Laplace transform of exponential
Understand the Problem
The question is asking for the Laplace transform of an exponential function. The Laplace transform is a technique used to convert a function of time, often denoted by f(t), into a function of a complex variable, usually denoted by s. The exponential function is commonly expressed in the form e^(at), where 'a' is a constant. The solution typically involves applying the definition of the Laplace transform to this exponential function.
Answer
$$ L\{e^{at}\} = \frac{1}{s - a}, \, s > a $$
Answer for screen readers
$$ L{e^{at}} = \frac{1}{s - a}, , s > a $$
Steps to Solve
- Recall the definition of the Laplace transform
The Laplace transform of a function $f(t)$ is defined as: $$ L{f(t)} = \int_0^\infty e^{-st} f(t) , dt $$ where $s$ is a complex variable.
- Identify the function to transform
For our case, we have the exponential function $f(t) = e^{at}$.
- Set up the integral for the Laplace transform
We substitute $f(t)$ into the Laplace transform definition: $$ L{e^{at}} = \int_0^\infty e^{-st} e^{at} , dt $$
- Combine the exponentials
We can combine the exponentials in the integral: $$ L{e^{at}} = \int_0^\infty e^{(a - s)t} , dt $$
- Evaluate the integral
To compute the integral, we need to find the antiderivative: $$ \int e^{(a - s)t} , dt = \frac{1}{a - s} e^{(a - s)t} $$
- Apply the limits of integration
Now we evaluate from $0$ to $\infty$: $$ L{e^{at}} = \left[ \frac{1}{a - s} e^{(a - s)t} \right]_0^\infty $$
- Evaluate the limits
As $t \to \infty$, $e^{(a - s)t}$ approaches zero if $s > a$, and at $t = 0$, it equals $1$: $$ \lim_{t \to \infty} e^{(a - s)t} = \begin{cases} 0 & \text{if } s > a \ \text{undefined} & \text{if } s \leq a \end{cases} $$
Therefore: $$ L{e^{at}} = \frac{1}{a - s}(0 - 1) = \frac{-1}{a - s} = \frac{1}{s - a} $$
- Final Result
Thus, the final expression for the Laplace transform is: $$ L{e^{at}} = \frac{1}{s - a}, , s > a $$
$$ L{e^{at}} = \frac{1}{s - a}, , s > a $$
More Information
The Laplace transform is widely used in engineering and physics, particularly for solving differential equations. The condition $s > a$ ensures that the integral converges and the Laplace transform is defined.
Tips
- Forgetting to specify the condition $s > a$ can lead to undefined results.
- Not combining the exponential terms correctly may cause errors in the integration step.
