Evaluating integral of delta function

In summary, the conversation is discussing how to evaluate a definite integral using a delta function. The suggested method is to use a dummy variable and substitute it back in after integration. The final goal is to simplify the function to have the form delta(t)*f(t)*dt, where the integral of this equals f(0) and in this case, the function should evaluate to e^3.
  • #1
elimenohpee
67
0

Homework Statement


Evaluate the integral:
gif.latex?\int_{\infty%20}^{\infty%20}\delta%20(t+3)e^{-t}dt.gif


Homework Equations


To integrate this, should one use a dummy variable to get the delta function only of t, then integrate, then substitute back in after integration?
 

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  • gif.latex?\int_{\infty%20}^{\infty%20}\delta%20(t+3)e^{-t}dt.gif
    gif.latex?\int_{\infty%20}^{\infty%20}\delta%20(t+3)e^{-t}dt.gif
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  • #2
elimenohpee said:

Homework Statement


Evaluate the integral:
gif.latex?\int_{\infty%20}^{\infty%20}\delta%20(t+3)e^{-t}dt.gif


Homework Equations


To integrate this, should one use a dummy variable to get the delta function only of t, then integrate,

The delta will be a function of the dummy variable, not t. But yes, that is what you want to do.

then substitute back in after integration?

It's a definite integral. Why would there be any need to substitute back after integration?
 
  • #3
Substitute u=t+3 and use the definition of the delta function.
 
  • #4
Ok I just want to make sure I'm thinking correctly and logically. If I let t1 = t + 3, then I need to change all the 't' in the function to t1. I know that the Delta function is defined as
gif.latex?\int_{-\infty%20}^{\infty}%20\delta(t)dt%20=%201.gif


So if I can get it into this form with t1:

gif.latex?\int_{-\infty%20}^{\infty}%20\delta(t_{1})dt_{1}%20=%201.gif


then the delta function should simplify as 1 correct?

The function would look like this:

gif.latex?\int_{-\infty%20}^{\infty}%20\delta(t_{1})(e^{-t_{1}+3})dt_{1}.gif


then simplify to this:

gif.latex?\int_{-\infty%20}^{\infty}%20(e^{-t_{1}+3})dt_{1}.gif


then separate the exponential, pull the exponential not containing t1 out of the integral, integrate the exponential containing t1, and should be left with just this:

gif.latex?e^3.gif


Does this look right? Or am I not thinking about this right.
 
Last edited by a moderator:
  • #5
The definition of the delta function is not that integral of delta(t)*dt=1. Many functions have that property. It's that the integral of f(t)*delta(t)*dt=f(0), if f is a continuous function.
 
Last edited:
  • #6
Oh ok I see what you mean. So if I have
gif.latex?\int_{-\infty%20}^{\infty}%20\delta(t_{1})(e^{-t_{1}+3})dt_{1}.gif

it should still evaluate to e^3 right?
 
  • #7
Yes. e^3. Sorry. I didn't read the post through to the end. But, once you have it in the form delta(t)*f(t)*dt you are done.
 
  • #8
Oh ok excellent! Thanks for the help, this has helped me 'de-mystify' the delta function :)
 

Related to Evaluating integral of delta function

1. What is the delta function and why is it used in integrals?

The delta function, also known as the Dirac delta function, is a mathematical function that is defined as 0 for all values except at the origin, where it is infinitely large. It is used in integrals because it acts as a spike that concentrates the integral at a single point, making it easier to evaluate.

2. How do you evaluate the integral of a delta function?

To evaluate the integral of a delta function, you need to determine the limits of integration and then use the properties of the delta function to simplify the integral. This usually involves rewriting the integral in terms of the delta function's argument and using the sifting property to eliminate other terms.

3. Can the delta function be integrated over a finite interval?

No, the delta function is only defined at a single point, so it cannot be integrated over a finite interval. However, you can use the properties of the delta function to evaluate an integral over a finite interval by dividing the interval into smaller intervals and integrating over each one.

4. What is the relationship between the delta function and the unit step function?

The unit step function, also known as the Heaviside function, is defined as 0 for negative values and 1 for positive values. It is often used in conjunction with the delta function because its derivative is equal to the delta function. This means that the integral of the delta function can be expressed as the difference between the unit step function evaluated at the upper and lower limits of integration.

5. Are there any practical applications of evaluating integrals of delta functions?

Yes, the delta function has many practical applications in fields such as engineering, physics, and signal processing. It is used to model impulsive forces, point sources, and impulse responses in systems. It is also used in the theory of distributions, which has applications in fields such as control theory and partial differential equations.

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