I have to find a fourier transform of $f(x) = \cos(x)/x$.

This is what I have:

$$ F(\omega) = \int_{-\infty}^\infty \frac{1}{2x}(e^{ix}+e^{-ix})e^{-i\omega x}dx$$ $$ F(\omega) = \frac{1}{2}\int_{-\infty}^{\infty}\frac{e^{i(1-\omega)x}}{x}dx + \frac{1}{2}\int_{-\infty}^{\infty}\frac{e^{i(-1-\omega)x}}{x}dx$$ Using integration in complex plane almost in the same manner as has been used in this example (I use fourier transform with minus sign), I obtain:

$I_1=i\pi$ for $\omega <1$ and $-i\pi$ for $\omega>1$

$I_2=i\pi$ for $\omega <-1$ and $-i\pi$ for $\omega >-1$

where $I_1$ and $I_2$ are first and second integral respectively. Combining these four results gives opposite sings than needed.

Does anyone have a clue what's wrong? I'm pretty sure I've done integration correctly since both integrals are exactly the same as in $f(x) = \frac{\sin(x)}{x}$ and I've got that one right.

  • $\begingroup$ You can write the result in this way $$\frac{1}{2} i \pi (\text{sgn}(\omega-1)+\text{sgn}(\omega+1))$$ $\endgroup$
    – Raffaele
    Aug 31 '17 at 17:55

You are right! but wolfram uses as Fourier transform the following definition: $$ F(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t)\mathrm e^{i\omega t}\mathrm d t $$

  • $\begingroup$ I thought that may be the reason but than I recalled transforming sin(x)/x with my definition gives the same answer as Wolfram does. That's probably because sin(x)/x is an even function and cos(x)/x odd one. $\endgroup$
    – userzg
    Sep 1 '17 at 19:48

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