Fourier transform of a complex exponential

Question

If we want to have the Fourier transform of a complex exponential $x(t) = e^{i\omega_0t}$ we could "guess" that it's $X(\omega)=2\pi\delta(\omega-\omega_0)$ and prove the equality: $$ x(t) = \frac{1}{2\pi}\int_{-\infty}^\infty 2\pi\delta(\omega-\omega_0)e^{i\omega t}d\omega = \int_{-\infty}^\infty \delta(\omega-\omega_0)e^{i\omega t}d\omega = e^{i\omega t} |_{\omega=\omega_0} = e^{i\omega_0t} $$

QUESTIONS:

1) What is the logic behind the above mentioned "guess"?

2) What is the correct way to get the Fourier transform of a complex exponential without "guessing"?

Thank you for your help.

Answer 1

Just apply the Fourier transformation to $e^{i\omega_0t}$:

$$\begin{align*} X(\omega) &= \int_{-\infty}^{\infty} e^{i\omega_0t} e^{-i\omega t} dt\\ &= \int_{-\infty}^{\infty} e^{-i(\omega-\omega_0)t} dt\\ &= 2\pi\delta(\omega - \omega_0) \end{align*}$$

Compare it with the Fourier transform of the constant function: (proof not given)

$$ 2\pi\delta(\omega) = \int_{-\infty}^{\infty} 1\cdot e^{-i\omega t} dt$$

Answer 2

Up to a factor of $2\pi$ the Fourier transformation can be seen as an expansion in terms of $e^{i\omega t}$. Clearly for $e^{i\omega_0 t}$ there is only one component in the expansion. In a discrete expansion this would mean that we have a Kronecker delta $\delta_{\omega_0}^{\omega}$ as component. But because we are doing a continuous transformation this becomes the Dirac delta. This (generalized) function will filter out that single component $e^{i\omega_0 t}$.