# How do I check if $e^{jwt}$ is absolutely integrable?

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?

• That integral should contain another factor of e if you are Fourier transforming it as $F(f(t))(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt$ – Triatticus Mar 4 '17 at 22:11
• 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. – Bungo Mar 5 '17 at 0:38
• @Bungo You're right I'm getting confused. – 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.