To compute the Laplace transform of $f(t) = t$, we use the integral:

$\mathcal{L}\{t\}(s) = \int_0^{\infty} t e^{-st} \, dt$

This requires integration by parts:

1. Set up for integration by parts:

   • Let $u = t$ and $dv = e^{-st} \, dt$.
   • Then $du = dt$ and $v = -\frac{e^{-st}}{s}$.

2. Apply integration by parts:

   • Using the formula $\int u \, dv = uv - \int v \, du$:

   $\int_0^{\infty} t e^{-st} \, dt = \left[ -\frac{t e^{-st}}{s} \right]_0^{\infty} + \int_0^{\infty} \frac{e^{-st}}{s} \, dt$

3. Evaluate the boundary terms:

   • As $t \to \infty$, $t e^{-st} \to 0$.
   • At $t = 0$, the term is also zero.
   • This results in $0 - 0 = 0$ in the first box.

4. Simplify the remaining integral:

   • We are left with $\frac{1}{s} \int_0^{\infty} e^{-st} \, dt$.
   • This integral evaluates to $\frac{1}{s^2}$.

5. Final Answer:

   • The Laplace transform of $t$ is $\frac{1}{s^2}$.

Here’s what goes in each box:

1. First box: 0
2. Second box: 0
3. Third box: 1
4. Fourth box: 0
5. Fifth box: 0

Would you like more details, or do you have any questions about this solution?

Related Questions

1. How do you perform integration by parts in general?
2. What is the significance of the Laplace transform in engineering?
3. Why does $t e^{-st}$ go to zero as $t \to \infty$?
4. Can the Laplace transform be applied to other types of functions?
5. How would the Laplace transform differ for a function like $f(t) = t^2$?

Tip

For functions involving $t$ multiplied by exponentials, integration by parts is typically a good approach.