Question

Given a function \(f(t) = e^{-at}\), where \(a\) is a constant. The Laplace transform of the function is \(\mathcal{L}[f(t)] = F(s)\). Which one of the following options is correct?

• A. \(F(s) = \dfrac{1}{s-a}\)
• B. \(F(s) = \dfrac{s}{s^2+a^2}\)
• C. \(F(s) = \dfrac{a}{s^2+a^2}\)
• D. \(F(s) = \dfrac{1}{s+a}\)

Hint:

Use the definition \(F(s)=\int_0^{\infty}e^{-st}f(t)\,dt\) and combine the exponents of \(e^{-st}\) and \(e^{-at}\) before integrating.

Answer 1

The Correct Option is D

Step 1: Recall the definition of the Laplace transform.
For a function \(f(t)\) defined for \(t \geq 0\), the Laplace transform is
\[ \mathcal{L}[f(t)] = F(s) = \int_0^{\infty} e^{-st} f(t)\, dt \] This integral turns a function of time \(t\) into a function of the complex variable \(s\), which is what makes it useful for solving differential equations.

Step 2: Substitute the given function.
Here \(f(t) = e^{-at}\), so
\[ F(s) = \int_0^{\infty} e^{-st} e^{-at}\, dt \] Since both terms have the same base \(e\), we add the exponents:
\[ F(s) = \int_0^{\infty} e^{-(s+a)t}\, dt \]

Step 3: Evaluate the integral.
This is a standard exponential integral. Using \(\int e^{kt}\,dt = \dfrac{e^{kt}}{k}\) with \(k = -(s+a)\),
\[ F(s) = \left[\frac{e^{-(s+a)t}}{-(s+a)}\right]_0^{\infty} \] Provided \(s+a > 0\), the term \(e^{-(s+a)t}\) goes to \(0\) as \(t \to \infty\), so only the lower limit survives:
\[ F(s) = 0 - \left(\frac{1}{-(s+a)}\right) = \frac{1}{s+a} \]

Step 4: Check the other options.
(A) \(1/(s-a)\): This is the transform of \(e^{+at}\), not \(e^{-at}\). Incorrect.
(B) \(s/(s^2+a^2)\): This is the standard Laplace transform of \(\cos(at)\). Incorrect.
(C) \(a/(s^2+a^2)\): This is the standard Laplace transform of \(\sin(at)\). Incorrect.

Step 5: Final conclusion.
The Laplace transform of \(e^{-at}\) is
\[ \boxed{F(s) = \frac{1}{s+a}} \]
Hence, the correct option is (D).