Integrating the Dirac Delta function

[Bugeye]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I think the answer should be \frac{5}{4} because \frac{2}{3}t-\frac{3}{2}=0 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.

 

[Dick]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I think the answer should be \frac{5}{4} because \frac{2}{3}t-\frac{3}{2}=0 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.

 

[Ray Vickson]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I think the answer should be \frac{5}{4} because \frac{2}{3}t-\frac{3}{2}=0 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.

 

[Bugeye]

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.