Integrating over a discontinuity

In summary, the equation shows that if the function is differentiable and rapidly goes to zero at infinity, then the integral over the space between -infinity and +infinity is equal to the derivative of the generalized function.
  • #1
Pacopag
197
4

Homework Statement


I would like to prove that [tex]{d \over {dx}} \sigma(x)=2\delta(x)[/tex],
where [tex]\sigma(x>0)=1[/tex]
[tex]\sigma(x=0)=0[/tex]
[tex]\sigma(x<0)=-1[/tex]

Homework Equations


The Attempt at a Solution


I think that I need to show that
[tex]\int_{-\infty}^{\infty}{d \over {dx}} \sigma(x)dx=2=\int_{-\infty}^{\infty}2 \delta(x)dx[/tex].
The integral looks rather harmless, and I would like to write
[tex]\int_{-\infty}^{\infty}{d \over {dx}} \sigma(x)dx=\sigma(x) |^{\infty}_{-\infty}=1-(-1)=2[/tex].
This looks like it works, but it seems to me that there should be some complication when integrating close to zero where we have discontiuous behaviour.
 
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  • #2
Consider this: is there a difference between integrating (some integrable function) over the closed interval [0,1] versus integrating it over the half-open interval (0, 1] ? If you're not sure, look up the concept of "measure zero".
 
Last edited:
  • #3
I think the answer to your question slider is no. But how do I integrate "over" the discontinuity with a clear conscience.
 
  • #4
What you are integrating is d(sigma(x))/dx. That's worse than being discontinuous, it doesn't even exist at x=0 in the usual sense. That is a complication. But the whole point to inventing delta functions is to make sense of problems like this. There is nothing wrong with what you are doing.
 
  • #5
What is bothering me is that it looks like I am applying the fundamental theorem of calculus. The the FTC is only applicable if the integrand is continuous over the interval of integration. Here, it clearly isn't so how can my calculation be valid (even if it is giving the right answer)??.
 
  • #6
When you extend the family of functions you are dealing with by including distributions like delta functions, the FTC remains valid provided the antiderivative is well defined at the endpoints. You do have to formally justify this stuff, but I don't think you are being asked to do this.
 
  • #7
Ok. I'm satisfied with that. I guess I would have to study the theory of generalized functions to see such justifications??
 
  • #8
Yep. Exactly. You are right, the usual FTC doesn't apply. The generalized one does.
 
  • #9
Great. Thank you very much.
 
  • #10
It turns out that what you really need to show is that, for any test function (i.e. infinitely differentiable, and 'rapidly' goes to zero at infinity) [itex]\varphi[/itex]:

[tex]
\int_{-\infty}^{+\infty} \left( \frac{d}{dx} \sigma(x) \right) \cdot \varphi(x) \, dx
=
\int_{-\infty}^{+\infty} \left( 2 \delta(x) \right) \cdot \varphi(x) \, dx
[/tex]

but you need to know more; the definition of the derivative of a generalized function is that it satisfies:

[tex]
\int_{-\infty}^{+\infty} \left( \frac{d}{dx} \sigma(x) \right) \cdot \varphi(x) \, dx
:=
-\int_{-\infty}^{+\infty} \sigma(x) \cdot \left( \frac{d}{dx} \varphi(x) \right) \, dx
[/tex]
 

Related to Integrating over a discontinuity

1. What is meant by "integrating over a discontinuity"?

Integrating over a discontinuity refers to the process of calculating the definite integral of a function that has a discontinuity, or a point where the function is not continuous. This requires special techniques and considerations compared to integrating a continuous function.

2. Why is integrating over a discontinuity important?

Integrating over a discontinuity allows for a more accurate representation of a function that may have sudden changes or breaks in its behavior. This is important in many fields of science, such as physics and engineering, where precise calculations are necessary.

3. How do you handle a discontinuity when integrating?

The method for handling a discontinuity when integrating depends on the type of discontinuity. This can include using different integration techniques, breaking the integral into smaller pieces, or applying mathematical properties such as the limit definition of a derivative.

4. Can a discontinuous function be integrated?

Yes, a discontinuous function can still be integrated. However, special techniques and considerations must be used in order to accurately calculate the integral. In some cases, the integral may not exist due to the nature of the discontinuity.

5. Are there any real-world applications of integrating over a discontinuity?

Yes, there are many real-world applications of integrating over a discontinuity. For example, in physics, the calculation of work done by a variable force often involves integrating over a discontinuous force function. In engineering, the calculation of stress and strain in materials with sudden changes in properties also requires integrating over discontinuities.

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