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(Please note that I'm aware there are already several questions asking about the difference between the two transforms. However, none of them that I could find touch on this specific issue of the affect of singularities.)

I was reading this answer which says that if,

the region of convergence is $Re\{s\}>0$ but there are singularities on the $j\omega$ axis [, then] both transforms exist but they have different forms. The Fourier transform has additional delta impulses. Consider the function $f(t)=e^{j\omega_0 t}u(t)$. From (1), its Laplace transform is given by

$$F(s)=\frac{1}{s-j\omega_0}$$ However, due to the singularity on the $j\omega$ axis, its Fourier transform is

$$F(j\omega)=\pi\delta(\omega-\omega_0)+\frac{1}{j\omega-j\omega_0}$$

The part I bolded is what has me confused. If I understand correctly, the Fourier transform integrates over the entire real number line and is the imaginary part of the the Laplace transform for functions which are zero for negative inputs (since Laplace integrates only on the nonnegative part of the real number line). So it seems like, for the example function in the above quote, Laplace and Fourier should have the same result when $s=j\omega$. Why does the singularity on the imaginary axis mean this isn't true? I don't understand why it adds a delta pulse to the Fourier transform but not the Laplace transform.

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  • $\begingroup$ offering this as a comment ( as Ahsan has a nice answer) for additional intuition if Fourier Series expansion is understood: Consider the Fourier Series expansion of a constant: it is an impulse a $f=0$ (DC). Now consider the Fourier Series expansion of a step: it would have all frequencies, going down at $1/f$. This is the difference between Fourier and (unilateral) Laplace: Fourier is a correlation with functions constant for all time, so would converge to an impulse. Laplace is a correlation with functions that step at $t=0$. That step smears out what would have otherwise been impulses. $\endgroup$ Jan 14 at 20:39

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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, $\frac{1}{j\omega-j\omega_0}$.

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    $\begingroup$ "the Fourier transform requires that the integral converges absolutely" -- I think that's the case for the Laplace transform, too, it's just that the constraint "the region of convergence (ROC) is the set of $s$ values for which this integral converges" satisfies that condition. I'm assuming that the fact that the integral is one-sided is what preserves necessary information when you do have to exclude values of $s$ to keep the integral finite. $\endgroup$
    – TimWescott
    Jan 14 at 17:25

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