1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.