Why do singularities on the imaginary axis affect the Fourier transform differently than the Laplace transform?

The Laplace transform of a function $f(t)$ is defined as:

$$F(s) = \int_{0^-}^{\infty} e^{-st} f(t) \, dt,$$

where $s$ is a complex variable $s = \sigma + j\omega$, and the region of convergence (ROC) is the set of $s$ values for which this integral converges. Also the lower limit $0^-$ is shorthand notation for:

$$\lim_{\epsilon \to 0^+} \int_{-\epsilon}^{\infty} \cdot$$

This limit emphasizes that any point mass located at 0 is entirely captured by the Laplace Transform.

The Fourier transform of a function $f(t)$ is defined as:

$$F(j\omega) = \int_{-\infty}^{\infty} e^{-j\omega t} f(t) \, dt.$$

For functions that are zero for $t < 0$, you can relate the Fourier transform to the Laplace transform by setting $s = j\omega$. However, the existence of the Fourier transform requires that the integral converges absolutely, which is a stricter condition than for the Laplace transform.

In the case you mentioned, the function $f(t)=e^{j\omega_0 t}u(t)$, where $u(t)$ is the unit step function, has a Laplace transform that is given by:

$$F(s) = \int_{0^-}^{\infty} e^{-st} e^{j\omega_0 t} u(t) \, dt = \int_{0^-}^{\infty} e^{-(s-j\omega_0)t} \, dt = \frac{1}{s-j\omega_0},$$

provided that $Re\{s\} > 0$, which ensures the convergence of the integral.

However, when you try to find the Fourier transform by setting $s = j\omega$, you're evaluating the integral at the singularity, which is the point $s = j\omega_0$. The Fourier transform integral does not converge in the conventional sense because of this singularity. The presence of the singularity means you can't directly apply the same formula you use for the Laplace transform to find the Fourier transform.

In such cases, distribution theory or generalized functions are used to interpret the Fourier transform. When you approach the singularity from the context of distributions, you end up with an additional term, which is the delta function, to account for the energy concentrated at the frequency $\omega_0$. The delta function, $\delta(\omega-\omega_0)$, represents an impulse at the frequency $\omega_0$. Hence, the Fourier transform includes a term $\pi\delta(\omega-\omega_0)$, which captures the singularity's effect on the transform, along with the principal value of the integral around the singularity, $\frac{1}{j\omega-j\omega_0}$.