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

[lys04]

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

 

[anuttarasammyak]

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}

 

[Orodruin]

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 $k = 1$.

 

[lys04]

This transformation seems incorrect because e−ikx is a complex number

Ohh alright yeah that makes sense. Ty!

 

[lys04]

Indeed, but more importantly it should also be realised that the primitive function is incorrect for k=1.

What does this mean?

 

[Orodruin]

What does this mean?

That your integration is wrong when k = 1.

 

[lys04]

That your integration is wrong when k = 1.

It's gonna be wrong for all k s.t e_-ikx has a real part?

 

[Orodruin]

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.

Indeed, but more importantly it should also be realised that the primitive function is incorrect for $k = 1$.

(my emphasis)

 

[Orodruin]

Oh, and on top of that, the last line is not equal to zero either. The sine terms have the same sign.