# Integral resulting in delta function.

• G01
In summary: I don't know if this is a limitation of the integral, or if this is a flaw in my understanding of the limits.In summary, the homework equation is equal to the heavy side step function multiplied by the delta function, but the limits on the equation are unknown.
G01
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## Homework Statement

Hi All. I am given this integral:

$$\int_{-\infty}^{\infty}A\Theta e^{i\omega t}dt$$

I need to show that it's equal to the following:

$$=A(\pi \delta(\omega)+\frac{i}{\omega})$$

## Homework Equations

Theta is the Heavyside step function.

## The Attempt at a Solution

The step function changes the lower bound of the integral to 0:

$$\int_{0}^{\infty}Ae^{i\omega t}dt$$

Elementary integration then gives:

$$\lim_{t=\infty}\frac{e^{i\omega t}}{i\omega} +\frac{i}{\omega}$$

I'm not sure how to relate that first limit to any of the delta function definitions I know. A hint was given to use the following definition:

$$\pi\delta(x)=\lim_{A=0}\frac{A}{x^2+A^2}$$

However, I'm not sure how to apply this limit to help get to the result.

Thanks for any hints/help!

The integral won't converge because the integrand doesn't vanish as t goes to infinity, so throw in a convergence factor e-λt and take the limit as λ→0+:

$$\lim_{\lambda \to 0^+} \int_0^\infty Ae^{(i\omega-\lambda)t}\,dt$$

You can transform the result into a form where you'll be able to apply the hint you were given.

(Side point: there are convergent integrals whose integrand fails to vanish at infinity)

Since the integral is apparently meant to be an integral of operators in $\omega$ rather than an integral of functions, one approach you might want to take is to show that it operates like the thing you're trying to prove it equal to. That is, compute

$$\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} A \Theta(\omega) e^{i \omega t} f(\omega) \, d\omega \, dt$$

I don't see how to use the hint if you tried this approach, though.

Another thing I'd be tempted to try is to invoke complex analysis; integrate around wedge-shaped regions whose corner is at the origin and whose edges are the positive real and imaginary axes, and take the limit as radius goes to infinity. I don't know if distributions and complex residues interact well, though.

Vela- Thanks! I was able to get to the result using your method.

Hurkyl- I thought about taking the complex analysis route as well. However, like you said the potential issues with residues and distribution scared me away. I didn't think it was worth the trouble, even though it may end up being more rigorous than vela's approach.

I think vela's approach will suffice for this class, however I am a little worried by some of the limits that appear. For instance, in order to get the correct result, we must have:

$$\lim_{\lambda=0^+}\lim_{t=\infty} (i\omega e^{(-\lambda + i\omega )t})=0$$

I am by no means certain that it is the case. i.e The exponential term becomes small as lambda goes to 0, but then grows again as you take t to infinity.

## 1. What is the definition of the delta function?

The delta function, denoted by δ(x), is a mathematical function that is zero everywhere except at x = 0, where it is infinitely large. It is often used in physics and engineering to describe the behavior of point-like objects or impulse signals.

## 2. How is the delta function related to the integral?

The delta function is often used in integrals to represent a point-like source or a spike in a signal. The integral of the delta function over any interval that includes x = 0 is equal to 1, and it is 0 for all other intervals. This property is known as the sifting property of the delta function.

## 3. Can the delta function be integrated with other functions?

Yes, the delta function can be integrated with other functions. In fact, the delta function is often used to simplify integrals involving complicated functions by "picking out" a specific value of the function at a point.

## 4. What is the significance of the delta function in solving differential equations?

The delta function is useful in solving differential equations because it can be used to represent impulse or sudden changes in a system. For example, in the study of electrical circuits, the delta function can be used to represent an instantaneous change in voltage or current.

## 5. How is the delta function used in Fourier analysis?

The delta function is used in Fourier analysis to represent a single frequency or a point-like signal in the frequency domain. It is often used to represent impulses or spikes in a signal, and it allows for the analysis of signals with discontinuities or sudden changes. The Fourier transform of the delta function is equal to 1, making it an important component in the analysis of signals and systems.

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