Question

Laplace transform of \( \frac{e^{-at} - e^{-bt}}{t} \) is:

• A. \( \log\left( \frac{s+a}{s+b} \right) \)
• B. \( \log\left( \frac{s+b}{s+a} \right) \)
• C. \( \frac{\pi}{2} - \tan^{-1}(s) \)
• D. \( \frac{\pi}{2} - \cot^{-1}(s) \)

Hint:

Memorize special Laplace transforms involving exponential terms divided by \( t \). They frequently result in logarithmic expressions.

Answer 1

The Correct Option is B

We are given the function: \[ f(t) = \frac{e^{-at} - e^{-bt}}{t} \] This is a standard Laplace transform result. Formula: \[ \mathcal{L}\left\{ \frac{e^{-at} - e^{-bt}}{t} \right\} = \log\left( \frac{s + b}{s + a} \right), \quad \text{for } s > -\min(a,b) \] This result comes from the integral identity: \[ \int_0^\infty \frac{e^{-at} - e^{-bt}}{t} e^{-st} dt = \log\left( \frac{s + b}{s + a} \right) \] Thus, the Laplace transform directly yields: \[ { \log\left( \frac{s + b}{s + a} \right) } \]