Discussion
Laplace transform F(s) of a function f(x) = 1, for \(0 \le x \le \infty\) is equal to :
Answer: A
The Laplace transform of f(x) = 1 is \(1/s\).
\[ \begin{aligned} F(s) &= \int_{0}^{\infty} e^{-sx}\cdot 1\,dx \\ &= \left[-\frac{e^{-sx}}{s}\right]_{0}^{\infty} = \frac{1}{s} \end{aligned} \]
valid for s > 0, where the exponential vanishes at the upper limit. \(1/s^{2}\) is the transform of f(x) = x, not of a constant. Neither s nor 2s can be a Laplace transform of an ordinary function at all, since a transform must tend to zero as s grows.
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