MHB Interval with Dirac function in a finite interval

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The integral $\int_{-1}^2\sin \left (\pi (t-1)\right )\delta (-t+1)\, dt$ evaluates to zero because the Dirac delta function $\delta(-t+1)$ is non-zero only at $t=1$, which lies within the integration limits. By the properties of the delta function, the integral simplifies to $\sin(\pi(1-1))$, resulting in $\sin(0) = 0. The discussion highlights the use of the delta function and confirms that no further steps are necessary since the integral has already been resolved. The participants clarify the application of the delta function's definition in this context.
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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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
[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
 
topsquark said:
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)
 
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