Laplace Transform of e^at: Formula, Proof

The Laplace transform of the exponential function e to the power at is 1/(s-a). In this article, we will learn how to prove this Laplace transform formula of exponential functions.

Laplace Transform of e^at Formula

The formula of the Laplace transform of e^at is 1/(s-a). Mathematically, we write it as

L{e^at} = 1/(s-a).

e^at Laplace Transform

The Laplace transform of e^at is equal to 1 divided by (s-a). That is,

L{e^at} = 1/(s-a).

Proof:

Recall, the definition of the Laplace transform of a function f(t).

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

Put f(t)=e^at.

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

= $\int_0^\infty$ e^-st+at dt

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

= lim_T→∞ $\left[\dfrac{e^{-(s-a)t}}{-(s-a)}\right]_0^T$

= lim_T→∞ $\left(\dfrac{e^{-(s-a)T}}{-(s-a)}-\dfrac{1}{-(s-a)}\right)$

= 0 – $\dfrac{1}{-(s-a)}$

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

So, the Laplace transform of e^at is 1/(s-a).

Now, replacing a with -a, we obtain the Laplace transform of e^-at which is 1/(s+a).

Summary: Denoting the Laplace transform of f(t) by L{f(t)}, we list the Laplace transform formulas for the exponential functions in the below table:

1	L{eat}	1/(s-a)
2	L{e-at}	1/(s+a)
3	L{et}	1/(s-1)
4	L{e-t}	1/(s+1)

Question-Answer

Question: What is the inverse Laplace transform of 1/(s-a)?

Answer:

From above, the Laplace transform of e^at is equal to 1/(s-a). That is,

L{e^at} = $\dfrac{1}{s-a}$

Now, we take inverse Laplace on both sides. As a result.we have that

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

Thus, the inverse Laplace transform of 1/(s-a) is equal to e^at. Similarly, the inverse Laplace of 1/(s+a) is equal to e^-at.

Read Also:

Concept of Laplace Transform: Definition, Table, Formulas, Properties & Examples

Laplace transform of sin(at): The Laplace transform of sin(at) is a/(s²+a²).

Laplace transform of cos(at): The Laplace transform of cos(at) is s/(s²+a²).

Laplace transform of constant: The Laplace transform of c is c/s.

Inverse Laplace transform of constant: The inverse Laplace transform of c is cδ(t), where δ(t) is the Dirac delta function.

Question: Find the Laplace transform of e^at+e^-at.

Solution:

By the linearity property of Laplace transform, we have

L{e^at+e^-at} = L{e^at} + L{e^-at}

= 1/(s-a) + 1/(s+a) from the above table

= (s+a+s-a)/(s-a)(s+a)

= 2s/(s²-a²)

So the Laplace transform of the sum of e^at and e^-at is 2s/(s²-a²).

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