Delta function at integral limit

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  • #1
jasonRF
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I have a question about delta functions. What I want to believe is the following
[tex]
\int_{-\infty}^{0} \, dt \, f(t) \delta(t) = \frac{1}{2} f(0).
[/tex]
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
[tex]\delta_n(t) = n/2[/tex] when [tex] -1/n \leq t \leq 1/n [/tex] and 0 otherwise.
Then
[tex]\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) [/tex]

Assuming f is continuous at 0, as n gets really large, we have
[tex]\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).
[/tex]

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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Answers and Replies

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

[tex]
\int_{a}^{b}{f(x) \, \delta(x - c) \, dx} = f(c) \, \Theta(c - a) \, \Theta(b - c)
[/tex]

where:

[tex]
\Theta(x) = \left\{\begin{array}{ll}
1 &, x > 0 \\

0 &, x < 0
\end{array}\right.
[/tex]

is the Heaviside step function. This function is not continuous at [itex]x = 0[/itex], not matter how you define [itex]\Theta(0)[/itex]. and the source of the factor 1/2 comes from its definition.
 
  • #3
jasonRF
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The 1/2 factor is a matter of convention. You should know the following rule:

[tex]
\int_{a}^{b}{f(x) \, \delta(x - c) \, dx} = f(c) \, \Theta(c - a) \, \Theta(b - c)
[/tex]

where:

[tex]
\Theta(x) = \left\{\begin{array}{ll}
1 &, x > 0 \\

0 &, x < 0
\end{array}\right.
[/tex]

is the Heaviside step function. This function is not continuous at [itex]x = 0[/itex], not matter how you define [itex]\Theta(0)[/itex]. 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 [tex] \Theta(x) [/tex] 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
 
  • #4
2,967
5
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:

[tex]
F(x) \equiv \int^{x}_{-\infty}{f(t) \, \delta(t) \, dt}
[/tex]

then

[tex]
F(-\epsilon) = 0
[/tex]

and

[tex]
F(+\epsilon) = f(0)
[/tex]

for any [itex]\epsilon > 0[/itex]. Thus, the left hand and right hand limits are different and the limit does not exist. No matter what value you define for [itex]F(0)[/itex], you cannot make it continuous.
 
  • #5
jasonRF
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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:

[tex]
F(x) \equiv \int^{x}_{-\infty}{f(t) \, \delta(t) \, dt}
[/tex]

then

[tex]
F(-\epsilon) = 0
[/tex]

and

[tex]
F(+\epsilon) = f(0)
[/tex]

for any [itex]\epsilon > 0[/itex]. Thus, the left hand and right hand limits are different and the limit does not exist. No matter what value you define for [itex]F(0)[/itex], 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
 
  • #6
Mute
Homework Helper
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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

[tex]F(x) = \int_{-\infty}^x dt f(t) \delta(t)[/tex]

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

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

Your argument that gives [itex]A = f(0)/2[/itex] 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

[tex]\int_{-\infty}^{0^+}dt f(t) \delta (t) = f(0).[/tex]

Similarly, the Laplace transform is typically defined such that the lower limit is actually [itex]0^-[/itex], in order to allow the delta function to be a Laplace-transformable function.
 
  • #7
jasonRF
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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
 
  • #8
2,967
5
Here is an argument (I know it isn't a proof). If I use the "delta-sequence" approach and define
[tex]\delta_n(t) = n/2[/tex] when [tex] -1/n \leq t \leq 1/n [/tex] and 0 otherwise.
Then
[tex]\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) [/tex]
Assume another type of sequence:

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

Then:

[tex]
\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
[/tex]

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

[tex]
\frac{1 + \epsilon}{2} \, f(0)
[/tex]

By choosing various values of [itex]\epsilon[/itex] 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.
 

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