Laplace transform of e^t | Laplace of e^t

The Laplace transform of e^t is equal to L{e^t} = 1/(s-1) whenever s>1. Here we find the Laplace of e^t by the definition of Laplace transforms.

The formula of the Laplace of e^t is given by

L{e^t} = $\dfrac{1}{s-1}$.

Find the Laplace Transform of e^t

By definition of Laplace transforms, the Laplace transform of a function f(t) is given by the integral

L{f(t)} = $\int_0^\infty$ e^-st f(t) dt

Let f(t) = e^t.

∴ L{e^t} = $\int_0^\infty$ e^t e^-st dt

= $\int_0^\infty$ e^-(s-1)t dt

= $\Big[ \dfrac{e^{-(s-1)t}}{-(s-1)}\Big]_0^\infty$

= lim_t→∞ $\Big[ \dfrac{e^{-(s-1)t}}{-(s-1)}\Big]$ $-\dfrac{e^{0}}{-(s-1)}$

= $0+ \dfrac{1}{s-1}$ as lim_t→∞ e^-(s-1)t=0 provided that s>1.

= $\dfrac{1}{s-1}$.

So the Laplace transform of e^t is 1/(s-1) whenever s>1.

L{et} = 1/(s-1).

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Laplace transform of 0

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