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Can anyone tell me how to prove the equation below using Fourier Transform for even function? I tried but I really have no idea.

$$\frac2\pi\int_0^\infty \frac{\sin\pi u \cos xu}u du = \begin{cases} 1,& (|x|\le\pi)\\ 0,& (|x|>\pi) \end{cases}$$

For Fourier Transform of an even function, what I was taught is as below:

If $f(x)$ is an even function the Fourier transform and the Inverse Fourier Transform is:

$F(\omega) = 2\int_0^\infty f(x)\cos \omega x dx$

$f(x) = \frac1\pi\int_0^\infty F(\omega)\cos \omega x d\omega$

Here is my though:

The left side is the Fourier transform of $f(u) = \frac{\sin\pi u}{\pi u}$

And I stuck here.

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  • $\begingroup$ Compute the FT of the function $f = 1, \ \lvert x \rvert \le \pi$ instead and invoke the inversion theorem. $\endgroup$
    – mattos
    Jun 20, 2018 at 8:12
  • $\begingroup$ @Mattos But then $F(\omega) = 2\int_0^\pi \cos\omega t dt = 0$ $\endgroup$
    – nghia95
    Jun 20, 2018 at 10:14
  • $\begingroup$ I'm not even sure how you got the integral above. That isn't $F(\omega)$ though. \begin{align} F(\omega) &= \frac{1}{2 \pi} \int_{\mathbb{R}} f \cdot e^{i \omega x} dx \\ &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} 1 \cdot e^{i \omega x} dx \\ &= \frac{1}{2 \pi} \frac{e^{i \omega x}}{i \omega} \bigg \lvert_{-\pi}^{\pi} \\&= \frac{\sin \pi \omega}{\pi \omega} \end{align} $\endgroup$
    – mattos
    Jun 20, 2018 at 11:09
  • $\begingroup$ To be honest in my university, what you wrote above is the inverse Fourier transform $\endgroup$
    – nghia95
    Jun 20, 2018 at 13:14
  • $\begingroup$ Then remove the $1/ 2\pi$ scaling and/or map $x \to -x$ and then do the computation. $\endgroup$
    – mattos
    Jun 20, 2018 at 13:47

1 Answer 1

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I can't provide the full answer right now, but a couple of hints:

Hint 1: The Fourier transform of the indicator function for the interval $x \in [-1/2, 1/2]$ (i.e. a function that is 1 inside that interval and 0 outside) is $\sin(\omega)/\omega$ (I'm possibly missing a constant factor somewhere, depending on the definition of Fourier transform you adhere to).

Hine 2: You could equally well consider the integral to be an inverse transform.

EDIT:

Full solution: Let $f(x)$ be the indicator function for the interval $[-\pi, \pi]$. According to the definition of Fourier transform you supplied (and sticking to $u$ instead of $\omega$ is the frequency variable), the transform of f(x) is $$F(u) = 2 \int_{0}^{\pi} \cos(ux) dx = 2 \frac{\sin(\pi u)}{u}.$$ Now, the inverse transform of $F(u)$ is, according to what you supplied, $$f(x) = \frac{1}{\pi} \int_{0}^{\infty} F(u) \cos(xu) du = \frac{2}{\pi} \int_{0}^{\infty} \frac{\sin(\pi u)\cos(xu)}{u} du,$$ where I in the last step just substituted $F(u)$ for it's expression. But now you know that this last step is the inverse transform of the transform of $f(x)$, which is $f(x)$, by definition.

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  • $\begingroup$ Thank you for the answer but I couldnt see that the Fourier transform of the function is sin(ω)/ω . Could you make it clearer? $\endgroup$
    – nghia95
    Jun 20, 2018 at 10:23
  • $\begingroup$ Full answer added above. $\endgroup$
    – roding
    Jun 20, 2018 at 11:21

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