I am having some trouble with some Fourier transform,

Suppose that $F(\omega)$ is the Fourier transform of $f(x)$, i.e. where

$$F(\omega)=\int_{-\infty}^{\infty}f(x)e^{-i\omega x}\,dx.$$

What is $$\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{F(\omega-a)}{i\omega}e^{i\omega x}\,d\omega ?$$ I was thinking of using the time integration property, since $i\omega$ appears at the denominator, but I am not sure how to start it.


One possible derivation:

Note that $$ \frac{d}{dx} \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{F(\omega-a)}{i\omega}e^{i\omega x}\,d\omega = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{F(\omega-a)}{i\omega}\frac{\partial}{\partial x}e^{i\omega x}\,d\omega\\ = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega-a)e^{i\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega')e^{i(\omega'+a) x}d\omega'\\ = e^{iax} \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega')e^{i\omega' x}d\omega' = e^{-iax}f(x) $$ So we must have $$ \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{F(\omega-a)}{i\omega}e^{i\omega x}\,d\omega = \int_{-\infty}^x e^{-iat}f(t)\,dt + C $$

  • $\begingroup$ Hi, thanks for your answer. But how did u derive your step? $\endgroup$ – freak_warrior Mar 6 '14 at 5:39
  • $\begingroup$ So what I'm really saying with all this is that you can find the answer by applying the frequency-shift property, and then applying the time integration property to the result. Does that answer your question? I'm not sure what step you're talking about, since there are several. $\endgroup$ – Omnomnomnom Mar 6 '14 at 5:48
  • $\begingroup$ Sorry, I meant your last step, where u have x in the limits of the integral. $\endgroup$ – freak_warrior Mar 6 '14 at 5:50
  • $\begingroup$ So, I should have an integration constant at the end there. I thought we should have $C = 0$ at first, but now I'm not so sure. $\endgroup$ – Omnomnomnom Mar 6 '14 at 5:54
  • $\begingroup$ We'd have $$ C = \lim_{x \to -\infty} \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{F(\omega-a)}{i\omega}e^{i\omega x}\,d\omega $$ which I suppose you could argue is zero for a nice enough function? Really not sure though. $\endgroup$ – Omnomnomnom Mar 6 '14 at 5:54

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