The existence question. The Laplace transform

\(X(s)=\int_{-\infty}^{\infty}x(t)e^{-st}\,dt\)

is meaningful only where the integral converges. The standard sufficient test is absolute convergence: the integral of the magnitude of the integrand must be finite.

Step 1 — take the magnitude of the kernel. With \(s=\sigma+j\omega\),

\(\left|e^{-st}\right|=\left|e^{-\sigma t}\right|\left|e^{-j\omega t}\right|=e^{-\sigma t}\)

because the oscillating factor has unit magnitude, \(\left|e^{-j\omega t}\right|=1\). Only the real part σ affects convergence — the imaginary part merely rotates the phase.

Step 2 — write the condition.

\(\int_{-\infty}^{\infty}\left|x(t)e^{-st}\right|dt=\int_{-\infty}^{\infty}\left|x(t)\right|e^{-\sigma t}\,dt \lt \infty\)

which is option 4.

Why the other options fail.

Option 2, \(\int|x(t)|dt \lt \infty\), is the existence condition for the Fourier transform (a Dirichlet condition). It is far more restrictive — it excludes u(t), the ramp and \(e^{+at}u(t)\), all of which have perfectly good Laplace transforms. It also omits σ entirely, so it cannot describe a region of convergence.

Options 1 and 3 use \(|x(t)|^{2}\), which is an energy condition (square integrability), not the absolute-integrability test that governs the convergence of this integral.

What this condition buys you — the ROC. The set of σ values satisfying the inequality forms the region of convergence, always a vertical strip or half-plane in the s-plane bounded by poles. For a right-sided signal \(e^{-at}u(t)\), for example, \(\int_0^{\infty}e^{-at}e^{-\sigma t}dt\) converges for \(\sigma \gt -a\). The damping factor is precisely what lets Laplace handle signals that grow without bound, provided σ is chosen large enough.

Hence, the Laplace transform is defined when \(\int_{-\infty}^{\infty}|x(t)|e^{-\sigma t}\,dt \lt \infty\).