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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 $0$?

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See…. – joriki Mar 22 '12 at 11:55
@joriki: Thanks! So it is just 1? – Ringo Mar 22 '12 at 12:15
As Carl's answer and the (partly contradictory) answers to the question I linked to indicate, it's not well-defined. If you tell us the context in which you're trying to use this, more might be said about what the appropriate definition might be for that context. – joriki Mar 22 '12 at 12:33
Joriki's comment is right. My answer uses the measure-theoretic interpretation of $\delta$ as a point mass measure. But if you used the interpretation via test functions instead, you should get $f(0)/2$ instead of $f(0)$. – Carl Mummert Mar 22 '12 at 12:42
See this answer. – Did Mar 24 '12 at 13:01

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".

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Thank you, Carl! – Ringo Mar 22 '12 at 15:00
What a nice answer! + 1 – Rudy the Reindeer Mar 25 '12 at 10:54

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.

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This is certainly an ambiguity, but I think there is another one as well. If we let $(g_n)$ be a sequence of even test functions such that $\lim_{n \to \infty} \int_{(-\infty,\infty)} f(x)g_n(x)\,dx = f(0)$ then, because each $g_n$ is even, if $f$ is also even we have $\lim_{n \to \infty} 2\int_{[0,\infty)} f(x)g_n(x)\,dx = f(0)$ and so $\lim_{n \to \infty} \int_{[0,\infty)} f(x)g_n(x)\,dx = f(0)/2$. Some people would write $\int_{[0,\infty)} f(x)\delta(x)\,dx$ to stand for the latter limit, and in particular they would have $\int_{[0,\infty)}\delta(x)\,dx = 1/2$ because of evenness. – Carl Mummert Mar 22 '12 at 13:21
Thank you, Brian! – Ringo Mar 22 '12 at 15:00
@Carl: Agreed. The original dirac delta was a limit of gaussians, defined in such a way. It is possible to define a point mass functional as a limit of 1-sided kernels as well. I think of these things in functional analytic terms: the $\delta$ defines a functional taking $f$ to its value at the location of the point mass. A limit of even functions is an odder beast, taking $f$ to $f(0)/2$ if the point is on the boundary and $f(0)$ or $0$ otherwise. Note that defining $\delta(x)$ as a limit of even functions has the annoying property that $\int_{(-\infty,0)} f(x) \delta(x) dx = f(0)/2$. – Brian B Mar 22 '12 at 16:07

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