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{e^{t}} = 1/(s-1). |

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## FAQs

**Q1: What is the Laplace transform of e**^{t}?

^{t}?

Answer: The Laplace transform of e^t is equal to 1/(s-1) whenever s>1.

**Q2: Find L{e**^{t}}.

^{t}}.

Answer: L{e^{t}} = 1/(s-1), provided that s>1.