Question

Let \[ f(t)= \begin{cases} \sin t, & 0\le t\le \pi,\\ 0, & t>\pi. \end{cases} \] The Laplace transform of \(f(t)\) is

• A. \(\dfrac{1}{s^2+1}\)
• B. \(\dfrac{e^{-\pi s}}{s^2+1}\;(s>0)\)
• C. \(\dfrac{e^{-\pi s}+1}{s^2+1}\;(s>0)\)
• D. \(\dfrac{e^{-\pi s}+e^{-\pi}}{s^2+1}\;(s>0)\)

Hint:

Whenever a function is zero after a finite interval, \[ \mathcal{L}\{f(t)\} = \int_0^{a}e^{-st}f(t)\,dt, \] where \(a\) is the point beyond which the function becomes zero.

Answer 1

The Correct Option is C

Concept: For a piecewise-defined function, \[ \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st}f(t)\,dt. \] Since \(f(t)=0\) for \(t>\pi\), \[ \mathcal{L}\{f(t)\} = \int_0^\pi e^{-st}\sin t\,dt. \] Also, \[ \int e^{-st}\sin t\,dt = \frac{e^{-st}(-s\sin t-\cos t)}{s^2+1}. \]

Step 1: Evaluate the Laplace transform. \[ F(s) = \int_0^\pi e^{-st}\sin t\,dt. \] Using the standard integral, \[ F(s) = \left[ \frac{e^{-st}(-s\sin t-\cos t)} {s^2+1} \right]_0^\pi. \]

Step 2: Substitute the limits. Since \[ \sin\pi=0,\qquad \cos\pi=-1, \] the upper limit becomes \[ \frac{e^{-\pi s}}{s^2+1}. \] Also, \[ \sin0=0,\qquad \cos0=1, \] the lower limit becomes \[ -\frac{-1}{s^2+1} = -\left(-\frac{1}{s^2+1}\right) = \frac{1}{s^2+1}. \] Therefore, \[ F(s) = \frac{e^{-\pi s}+1}{s^2+1}. \] Hence, \[ \boxed{ \mathcal{L}\{f(t)\} = \frac{e^{-\pi s}+1}{s^2+1} }. \] Therefore, the correct option is \[ \boxed{(C)\; \dfrac{e^{-\pi s}+1}{s^2+1}.} \]