Delta Function Integration: Justified or Fudging?

In summary, the person is trying to solve a homework equation but is having trouble figuring out the answer.
  • #1
BOAS
552
19
Hello,

I feel like I am fudging these integrals a bit and would like some concrete guidance about what's going on.

1. Homework Statement

Evaluate ##I = \int_{-1}^{1} dx \delta'(x)e^3{x} ##

Homework Equations

The Attempt at a Solution


[/B]
I use integration by parts as follows,
##u = e^{3x}##
##du = 3e^{3x} dx##

##dv = \delta'(x)##
##v = \delta(x)##

##I = uv|^{1}_{-1} - \int^{1}_{-1} v du##

Evaluating ##uv## gives zero because the delta function is zero at both limits.

##I = - \int^{1}_{-1} dx \delta(x) 3e^{3x}##

Clearly, the only non zero value of this function is at ##x = 0##, making me think the answer to my problem is ##I = -3##, but this is the step that I think I'm 'fudging'. I feel like I'm disregarding the integration sign...

Is this justified?

Thanks for any guidance you can give.
 
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  • #2
That's justified because ##\delta(x)## is centered in ##x=0##, which is contained by the integration limits.
 
  • #3
That looks correct. It sometimes helps to think of a finite spiked function instead of delta. Let's say it's 0 outside ##[-a, a]## for some small a, and has a value of ##1/2a## in that interval. The integral is then approx: ##2a \frac{1}{2a} e^0 = e^0##
 
  • #4
PeroK said:
That looks correct. It sometimes helps to think of a finite spiked function instead of delta. Let's say it's 0 outside ##[-a, a]## for some small a, and has a value of ##1/2a## in that interval. The integral is then approx: ##2a \frac{1}{2a} e^0 = e^0##

I don't quite follow.

Where do you evaluate the limits?

I see that the spiked function is ##\frac{1}{2a}## inside this interval, and since (I assume) it's centered on ##0##, I see where ##e^0## comes from, but not the ##2a##.
 
  • #5
BOAS said:
I don't quite follow.

Where do you evaluate the limits?

I see that the spiked function is ##\frac{1}{2a}## inside this interval, and since (I assume) it's centered on ##0##, I see where ##e^0## comes from, but not the ##2a##.

2a is the width of the interval. An integral is the area under a curve!
 
  • #6
PeroK said:
2a is the width of the interval. An integral is the area under a curve!

Ah!

I'm with you now. So an integral of a function multiplied by the dirac delta function is like evaluating the function itself at the point where the delta function is non-zero?
 
  • #7
BOAS said:
Ah!

I'm with you now. So an integral of a function multiplied by the dirac delta function is like evaluating the function itself at the point where the delta function is non-zero?

That's exactly what it is. You can think of the delta function as the limit of a sequence of rectangular functions with area 1. That gives some justification for things.
 
  • #8
PeroK said:
That's exactly what it is. You can think of the delta function as the limit of a sequence of rectangular functions with area 1. That gives some justification for things.

Cool - Thanks for your help!
 

1. What is a Delta Function Integral?

A Delta Function Integral is a mathematical tool used to describe the behavior of a point-like source or impulse in a continuous system. It is a special type of integral that involves the Dirac Delta Function, which is a mathematical function that is zero everywhere except at a single point where it is infinite.

2. How is a Delta Function Integral represented mathematically?

A Delta Function Integral is represented mathematically as ∫(f(x)δ(x-a))dx, where f(x) is the function being integrated and δ(x-a) is the Dirac Delta Function centered at the point a. This integral evaluates to the value of the function f(x) at the point a.

3. What is the significance of the Dirac Delta Function in Delta Function Integrals?

The Dirac Delta Function is crucial in Delta Function Integrals because it allows us to describe the behavior of a point-like source or impulse in a continuous system. It acts as a mathematical representation of a point or impulse, and its properties allow us to perform calculations and solve problems involving these types of sources.

4. How are Delta Function Integrals used in physics?

Delta Function Integrals are widely used in physics to describe physical phenomena involving point-like sources or impulses. They are used in fields such as electromagnetism, quantum mechanics, and fluid mechanics to model and analyze systems with point sources or impulses.

5. Are there any real-world applications of Delta Function Integrals?

Yes, there are several real-world applications of Delta Function Integrals. They are used in signal processing to analyze signals with sharp changes or spikes, in image processing to sharpen edges and remove noise, and in finance to model and analyze market shocks. They are also used in engineering to analyze systems with sudden changes or impulses, such as in control systems and circuit analysis.

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