- #1

- 35

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter jennyjones
- Start date

- #1

- 35

- 0

- #2

BruceW

Homework Helper

- 3,611

- 121

- #3

BruceW

Homework Helper

- 3,611

- 121

- #4

vela

Staff Emeritus

Science Advisor

Homework Helper

Education Advisor

- 15,439

- 2,063

$$\int_0^\infty \frac{\sin ax\cos \omega x}{x}\,dx = \frac{\pi}{2}.$$ When ##\lvert\omega\rvert > a##, you have

$$\int_0^\infty \frac{\sin ax\cos \omega x}{x}\,dx = 0.$$ Somehow, you have to figure out what happens when ##\lvert\omega\rvert = a##.

- #5

- 35

- 0

- #6

vela

Staff Emeritus

Science Advisor

Homework Helper

Education Advisor

- 15,439

- 2,063

- #7

BruceW

Homework Helper

- 3,611

- 121

Ah whoops, ignore this. I thought you wrote an ##a## for the upper limit, but it is an ##\infty##, as it should be.

right. yes, you are using the same definition of the Fourier transform as they are. Which is good :)jennyjones said:bruce W i'm using the cosine transform, i made a picture of this formula for my textbook.

I think you meant to say the average of the values of the Fourier transform on either side of the point ##\omega=a##. If this is what you meant, then yes that's right. Was it a guess? You have good intuition if it was. Yeah, there is a specific theorem (which is pretty hard to find on the internet), as vela is hinting at. This theorem works for certain kinds of function, like the rectangular function.jennyjones said:Vela, do you know if i can than say |ω|= (∏/2+0)/2=∏/4 ?

the average?

- #8

- 35

- 0

Yey! thank you, than i solve the problem now!

Do you maybe know the name of this theorem?

jenny

Do you maybe know the name of this theorem?

jenny

- #9

BruceW

Homework Helper

- 3,611

- 121

Share: