Interval with Dirac function in a finite interval

In summary, the conversation was about calculating an integral using the Dirac delta function and the Heaviside step function. The speaker had already set up the integral and determined the result to be 0. They were then wondering how to continue, but after a suggestion from the other person, they realized they could use the definition of the Dirac delta function to simplify the integral further.
  • #1
evinda
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Hello! (Wave)

I want to calculate the integral $\int_{-1}^2\sin \left (\pi (t-1)\right )\delta (-t+1)\, dt$. I have done the following so far:

$$\int_{-\infty}^{+\infty}\sin \left (\pi (t-1)\right )\delta (-t+1)\, dt=\int_{-\infty}^1\sin \left (\pi (t-1)\right )\delta (-t+1)\, dt+\int_{-1}^2\sin \left (\pi (t-1)\right )\delta (-t+1)\, dt+\int_2^{+\infty}\sin \left (\pi (t-1)\right )\delta (-t+1)\, dt$$

We have that $\int_{-\infty}^{+\infty}\sin \left (\pi (t-1)\right )\delta (-t+1)\, dt=\int_{-\infty}^{+\infty}\sin \left (\pi (t-1)\right )\delta (t-1)\, dt=\sin (-\pi)=0$.

How can we continue? (Thinking) Do we use somehow the Heaviside step function? (Thinking)
 
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  • #2
evinda said:
Hello! (Wave)

I want to calculate the integral $\int_{-1}^2\sin \left (\pi (t-1)\right )\delta (-t+1)\, dt$. I have done the following so far:

$$\int_{-\infty}^{+\infty}\sin \left (\pi (t-1)\right )\delta (-t+1)\, dt=\int_{-\infty}^1\sin \left (\pi (t-1)\right )\delta (-t+1)\, dt+\int_{-1}^2\sin \left (\pi (t-1)\right )\delta (-t+1)\, dt+\int_2^{+\infty}\sin \left (\pi (t-1)\right )\delta (-t+1)\, dt$$

We have that $\int_{-\infty}^{+\infty}\sin \left (\pi (t-1)\right )\delta (-t+1)\, dt=\int_{-\infty}^{+\infty}\sin \left (\pi (t-1)\right )\delta (t-1)\, dt=\sin (-\pi)=0$.

How can we continue? (Thinking) Do we use somehow the Heaviside step function? (Thinking)
We have, by definition
\(\displaystyle \int_a ^b f(x) \delta (x - c) ~ dx = f(c)\) if \(\displaystyle c \in [a, b]\) and 0 else.

In your case \(\displaystyle t = 1 \in [-1, 2]\), so \(\displaystyle \int_{-1}^2 sin( \pi (t - 1) ) ~ \delta(-1 + t) ~ dx = sin( \pi (1 - 1)) = 0\)
as you said. I don't understand what there is to continue?

-Dan
 
  • #3
topsquark said:
We have, by definition
\(\displaystyle \int_a ^b f(x) \delta (x - c) ~ dx = f(c)\) if \(\displaystyle c \in [a, b]\) and 0 else.

In your case \(\displaystyle t = 1 \in [-1, 2]\), so \(\displaystyle \int_{-1}^2 sin( \pi (t - 1) ) ~ \delta(-1 + t) ~ dx = sin( \pi (1 - 1)) = 0\)
as you said. I don't understand what there is to continue?

-Dan

I didn't have in mind this definition. Thanks a lot! (Smile)
 

FAQ: Interval with Dirac function in a finite interval

1. What is an interval with Dirac function in a finite interval?

An interval with Dirac function in a finite interval refers to a mathematical concept where a Dirac delta function is defined within a specific interval. The Dirac delta function is a generalized function that is used to represent a point mass or impulse at a specific location within the interval.

2. How is the Dirac function defined in an interval?

The Dirac function is defined as a function that is zero everywhere except at a single point, where it is infinite. In an interval, the Dirac function is defined as a spike or impulse at a specific point within the interval, with a magnitude equal to the inverse of the length of the interval.

3. What is the significance of the Dirac function in an interval?

The Dirac function in an interval is significant because it allows for the representation of point masses or impulses within a continuous function. This is useful in many areas of science and engineering, such as signal processing, physics, and mathematics.

4. How is the Dirac function used in practical applications?

The Dirac function is used in practical applications to model and analyze systems that involve point masses or impulses. For example, in signal processing, the Dirac function can be used to represent sudden changes in a signal, and in physics, it can be used to represent the force exerted by a point mass on a system.

5. What are some limitations of using the Dirac function in an interval?

One limitation of using the Dirac function in an interval is that it is not a true function, as it is not defined at the point where it is infinite. This can lead to difficulties in integration and other mathematical operations. Additionally, the Dirac function can only represent point masses or impulses at a single point within the interval, which may not accurately represent more complex systems.

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