# Integrating the Dirac Delta Function

• opticaltempest
In summary, the integral is evaluated by changing the variable to u = 2t - 3 and solving for t to get t = (u + 3)/2. The limits of integration also need to be adjusted accordingly. After substituting and evaluating the integral using the formula int(f(u)*dirac(u))=f(0), the correct answer is -1/2. For the second integral, the variable change is u = -1/3t and the limits need to be reversed. The final answer for this integral is -3.
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?

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.

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.

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

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

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opticaltempest said:
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.

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Thanks

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?

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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.

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?

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

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)

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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.

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)

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.

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.

## 1. What is the Dirac Delta Function?

The Dirac Delta Function, denoted as δ(x), is a mathematical function that is used to describe the distribution of point masses or impulses. It is defined as zero everywhere except at x=0, where it is infinite, and it has an area of 1 under the curve.

## 2. How is the Dirac Delta Function integrated?

The Dirac Delta Function can be integrated by using its defining property, also known as the sifting property. This property states that the integral of the Dirac Delta Function multiplied by any continuous function f(x) is equal to the value of f(x) at x=0. In other words, ∫ δ(x) * f(x) dx = f(0).

## 3. What is the purpose of integrating the Dirac Delta Function?

The Dirac Delta Function is often used in physics and engineering to represent an impulse or concentrated force. By integrating the Dirac Delta Function, we can find the effect of this impulse on a system, such as the displacement or velocity of a particle.

## 4. Can the Dirac Delta Function be integrated with other functions?

Yes, the Dirac Delta Function can be integrated with other functions. However, it is important to note that the integral of the Dirac Delta Function multiplied by a function must be evaluated at x=0, as stated by the sifting property. This means that the function must be continuous at x=0 in order for the integral to be well-defined.

## 5. Are there any other properties of the Dirac Delta Function that are useful when integrating?

Yes, there are other properties of the Dirac Delta Function that can be useful when integrating. These include linearity, translation, and scaling properties, which allow us to manipulate the function and integrate it with other functions more easily.

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