📂Odinary Differential Equations

Laplace Transform of Exponential Functions

Formulas¹

$$\mathcal {L} \left\{ e^{at} \right\} = \dfrac{1}{s-a},\quad s>a$$

Description

Let’s compare this with the result of the Laplace transform of a constant function (../745).

$$\mathcal{L} \left\{ 1 \right\} =\dfrac{1}{s}$$

The Laplace transform result of $e^{at}$ is the same as when $F(s)$ is shifted by $a$, when $f(t)=1$. This is inevitable because when $e^{at}$ is multiplied by the original function, $\displaystyle \int e^{-st}f(t) dt$ becomes $\displaystyle \int e^{-(s-a)t} f(t) dt$. Except that $s$ changes to $s-a$, there is no difference, so the result changes from $F(s)$ to $F(s-a)$.

Derivation

$$\begin{align*} \mathcal{L} \left\{ e^{at} \right\} &= \int_{0}^\infty e^{-st} e^{at} dt \\ &= \int _{0} ^\infty e^{-(s-a)t}dt \\ &= \lim_{A \to \infty } \left[ -\dfrac{1}{s-a}e^{-(s-a)t} \right]_{0}^A \\ &= \dfrac{1}{s-a} \end{align*}$$

Provided that $\lim \limits_{A \to \infty} e^{-(s-a)A}$ converges to $0$, thus $s>a$

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See Also

• Laplace Transform Table