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: Use the first shifting property instead of integrating from scratch.
A well known result is that the Laplace transform of the constant function $f(t)=1$ is $\mathcal{L}[1]=\dfrac{1}{s}$. This comes from $\int_0^{\infty}e^{-st}\,dt=\dfrac{1}{s}$ for $s>0$.

Step 2: Recall the first shifting theorem.
The first shifting theorem says that multiplying a function $f(t)$ by $e^{-at}$ in the time domain shifts the variable $s$ to $s+a$ in the transform domain:
\[ \mathcal{L}[e^{-at}f(t)] = F(s+a), \quad \text{where } F(s)=\mathcal{L}[f(t)] \]

Step 3: Apply the theorem to our function.
Here $f(t)=1$, so $F(s)=\dfrac{1}{s}$. Shifting $s$ to $s+a$ gives the transform of $e^{-at}\cdot 1 = e^{-at}$:
\[ \mathcal{L}[e^{-at}] = F(s+a) = \frac{1}{s+a} \]

Step 4: Cross check with the exponential rule of thumb.
A quick memory check: $\mathcal{L}[e^{ct}]=\dfrac{1}{s-c}$ for any constant $c$, valid when $s>c$. Setting $c=-a$ directly gives $\dfrac{1}{s-(-a)}=\dfrac{1}{s+a}$, matching Step 3.

Step 5: Conclude.
Both routes agree.
\[ \boxed{F(s)=\dfrac{1}{s+a}} \]
This is option (D).