Integrating the Dirac Delta function

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Bugeye
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Homework Statement


I am trying to integrate the function
[itex]\int _{-\infty }^{\infty }(t-1)\delta\left[\frac{2}{3}t-\frac{3}{2}\right]dt[/itex]

Homework Equations



The Attempt at a Solution


I think the answer should be [itex]\frac{5}{4}[/itex] because [itex]\frac{2}{3}t-\frac{3}{2}=0[/itex] when t=9/4. then (9/4-1) = 5/4. However, when I put the equation into Mathematica, it gives me an anser of 15/8.
 
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Bugeye said:

Homework Statement


I am trying to integrate the function
[itex]\int _{-\infty }^{\infty }(t-1)\delta\left[\frac{2}{3}t-\frac{3}{2}\right]dt[/itex]

Homework Equations



The Attempt at a Solution


I think the answer should be [itex]\frac{5}{4}[/itex] because [itex]\frac{2}{3}t-\frac{3}{2}=0[/itex] when t=9/4. then (9/4-1) = 5/4. However, when I put the equation into Mathematica, it gives me an anser of 15/8.

There's more to it than that. Do a change of variables u=2t/3-3/2. Don't forget du isn't the same as dt.
 
Bugeye said:

Homework Statement


I am trying to integrate the function
[itex]\int _{-\infty }^{\infty }(t-1)\delta\left[\frac{2}{3}t-\frac{3}{2}\right]dt[/itex]

Homework Equations



The Attempt at a Solution


I think the answer should be [itex]\frac{5}{4}[/itex] because [itex]\frac{2}{3}t-\frac{3}{2}=0[/itex] when t=9/4. then (9/4-1) = 5/4. However, when I put the equation into Mathematica, it gives me an anser of 15/8.

You need to be really, really careful when dealing with things like δ(f(t)). In your case, just change variables to x = (2/3)t.
 
Dick said:
There's more to it than that. Do a change of variables u=2t/3-3/2. Don't forget du isn't the same as dt.

Great, I got it, thanks a lot.