Delta function at integral limit

[jasonRF]

I have a question about delta functions. What I want to believe is the following
<br /> \int_{-\infty}^{0} \, dt \, f(t) \delta(t) = \frac{1}{2} f(0).<br />
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) <br /> = \lim_{n \rightarrow \infty} \int_{-\infty}^{0} \, dt \, f(t) \delta_n(t) <br /> = \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).<br />

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

 

[Dickfore]

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

<br /> \int_{a}^{b}{f(x) \, \delta(x - c) \, dx} = f(c) \, \Theta(c - a) \, \Theta(b - c)<br />

where:

<br /> \Theta(x) = \left\{\begin{array}{ll}<br /> 1 &amp;, x &gt; 0 \\<br /> <br /> 0 &amp;, x &lt; 0<br /> \end{array}\right.<br />

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.

 

[jasonRF]

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

<br /> \int_{a}^{b}{f(x) \, \delta(x - c) \, dx} = f(c) \, \Theta(c - a) \, \Theta(b - c)<br />

where:

<br /> \Theta(x) = \left\{\begin{array}{ll}<br /> 1 &amp;, x &gt; 0 \\<br /> <br /> 0 &amp;, x &lt; 0<br /> \end{array}\right.<br />

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

 

[Dickfore]

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:

<br /> F(x) \equiv \int^{x}_{-\infty}{f(t) \, \delta(t) \, dt}<br />

then

<br /> F(-\epsilon) = 0<br />

and

<br /> F(+\epsilon) = f(0)<br />

for any \epsilon &gt; 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.

 

[jasonRF]

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:

<br /> F(x) \equiv \int^{x}_{-\infty}{f(t) \, \delta(t) \, dt}<br />

then

<br /> F(-\epsilon) = 0<br />

and

<br /> F(+\epsilon) = f(0)<br />

for any \epsilon &gt; 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]

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.

 

[jasonRF]

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

 

[Dickfore]

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) <br /> = \lim_{n \rightarrow \infty} \int_{-\infty}^{0} \, dt \, f(t) \delta_n(t) <br /> = \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 &lt; \epsilon &lt; 1 and 0 otherwise.

Then:

<br /> \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 &lt; \theta_{n} &lt; 1<br />

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

<br /> \frac{1 + \epsilon}{2} \, f(0)<br />

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.