Question

The Laplace transform of \( e^{-at} \) is:

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

Hint:

Integrate \( \int_0^\infty e^{-(s+a)t}dt \), or shift \( \mathcal{L}\{1\}=1/s \) by \( a \). The result is \( 1/(s+a) \).

Answer 1

The Correct Option is B

Step 1: Recall the elementary transform $\mathcal{L}\{1\}=\dfrac{1}{s}$ for $s>0$, obtained directly from $\int_0^\infty e^{-st}\,dt=1/s$.

Step 2: Invoke the first shifting (frequency-shift) theorem: if $\mathcal{L}\{g(t)\}=G(s)$, then
\[ \mathcal{L}\{e^{-at}g(t)\} = G(s+a). \]

Step 3: Take $g(t)=1$ so that $G(s)=\dfrac{1}{s}$.

Step 4: Replace $s$ by $s+a$ in $G(s)$:
\[ \mathcal{L}\{e^{-at}\cdot 1\} = G(s+a) = \frac{1}{s+a}. \]

Step 5: This matches the direct integration, so the transform of $e^{-at}$ is $\dfrac{1}{s+a}$, converging for $\operatorname{Re}(s)>-a$.
\[ \boxed{\, \mathcal{L}\{e^{-at}\} = \dfrac{1}{s+a} \,} \]