Question

The Laplace transform of \[ \int_{0}^{t} p\sin(t-p)\,dp \] is

• A. \(\dfrac{s}{s^2+1}\)
• B. \(\dfrac{s+1}{s^2+1}\)
• C. \(\dfrac{1}{s}-\dfrac{s}{s^2+1}\)
• D. \(\dfrac{1}{(s-1)(s+1)}\)

Hint:

If \[ (f*g)(t)=\int_0^t f(p)g(t-p)\,dp, \] then \[ \boxed{ \mathcal{L}\{f*g\} = \mathcal{L}\{f\}\, \mathcal{L}\{g\}. } \] This is known as the Convolution Theorem.

Answer 1

The Correct Option is C

Step 1: Identify the given integral as a convolution. The given integral is \[ \int_0^t p\sin(t-p)\,dp, \] which is the convolution of \[ f(t)=t \] and \[ g(t)=\sin t. \] Hence, \[ f*g=\int_0^t f(p)g(t-p)\,dp. \]

Step 2: Apply the convolution theorem. Using the convolution theorem, \[ \mathcal{L}\{f*g\} = \mathcal{L}\{f\}\, \mathcal{L}\{g\}. \] Now, \[ \mathcal{L}\{t\} = \frac1{s^2}, \] and \[ \mathcal{L}\{\sin t\} = \frac1{s^2+1}. \] Therefore, \[ \mathcal{L}\left\{\int_0^t p\sin(t-p)\,dp\right\} = \frac1{s^2(s^2+1)}. \]

Step 3: Express the result in the required form. Using partial fractions, \[ \frac1{s^2(s^2+1)} = \frac1s-\frac{s}{s^2+1}. \] Therefore, \[ \boxed{\frac1s-\frac{s}{s^2+1}} \] is the correct answer. Thus, \[ \boxed{(C)} \] is the correct answer.