Interval with Dirac function in a finite interval

[evinda]

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)

 

[topsquark]

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
[math]\int_a ^b f(x) \delta (x - c) ~ dx = f(c)[/math] if [math]c \in [a, b][/math] and 0 else.

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

-Dan

 

[evinda]

We have, by definition
[math]\int_a ^b f(x) \delta (x - c) ~ dx = f(c)[/math] if [math]c \in [a, b][/math] and 0 else.

In your case [math]t = 1 \in [-1, 2][/math], so [math]\int_{-1}^2 sin( \pi (t - 1) ) ~ \delta(-1 + t) ~ dx = sin( \pi (1 - 1)) = 0[/math]
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)