I'm trying to see if the Fourier transform of $e^{jwt}$ exists, so I am trying to evaluate this integral: $\int_{-\infty}^\infty|e^{jwt}|$ but I am not getting anywhere and $|\int_{-\infty}^\infty e^{jwt}|$ says nothing. How do I directly integrate this?

Also, just to confirm, its fourier transform doesn't exist right?

  • $\begingroup$ That integral should contain another factor of e if you are Fourier transforming it as $F(f(t))(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt$ $\endgroup$ – Triatticus Mar 4 '17 at 22:11
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    $\begingroup$ Note that $|e^{i\omega t}| = 1$, so $\int_{-\infty}^{\infty}|e^{i \omega t}|\ dt = \int_{-\infty}^{\infty} 1\ dt = \infty$, hence your function is not ($L^1$) integrable. $\endgroup$ – Bungo Mar 5 '17 at 0:38
  • $\begingroup$ @Bungo You're right I'm getting confused. $\endgroup$ – Goldname Mar 5 '17 at 3:16

The Fourier Transform of $1$ is

$$\mathscr{F}\{1\}(\omega)=\int_{-\infty}^\infty (1)e^{j\omega t}\,dt \tag 1$$

As an improper Riemann integral or as a Lebesgue integral, the integral in $(1)$ does not exist. However, interpreted as a Distribution, the Fourier Transform of $1$ is

$$\mathscr{F}\{1\}(\omega)=2\pi \delta(\omega)$$

where $\delta$ is the Dirac Delta, which is a distribution (or generalized function) and not a function.

  • $\begingroup$ Please let me know how I can improve this answer too. As always, I really want to give you the best answer I can. If the answer was not useful, I am happy to delete it. -Mark $\endgroup$ – Mark Viola Jun 16 at 2:47

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