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: By definition, the Laplace transform is
\[ \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t)\, dt. \]

Step 2: Put \( f(t)=e^{-at} \):
\[ \mathcal{L}\{e^{-at}\} = \int_{0}^{\infty} e^{-st}\,e^{-at}\, dt = \int_{0}^{\infty} e^{-(s+a)t}\, dt. \]

Step 3: Integrate the exponential (valid for \( s+a > 0 \)):
\[ \int_{0}^{\infty} e^{-(s+a)t}\, dt = \left[ \frac{e^{-(s+a)t}}{-(s+a)} \right]_{0}^{\infty}. \]

Step 4: Evaluate the limits: at \( t\to\infty \) the term vanishes, at \( t=0 \) it is \( \dfrac{1}{-(s+a)} \):
\[ = 0 - \left(\frac{1}{-(s+a)}\right) = \frac{1}{s+a}. \]
\[ \boxed{\, \mathcal{L}\{e^{-at}\} = \dfrac{1}{s+a} \,} \]