Evaluating integral of delta function

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Homework Help Overview

The discussion revolves around evaluating an integral involving the delta function, specifically focusing on the integration process and the properties of the delta function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of a dummy variable to isolate the delta function and the implications of substituting back after integration. There are questions about the necessity of substitution in definite integrals and the correct interpretation of the delta function's properties.

Discussion Status

Some participants have offered guidance on the integration process and the definition of the delta function. There is an exploration of different interpretations regarding the integration steps and the simplification of the delta function.

Contextual Notes

Participants are considering the implications of changing variables in the context of the delta function and its integral properties. There is an emphasis on ensuring logical consistency in the approach taken.

elimenohpee
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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?
 

Attachments

  • 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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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?
 
Substitute u=t+3 and use the definition of the delta function.
 
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:
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:
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?
 
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.
 
Oh ok excellent! Thanks for the help, this has helped me 'de-mystify' the delta function :)
 

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