# Integrating the Dirac Delta function

• Bugeye
In summary, the conversation discusses the process of integrating the function (t-1)δ[(2/3)t-(3/2)] and the discrepancy between the expected answer of 5/4 and the actual answer of 15/8 when using Mathematica. The solution involves a change of variables to u=(2/3)t-3/2.
Bugeye

## Homework Statement

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

## 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.

Bugeye said:

## Homework Statement

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

## 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.

Bugeye said:

## Homework Statement

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

## 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.

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.

## 1. What is the Dirac Delta function?

The Dirac Delta function, denoted by δ(x), is a mathematical function that is defined as zero everywhere except at x = 0, where it is infinite. It is used to model point-like sources of energy, such as a point charge or a point mass.

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

The Dirac Delta function is integrated using a special technique called the sifting property. This property states that when the Dirac Delta function is multiplied by any continuous function f(x), the integral of the product is equal to the value of f(x) at x = 0. In other words, ∫f(x)δ(x)dx = f(0).

## 3. What are the applications of the Dirac Delta function?

The Dirac Delta function has many applications in physics and engineering. It is used to represent point sources of force or energy, such as in the analysis of electrical circuits or the behavior of particles in quantum mechanics. It is also used in signal processing to model impulses or sharp changes in a signal.

## 4. Can the Dirac Delta function be generalized to higher dimensions?

Yes, the Dirac Delta function can be extended to higher dimensions. In three-dimensional space, it is denoted by δ(x, y, z) and is defined as zero everywhere except at the origin (x = y = z = 0), where it is infinite. It can also be generalized to n-dimensional space, where it is denoted by δ(x1, x2, ..., xn) and is defined as zero everywhere except at the origin (x1 = x2 = ... = xn = 0), where it is infinite.

## 5. What is the relationship between the Dirac Delta function and the Kronecker Delta function?

The Dirac Delta function and the Kronecker Delta function are closely related, but they are not the same. The Kronecker Delta function, denoted by δij, is defined as 1 if i = j and 0 otherwise. It is used in discrete mathematics to represent the identity matrix. The Dirac Delta function is the continuous analog of the Kronecker Delta function, representing a point source in a continuous space. In some cases, the Kronecker Delta function can be thought of as the discrete version of the Dirac Delta function.

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