What is the Fourier transform of sin(x) with non-zero terms?

AI Thread Summary
The discussion centers on the Fourier transform of sin(x) and highlights that the transformation appears incorrect due to the complex nature of e^{-ikx}. Substituting sin(x) with its exponential form reveals issues, particularly when k equals 1, indicating that the integration is flawed. Participants emphasize that the primitive function is incorrect for k=1 and that this error extends to other values of k where e^{-ikx} has a real part. Additionally, it is noted that the final line of the transformation is not equal to zero, as the sine terms share the same sign. Overall, the conversation underscores the need for careful analysis in the Fourier transform process.
lys04
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Homework Statement
Why is what I'm doing here not correct? (Because last line seems to give me 0) Could someone point me in the right direction please.
Relevant Equations
Fourier transform
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This transformation seems incorrect because ##e^{-ikx}## is a complex number. How about checking it by substitution of \sin x = \frac{e^{ix}-e^{-ix}}{2i}
 
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anuttarasammyak said:
View attachment 352508

This transformation seems incorrect because ##e^{-ikx}## is a complex number. How about checking it by substitution of \sin x = \frac{e^{ix}-e^{-ix}}{2i}
Indeed, but more importantly it should also be realised that the primitive function is incorrect for ##k = 1##.
 
anuttarasammyak said:
This transformation seems incorrect because e−ikx is a complex number
Ohh alright yeah that makes sense. Ty!
 
Orodruin said:
Indeed, but more importantly it should also be realised that the primitive function is incorrect for k=1.
What does this mean?
 
lys04 said:
What does this mean?
That your integration is wrong when k = 1.
 
Orodruin said:
That your integration is wrong when k = 1.
It's gonna be wrong for all k s.t e_-ikx has a real part?
 
lys04 said:
It's gonna be wrong for all k s.t e_-ikx has a real part?
You are misunderstanding my comment. I said that apart from the fact pointed out by @anuttarasammyak in post #2, the integral is also wrong for ##k = 1##. This will be crucial to correct also when you have corrected the imaginary part error.
Orodruin said:
Indeed, but more importantly it should also be realised that the primitive function is incorrect for ##k = 1##.
(my emphasis)
 
Oh, and on top of that, the last line is not equal to zero either. The sine terms have the same sign.
 
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