Integrating the Dirac Delta Function

[opticaltempest]

I am trying to evaluate the following integral.

\int_{-\infty}^{\infty}{\delta(2t-3)\sin(\pi t) dt}

where delta represents the Dirac delta function.


I am told that the answer is -1. However, when I evaluate it in MATLAB and Maple 11, I get an answer of -1/2. What is the correct way to evaluate this integral by hand? Which answer is correct?

 

[CompuChip]

The delta function just picks out the value in which the argument of the delta is zero.
So the integral will just give
sin(\pi t_0)
where t_0 is the solution to 2t - 3 = 0.

 

[opticaltempest]

When I evaluate this integral in MATLAB I type:

syms t;
int(dirac(2*t-3)*sin(pi*t),-inf,inf)

and it returns -1/2.

 

[opticaltempest]

Wouldn't it be -1/2 by using equation (5) on

http://mathworld.wolfram.com/DeltaFunction.html"

 

[Dick]

Wouldn't it be -1/2 by using equation (5) on

http://mathworld.wolfram.com/DeltaFunction.html"

Yes, but you don't necessarily have to trust that formula. Change variables to u=2t-3. du=2dt (there's the missing 2). So the integral becomes (1/2)*delta(u)*sin(pi(u+3/2))*du. Now you can just put u=0.

 

[opticaltempest]

Thanks

 

[opticaltempest]

What would happen if I had

\int_{-\infty}^{\infty}{\delta \bigg(-\frac{1}{3} t \bigg) dt} ?

Using change of variables we could have

u=-\frac{1}{3}t

Then

du=-\frac{1}{3}dt \implies dt=-3du

Our original integral is then equivalent to

\int_{-\infty}^{\infty}{\delta (u) \cdot -3 du}= -3

This answer doesn't agree with equation (5) listed on
http://mathworld.wolfram.com/DeltaFunction.html"

According to (5) I should get 3 not -3. What is wrong?

 

[Dick]

For that change of variables you have to reverse the limits as well. If t goes -infinity -> +infinity, u goes +infinity -> -infinity. Gives you an extra sign.

 

[opticaltempest]

I recall changing limits when doing u-substitutions in previous calculus classes, but how do I know when to change the limits when dealing with infinite limits? Why do we need to 'flip' the limits on this integral with this change of variables?

 

[Dick]

I just told you. If u=(-1/3)*t, then if t~+infinity then u~-infinity. So the upper limit becomes -infinity instead of +infinity. \int_{\infty}^{-\infty}{\delta (u) \cdot -3 du}= -\int_{-\infty}^{\infty}{\delta (u) \cdot -3 du}=3.

 

[ryukyu]

When you do the change of variable in the OP's first problem, do the limits of integration change. If have a similar problem where the delta is 4t-3 instead of 2t-3. So I know

u=4t-3
du-4dt
t=(u+3)/4
dt=(1/4) * du

so I have 1/4 * int(sin(pi*(u+3)/4)*dirac(u)du

I have no idea how to move forward if the limits are +/- inf

I know my final result is supposed to be (1/8)sqrt(2)

 

[Dick]

The direction of the infinity limits doesn't change if u=4t-3. It looks to me like your main problem is that t=(u+3)/4 NOT t=(u-3)/4.Try solving for t again.

 

[ryukyu]

I didn't think they limits changed. I just needed to be sure. I did fix the t in my original post. I didn't pay close enough attention to what I actually had written correctly on my homework.

I guess where I get confused is in placing u = 0. I know the goal was to find a t that would make the dirac argument equal 0 and that this is what we actually did in the substitution.

If we do this though doesn't dirac(0) = inf ?

making our integral (1/4)*int(sin(3pi/4)*inf du , -inf, inf)

 

[Dick]

No, no. dirac(0) isn't really even inf, it's just plain not defined. The rule to remember is that integral(f(u)*dirac(u))=f(0) if zero is inside the limits of integration and f is continuous. Notice there is no 'integral' on the right side. Reread any material you have on the dirac delta function.

 

[ryukyu]

Great, I've got it now. Thanks!

I did know that int(f(x)dirac(a)) = f(a) if a is in the limits. I don't know why I have trouble spotting the simple stuff like this. Our textbook is the bane of signals and systems (isbn: 0073309508) so it's no much help. Between the textbook and the instructor we have very few examples of how to do things.