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How does one find the Fourier transform of $f(x):=\mathbf 1_{[0,2\pi]}(x)\sin(x)$?

I have tried to use the definition from my text:

\begin{align*} \hat f(\xi) & = \frac{1}{\sqrt{2\pi}} \int_0^{2\pi}\sin(x)e^{-ix\xi} \ dx \\ & = \frac{1}{\sqrt{2\pi}} \int_0^{2\pi}\frac{1}{2i}\left(e^{ix}-e^{-ix}\right)e^{-ix\xi} \ dx \end{align*}

It doesn't seem to lead anywhere. (Anywhere in the sense that, since we are decomposing sine, I was expecting the transform to be something simple). Is this the right way to go?

Wolfram alpha returns:

$$i \sqrt\frac{\pi}{2}\delta(\omega-1)-i \sqrt\frac{\pi}{2}\delta(\omega+1)$$

If this is relevant to the problem: how does Dirac delta function come into play?

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  • $\begingroup$ In the title of the question and in the WolframAlpha input, you ask for the Fourier transform of $\sin(x)$. But in the body of the question and in the integral you set up, you ask for the Fourier transform of $\mathbf 1_{[0,2\pi]}(x)\sin(x)$. These are different functions, and they have different Fourier transforms. Which one are you interested in? $\endgroup$ Commented Apr 2, 2014 at 22:59
  • $\begingroup$ @ChrisCulter: I'm interested in $\mathbf 1_{[0,2\pi]}(x)\sin(x)$. So would this be the correct answer? $\endgroup$
    – Leo
    Commented Apr 2, 2014 at 23:07
  • $\begingroup$ Probably! The Fourier transform of a rectangle is a sinc, and the Fourier transform of a sine is a pair of Dirac deltas. So the Fourier transform of a rectangle times a sine is a sinc convolved with a pair of deltas, which equals a pair of sincs. From there, it's a matter of getting all the constants right. $\endgroup$ Commented Apr 2, 2014 at 23:21
  • $\begingroup$ @ChrisCulter: Could you take a look at my comment to the answer, I don't quite see how to work it out her hand. (And WA seems to give something entirely different.) $\endgroup$
    – Leo
    Commented Apr 2, 2014 at 23:32

1 Answer 1

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$$ \begin{align*} \hat f(\xi) & = \frac{1}{\sqrt{2\pi}} \int_0^{2\pi}\frac{1}{2i}\left(e^{ix}-e^{-ix}\right)e^{-ix\xi} \ dx \\ &=\frac{1}{\sqrt{2\pi}} \frac{1}{2i}\int_0^{2\pi}e^{ix-ix\xi}-e^{-ix-ix\xi} \ dx \\ &=\frac{1}{\sqrt{2\pi}} \frac{1}{2i}\int_0^{2\pi}e^{(i-i\xi)x}-e^{(-i-i\xi)x} \ dx \\ \end{align*} $$

Now simply use

$$ \int_0^{2\pi}e^{\alpha x}= \frac{e^{\alpha x}}{\alpha}|_0^{2 \pi}=\frac{e^{2 \pi \alpha}-1}{\alpha}$$

The dirac function comes from the fact that you asked WA to calculate the FT of $\sin(x)$ over the reals not restricted to $[0, 2 \pi]$.

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  • $\begingroup$ I did this and ended up with $$\frac{1}{\sqrt{2\pi}} \frac{1}{2i}\left(\frac{e^{2\pi(i-i\xi)}-1}{i-i\xi}-\frac{e^{2\pi(-i-i\xi)}-1}{-i-i\xi}\right)$$ and didn't know what to do next. WA can't simplify it either. (That's basically, why I asked the question, sorry for not being clear enough.) $\endgroup$
    – Leo
    Commented Apr 2, 2014 at 23:25
  • $\begingroup$ I'm not sure if it'll get much simpler! Each of the two terms could be written as a product of a complex exponential and a sinc function, if you prefer. $\endgroup$ Commented Apr 2, 2014 at 23:53
  • $\begingroup$ The only other possible simplification is bringing the two fractions to common denominator. The expression looks a tad simpler... $\endgroup$
    – N. S.
    Commented Apr 3, 2014 at 0:14

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