The Laplace transform of a constant 1 is
1/s
The Laplace transform of a function f(t) is defined as \(F(s) = \int_0^{\infty} f(t)\, e^{-st}\, dt\).
For the constant f(t) = 1: \(\mathcal{L}\{1\} = \int_0^{\infty} e^{-st}\, dt = \left[ \dfrac{e^{-st}}{-s} \right]_0^{\infty}\).
For Re(s) > 0 the term \(e^{-st}\) tends to zero as t → ∞, so the bracket evaluates to \(0 - \left(\dfrac{1}{-s}\right) = \dfrac{1}{s}\).
This is the transform of the unit step u(t) as well, since u(t) = 1 for t ≥ 0. By comparison, \(1/s^2\) is the transform of the ramp t, the constant 1 in the s-domain corresponds to the impulse δ(t), and \(e^{-st}\) is the kernel of the integral, not a transform.
Hence, the answer is 1/s.
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