What is the Fourier Transform of sin?

asi123
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Homework Statement



Hey guys.
I need to find the Fourier transform of sin, is this right?

http://img156.imageshack.us/img156/5531/scan0004r.jpg

I searched the internet but all I could find is the answer with the dirac delta and I don't need that.

Thanks.


Homework Equations





The Attempt at a Solution

 
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The Fourier transform of sin(t) involves the Dirac delta function. What do you mean by "I don't need that"? And why did you change the limits from -∞ to ∞ to -π to π in your integral?
 
dx said:
The Fourier transform of sin(t) involves the Dirac delta function. What do you mean by "I don't need that"? And why did you change the limits from -∞ to ∞ to -π to π in your integral?

Oh, sorry, I need to find it from -pi to pi.
Is there something wrong with what I did?

Thanks.
 
I didn't read your whole solution, but there is a mistake in your first step. The Fourier transform integral goes from -∞ to ∞. Why did you change those limits?
 
dx said:
I didn't read your whole solution, but there is a mistake in your first step. The Fourier transform integral goes from -∞ to ∞. Why did you change those limits?

Yeah, I need to find it from -pi to pi.
Is that way it doesn't involves Dirac function?

Thanks.
 
No! It's not from -pi to pi. It's -∞ to ∞.
 
dx said:
No! It's not from -pi to pi. It's -∞ to ∞.

:smile:

But that is the question.
Find Fourier transform of sin in -pi<t<pi.

What do you mean?

Thanks.
 
Your question is to transform the function f(t) = \left\{ \begin{matrix} \sin t &amp; \mathrm{if}\; -\pi &lt; t &lt; \pi \\ 0 &amp; \mathrm{otherwise} \end{matrix} \right ?
 
Cyosis said:
Your question is to transform the function f(t) = \left\{ \begin{matrix} \sin t &amp; \mathrm{if} -\pi &lt; t &lt; \pi \\ 0 &amp; \mathrm{otherwise} \end{matrix} \right ?

Yeah, sorry for the misconfusion.
 
  • #10
Then your approach is correct since the function is zero outside -pi<t<pi anyway so you may as well integrate from -pi to pi.
 
  • #11
Ah, now it makes sense! Thanks Cyosis!
 
  • #12
You're welcome.
 
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