Question

Laplace transform of $\int_0^t u^2 \sin(t - u) \, du$ is:

• A. $\frac{2}{s(s+1)}$
• B. $\frac{1}{s^2(s^3+1)}$
• C. $\frac{2}{s^3(s^2+1)}$
• D. $\frac{2}{s(s^2+1)}$

Hint:

Whenever you see an integral of the form $\int_0^t f(u)g(t-u)du$, avoid performing manual integration. Instantly treat it as a convolution and multiply their standard Laplace transforms.

Answer 1

The Correct Option is C

Concept: The given integral has the standard form of a convolution integral. The convolution of two functions $f(t)$ and $g(t)$, denoted by $f(t) * g(t)$, is defined as: \[ f(t) * g(t) = \int_0^t f(u)g(t - u) \, du \] By the Convolution Theorem for Laplace Transforms: \[ \mathcal{L}\{f(t) * g(t)\} = \mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\} \] In this problem, the integrand contains $u^2$ and $\sin(t - u)$. Therefore, we can identify: \[ f(t) = t^2 \quad \text{and} \quad g(t) = \sin t \]

Step 1: Find the Laplace transform of $f(t) = t^2$.
Using the standard formula $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$: \[ \mathcal{L}\{t^2\} = \frac{2!}{s^{2+1}} = \frac{2}{s^3} \]

Step 2: Find the Laplace transform of $g(t) = \sin t$.
Using the standard formula $\mathcal{L}\{\sin(at)\} = \frac{a}{s^2 + a^2}$ with $a = 1$: \[ \mathcal{L}\{\sin t\} = \frac{1}{s^2 + 1^2} = \frac{1}{s^2 + 1} \]

Step 3: Apply the Convolution Theorem.
Multiply the individual Laplace transforms obtained in Step 1 and Step 2: \[ \mathcal{L}\left\{ \int_0^t u^2 \sin(t - u) \, du \right\} = \mathcal{L}\{t^2\} \cdot \mathcal{L}\{\sin t\} \] \[ = \frac{2}{s^3} \cdot \frac{1}{s^2 + 1} = \frac{2}{s^3(s^2 + 1)} \]