# Quick Question on the Dirac Delta Function

1. Mar 1, 2008

### G01

The Dirac delta function, $$\delta (x)$$ has the property that:

(1) $$\int_{-\infty}^{+\infty} f(x) \delta (x) dx = f(0)$$

Will this same effect happen for the following bounds on the integral:

(2) $$\int_{0}^{+\infty} f(x) \delta (x) dx = f(0)$$

My intuition tells me that it should, but the fact that the peak of the delta function lies on one of the bounds makes me think I should double check my reasoning. So, does anyone know if (2) above is correct?

Thanks for any advice you can offer.

Last edited: Mar 1, 2008
2. Mar 1, 2008

### Vid

According to mathematica integral from 0 to infinity and the integral from negative infinity to 0 are equal to f(0)/2.

3. Mar 1, 2008

### jostpuur

I think I've seen sometimes integration limits like this

$$\lim_{\epsilon\to 0^+}\int\limits_{0-\epsilon}^{\infty}$$

to avoid this problem, which suggests that there probably is not very simple answer to the question. For example

$$\lim_{n\to\infty} \int\limits_0^{\infty} f(x) \sqrt{\frac{n}{\pi}}e^{-nx^2} dx = \frac{1}{2}f(0)$$

but suppose you modify the kernel like this

$$\sqrt{\frac{n}{\pi}} e^{-n(x-x_0(n))^2}$$

where $x_0:\mathbb{N}\to\mathbb{R}$ is some sequence such that $x_0(n)\to 0$ as $n\to\infty$. That still behaves as a delta function if the zero is contained in an open integration interval, but if the zero is on the boundary of the integration interval, the answer could be anything, depending on the chosen $x_0$.

4. Mar 1, 2008

### tiny-tim

$$\delta (x)$$ isn't really a function.

It only has meaning inside $$\int_{-\infty}^{+\infty} f(x) \delta (x) dx$$.

When it is on its own as part of a calculation, that is because it will be put inside such an integral in the next step of the calculation.

This is because it is a symbol which is defined by the property $$\int_{-\infty}^{+\infty} f(x) \delta (x) dx = f(0)$$

Inside $$\int_{0}^{+\infty} f(x) \delta (x) dx$$ , it's a symbol without any meaning.

(Though of course, you can write $$\int_{-\infty}^{+\infty} \theta (x) f(x) \delta (x) dx$$, which looks pretty similar.)

5. Mar 3, 2008

### Rainbow Child

The answer comes by joining the posts of jostpuur and tiny-tim.

The second integral is
$$I=\int_{0}^{+\infty} f(x) \delta (x) dx = \int_{-\infty}^{+\infty} \theta (x) f(x) \delta (x) dx$$
where $\theta(x)$ is the step function. Thus $I$ equals

$$I=\theta(0)\,f(0)[/itex] and the question now is what's the definition of the step function. The one which is most commonly used is that $\theta(0)=\frac{1}{2}$, thus [tex]I=\frac{1}{2}\,f(0)$$

6. Mar 3, 2008

### jostpuur

This doesn't fully make sense. If f is a function, then

$$\int\limits_0^{\infty} f(x) dx = \int\limits_{-\infty}^{\infty} \theta(x)f(x) dx$$

is true with arbitrary value $\theta(0)$. Why should the $\theta(0)=1/2$ be chosen when there is $f(x)\delta(x)$ in the integrand instead?

I think the original question is of similar nature as the question about what 0/0 is supposed to be. As I showed in my first response, if you try to approach this integral with some limits, you can get different results with different ways of taking the limit. Thus, we say that the limits cannot be used to give proper meaning for the this integral.

7. Mar 3, 2008

Because when the integrand contains a Dirac delta, it is not a function, and so the condition "if f is a function" does not apply.

Well, it's more like asking what half of infinity is. But, yeah, to make Dirac deltas at all consistent (let alone rigorous), you need to enforce a definition of them as a limit of some sequence of well-defined functions. As you demonstrated, which exact sequence you choose will affect the answers you get if you ask questions like the one in this thread. That said, most people who are that concerned about consistency forgo the Dirac delta entirely and instead use distributions. The rest of us just live with the fact that certain operations (again, like the OP) aren't defined.

8. Mar 3, 2008

### jostpuur

I'll be more explicit!

Consider the sequence of following functions

$$\delta_{n,1}(x) = \sqrt{\frac{n}{\pi}} \exp\big(-n(x-n^{-1/4})^2\big) = \sqrt{\frac{n}{\pi}} \exp\big(-(\sqrt{n}x -n^{1/4})^2\big)$$

It is easy to believe, that the distributions, represented by these functions, converge towards the delta distribution, because this is just the usual Gaussian peak representation, but displaced slightly. Let a and b be such that $a\leq 0 < b$. Let us calculate the next integral.

$$\lim_{n\to\infty}\int\limits_a^b f(x)\delta_{n,1}(x) dx = \lim_{n\to\infty} \sqrt{\frac{n}{\pi}} \int\limits_a^b f(x)\exp\big(-(\sqrt{n}x-n^{1/4})^2\big) dx = \lim_{n\to\infty} \frac{1}{\sqrt{\pi}} \int\limits_{\sqrt{n}a-n^{1/4}}^{\sqrt{n}b-n^{1/4}} f\big(\frac{y}{\sqrt{n}} + n^{-1/4}\big) e^{-y^2} dy$$
$$= \frac{1}{\sqrt{\pi}} \int\limits_{-\infty}^{\infty} f(0) e^{-y^2} dy = f(0)$$

However, notice that this is indeed true even with a=0.

If we instead chose the sequence

$$\delta_{n,2}(x) = \sqrt{\frac{n}{\pi}} e^{-nx^2}$$

we would have

$$\lim_{n\to\infty} \int\limits_a^b f(x)\delta_{n,2}(x) dx = f(0),\quad\quad a<0<b$$

and

$$\lim_{n\to\infty} \int\limits_a^b f(x)\delta_{n,2}(x) dx = \frac{1}{2}f(0),\quad\quad a=0<b.$$

This means, that the distributions, represented by functions $\delta_{n,1}$ and $\delta_{n,2}$, converge towards the same delta distribution. This is so, because if you want to calculate

$$\int\limits_a^b f(x)\delta(x) dx,\quad\quad a<0<b$$

you can choose either of these sequences, and they give the same result. But when you calculate

$$\int\limits_0^b f(x)\delta(x) dx,$$

choosing different representations for the delta distribution, you get different numbers out of the integral. Thus, the delta distribution does not contain enough information to calculate this integral.

Last edited: Mar 3, 2008
9. Mar 3, 2008

### jostpuur

I was not applying any function assumption to any calculation with distributions. I merely noted, that with integrals of functions, the value of $\theta(0)$ does not matter, and then asked that why should some specific value for it be chosen when we integrate distributions.

Not precisely true. Distributions can be defined without sequences of functions, but the sequences of functions still remain as an important way to handle distributions.

Yes. This was my point. I explained it even more explicitly in my last post #8.

10. Mar 4, 2008

### tiny-tim

The endearing quality of distributions

Because that's how distributions are defined, and what they were created for!

It is distributions' most endearing (and, in my opinion, only endearing ) quality.

11. Mar 4, 2008

### arildno

Not at all.

Let F be some function space over R, I some open interval in R, and define the functional D as follows:
$$D(f,I)=f(0), 0\in{I},f\in{F},D(f,I)=0,0\notin{I},f\in{F}$$
This is the delta functional, and it can be shown to be linear, i.e, a distribution.
This is how you rigorously define the Dirac Delta "function".

12. Mar 4, 2008

### Rainbow Child

But that's the whole point! $\delta(x)$ is a distribution thus it does not behaves like ordinary functions. Your point of view is that the integral does not exist, because it can take multiple values, but I say that it can be defined if you choose the value of $\theta(0)$ yielding to

$$I=\int\limits_0^{\infty} f(x)\,\delta(x) dx = \theta(0)\,f(x)$$

13. Mar 4, 2008

### jostpuur

When you choose the value for theta(0), in effect you are choosing the value of the entire integral. The number, that is supposed to come out of the integral, is not something that should be chosen. It should come out from some calculation. Right now the calculations don't give a unique number, and I don't feel like choosing some unique number by force would be a very good way to deal with this.