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Calculus and Beyond Homework Help News on Phys.org

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$$\int_0^\infty \frac{\sin ax\cos \omega x}{x}\,dx = \frac{\pi}{2}.$$ 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##.

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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?

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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

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