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

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

The formula of the Laplace of et is given by

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

## Find the Laplace Transform of et

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) = et.

∴ L{et} = $\int_0^\infty$ et e-st dt

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

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

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

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

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

So the Laplace transform of et is 1/(s-1) whenever s>1.

Related Topics:

Laplace transform of 0

Laplace transform of t

Laplace transform of t2

Laplace transform of 2

Laplace transform of sint

## FAQs

### Q1: What is the Laplace transform of et?

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

### Q2: Find L{et}.

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