sentence of two saying "this is formally wrong but it works as long as you do it in the way shown here".

Answer 2

You are something of a victim of ambiguity in notation. The definition of the integral $$ \int_A f \; da $$ is taken over the open or closed (or neither) object $A$. As Carl M notes, we have that $$ \int_{\{0\}} f(x) \delta(x) \; dx = f(0) $$ by construction of the delta functional, while $$ \int_{(0,\infty)} f(x) \delta(x) \; dx = \int_{(-\infty,0)} f(x) \delta(x) \; dx = 0. $$ More generally, $$ \int_A f(x) \delta(x) \; dx = \begin{cases}f(0)\qquad&0 \in A\\0\qquad&\text{otherwise.}\end{cases} $$ When someone writes $$ \int_0^\infty g(x) \; dx $$ they are not being clear what the domain of integration really is.