Integral involving Dirac Delta generalized function

1. Feb 12, 2015

Legend101

• Member warned about posting with no effort
1. The problem statement, all variables and given/known data
Evaluate the integrals in the attached image

2. Relevant equations

3. The attempt at a solution

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2. Feb 12, 2015

RUber

The dirac delta function has the great property that $\int_{-\infty}^{\infty} \delta(x) dx = 1$.
Couple that with what you know about when it is not equal to zero, and these integrals are quick and easy.
In the future, please put a little more meat into you post, so we can better understand what you already know and what you need help fixing.

3. Feb 12, 2015

Legend101

thanks . I already know this property and many others (including the one in this new attached picture) . If I apply that rule (image) , i would get 1 in the first integral while many others told me 0 . I have problem with the first 2 integrals only .

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4. Feb 12, 2015

RUber

Using the rule you attached, you should get 1 in the first integral. What about the second?

5. Feb 12, 2015

Legend101

The second one would be 0 since -0.5 is not in the interval of integration .
I found a solution to the first integral which totally confused me ( in the attached image) claiming that the answer should be 0 for 1st integral .

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6. Feb 12, 2015

Matterwave

Where did you get that image? That is not how the Dirac Delta is defined...

7. Feb 12, 2015

Legend101

It is a Chegg solution . And it confused me .
I believe that the first integral should be 1 , i just need more proof

8. Feb 12, 2015

Dick

I think the proof is the rule you quoted in post 3. The Chegg solution appears to be for a unit step function, not for a delta function. It's wrong.

9. Feb 12, 2015

Matterwave

Using the fact that $$\int_a^b f(x)\delta(x-c)dx = f(c),~\text{iff}~~a<c<b$$ we can get immediately $$\int_0^5 \cos(2\pi t)\delta(t-2)dt=\left.\cos(2\pi t)\right|_{t=2}=\cos(4\pi)=1$$

The Chegg solution is clearly wrong. The function used there is not a dirac delta function.

10. Feb 13, 2015

stevendaryl

Staff Emeritus
There is a related function, called the "Heaviside step function" or just the "step function":

$x < 0 \Rightarrow \Theta(x) = 0$

$x = 0 \Rightarrow \Theta(x) = 1/2$

$x > 0 \Rightarrow \Theta(x) = 1$

The definition in that image seems correct for the step function, but not the delta function. The two are related by:

$\Theta(x) = \int_{-\infty}^x \delta(s) ds$

11. Feb 13, 2015

RUber

I agree with Dick and Matterwave. They were defining the Heavyside function which is related to the dirac delta by:
$H(x) = \int_{-\infty}^x \delta(t) dt$ or $\frac{d}{dx} H(x) = \delta(x)$

Edit: Got scooped by stevendaryl...Exactly.

12. Feb 13, 2015

Legend101

Thank you guys for your contribution :)
I'll be posting a new topic regarding convolution integral :)

13. Feb 13, 2015

Staff: Mentor

@Legend101, if you start another thread with no effort shown, it will be deleted.

14. Feb 13, 2015

Legend101

No worries . I showed my efforts in attached file

15. Feb 13, 2015

Staff: Mentor

You need to show what you have tried in the first post of the thread. And it would be better to include the work directly in the post, rather than an image of the work.

16. Feb 13, 2015

Legend101

The work involves many integrals and mathematical notations . It would be hard to show all the work directly written on the post especially if someone is using a smartphone .

17. Feb 13, 2015

Staff: Mentor

See my post in that thread.