# Delta function at integral limit

Gold Member
I have a question about delta functions. What I want to believe is the following
$$\int_{-\infty}^{0} \, dt \, f(t) \delta(t) = \frac{1}{2} f(0).$$
It even shows up on Wikipedia (so it must be true!)

Here is an argument (I know it isn't a proof). If I use the "delta-sequence" approach and define
$$\delta_n(t) = n/2$$ when $$-1/n \leq t \leq 1/n$$ and 0 otherwise.
Then
$$\int_{-\infty}^{0} \, dt \, f(t) \delta(t) = \lim_{n \rightarrow \infty} \int_{-\infty}^{0} \, dt \, f(t) \delta_n(t) = \lim_{n \rightarrow \infty} n/2 \int_{-1/n}^{0} \, dt \, f(t)$$

Assuming f is continuous at 0, as n gets really large, we have
$$\int_{-\infty}^{0} \, dt \, f(t) \delta(t) \approx \lim_{n\rightarrow \infty} n/2 f(0) \int_{-1/n}^{0} \, dt = \frac{1}{2} f(0).$$

However, it seems that all the books I have make this undefined. In particular, this would mean that the Heaviside step function has a value of 1/2 at 0, but most treatments let the setup function be undefined at 0, indicating that the above integral is also undefined.

I am an engineer, not a mathematician, so I'm sure the books I am looking in are not what mathematicians looks at. But is seems that there is more then one way to define this integral? I did take an applied math course that spent a few weeks on distributions (used Strichartz's book) but we didn't get into this exact question, and the fellow I loaned my book to has disappeared!

Is there a consensus on the value of the integral above, or does it depend on how one defines things and builds up the theory of distributions?

Thanks!

jason

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The 1/2 factor is a matter of convention. You should know the following rule:

$$\int_{a}^{b}{f(x) \, \delta(x - c) \, dx} = f(c) \, \Theta(c - a) \, \Theta(b - c)$$

where:

$$\Theta(x) = \left\{\begin{array}{ll} 1 &, x > 0 \\ 0 &, x < 0 \end{array}\right.$$

is the Heaviside step function. This function is not continuous at $x = 0$, not matter how you define $\Theta(0)$. and the source of the factor 1/2 comes from its definition.

Gold Member
The 1/2 factor is a matter of convention. You should know the following rule:

$$\int_{a}^{b}{f(x) \, \delta(x - c) \, dx} = f(c) \, \Theta(c - a) \, \Theta(b - c)$$

where:

$$\Theta(x) = \left\{\begin{array}{ll} 1 &, x > 0 \\ 0 &, x < 0 \end{array}\right.$$

is the Heaviside step function. This function is not continuous at $x = 0$, not matter how you define $\Theta(0)$. and the source of the factor 1/2 comes from its definition.
yes, I realize that the step function is not continuous; it essentially is the integral I posted with f(t) = 1. It is just that I have read many treatments that explicitly state $$\Theta(x)$$ is undefined at 0. These same sources also typically don't address the integral I posted, which is only slightly more general than the step function. Perhaps I am just looking at the wrong sources (signal processing, applied math and math methods for physics type books)? I don't feel like my question is profound or anything, I just want to understand the proper way to understand this. If it just comes down to how one choses to define the integral above, I can live with that, although intellectually it isn't so satisfying. To my naive brain it seems that my argument should be able to be turned into a proof that could settle this ?

thanks,

jason

Perhaps you are accustomed to view definite integrals with variable upper (or lower) bounds as continuous functions. This is true when the integrand is a bounded function. Since the integrand in our case contains a function that is unbounded, this theorem does not hold and the integral is discontinuous. I would say that if we define:

$$F(x) \equiv \int^{x}_{-\infty}{f(t) \, \delta(t) \, dt}$$

then

$$F(-\epsilon) = 0$$

and

$$F(+\epsilon) = f(0)$$

for any $\epsilon > 0$. Thus, the left hand and right hand limits are different and the limit does not exist. No matter what value you define for $F(0)$, you cannot make it continuous.

Gold Member
Perhaps you are accustomed to view definite integrals with variable upper (or lower) bounds as continuous functions. This is true when the integrand is a bounded function. Since the integrand in our case contains a function that is unbounded, this theorem does not hold and the integral is discontinuous. I would say that if we define:

$$F(x) \equiv \int^{x}_{-\infty}{f(t) \, \delta(t) \, dt}$$

then

$$F(-\epsilon) = 0$$

and

$$F(+\epsilon) = f(0)$$

for any $\epsilon > 0$. Thus, the left hand and right hand limits are different and the limit does not exist. No matter what value you define for $F(0)$, you cannot make it continuous.
I fully realize that this cannot be continuous. That is not what I am attempting to ask about. My question is, what is the value of the integral I originally posted? I believe it should be f(0)/2 and presented a crude argument in support of that assertion; if it were continuous then I would have no need to ask a question. The books I have fail to address this so I thought I'd post to learn if I was correct or incorrect, or to find out if "it depend ..."

thanks,

jason

Mute
Homework Helper
I fully realize that this cannot be continuous. That is not what I am attempting to ask about. My question is, what is the value of the integral I originally posted? I believe it should be f(0)/2 and presented a crude argument in support of that assertion; if it were continuous then I would have no need to ask a question. The books I have fail to address this so I thought I'd post to learn if I was correct or incorrect, or to find out if "it depend ..."

thanks,

jason
Dickfore's point is that the integral you posted has no well-defined value. As he already said, the function

$$F(x) = \int_{-\infty}^x dt f(t) \delta(t)$$

is discontinuous at x = 0 as the limits approaching zero from above or below are not the same. This means that I could define the function

$$F(x) = \left\{\begin{array}{c c} \int_{-\infty}^x dt f(t) \delta(t),~x \neq 0 \\ A,~x=0 \end{array}\right\.$$
where A is anything I want it to be.

Your argument that gives $A = f(0)/2$ suggests a nice value to choose A to be, but as you said, it is not a proof. I can pick anything I want for A.

Typically to get around this ambiguity, your integral, for example, would be define to be

$$\int_{-\infty}^{0^+}dt f(t) \delta (t) = f(0).$$

Similarly, the Laplace transform is typically defined such that the lower limit is actually $0^-$, in order to allow the delta function to be a Laplace-transformable function.

Gold Member
Thank you both for spelling this out for me - I can be pretty slow at times! So mathematically this integral is undefined. I guess, since the "value" of a distribution at a point isn't a concept that means anything, and a special case of my integral is the Heaviside step distribution, the value at at the origin doesn't mean anything?

Of course, it is often the case that when I use a delta function it is an idealized model of something physical; usually something that looks like a Gaussian (or other such function) with extremely narrow width and very tall height. In these cases, my naive type of argument should yield the value of the integral that I am require.

Thanks again!

jason

Here is an argument (I know it isn't a proof). If I use the "delta-sequence" approach and define
$$\delta_n(t) = n/2$$ when $$-1/n \leq t \leq 1/n$$ and 0 otherwise.
Then
$$\int_{-\infty}^{0} \, dt \, f(t) \delta(t) = \lim_{n \rightarrow \infty} \int_{-\infty}^{0} \, dt \, f(t) \delta_n(t) = \lim_{n \rightarrow \infty} n/2 \int_{-1/n}^{0} \, dt \, f(t)$$
Assume another type of sequence:

$$\delta_n(t; \epsilon) = n/2$$ when $$(-1 - \epsilon)/n \leq t \leq (1 - \epsilon)/n, -1 < \epsilon < 1$$ and 0 otherwise.

Then:

$$\int_{-\infty}^{0}{f(t) \, \delta_{n}(t; \epsilon) \, dt} = \frac{n}{2} \, \int_{-\frac{1 + \epsilon}{n}}^{0}{f(t) \, dt} = \frac{n}{2} \, f(\frac{(1 + \epsilon) (\theta_{n} - 1)}{n}) \, \frac{1 + \epsilon}{n}, \; 0 < \theta_{n} < 1$$

according to the mean - value theorem. Then, as we tend $n \rightarrow \infty$, we get the limit of the above sequence to be:

$$\frac{1 + \epsilon}{2} \, f(0)$$

By choosing various values of $\epsilon$ we get different limits for the integral, so you cannot claim that the limit is unique and independent of the sequence of functions you select.