The integral presented is:

$\int_{0}^{\infty} t e^{-3t} \sin(t) \, dt$

To evaluate this integral, we can use the Laplace transform approach or integration by parts. Here's the step-by-step solution:

Step 1: Recall the integral formula for Laplace Transforms

The Laplace Transform of $t^n e^{-at} \sin(bt)$ is: $\mathcal{L} \left[ t^n e^{-at} \sin(bt) \right] = \frac{n! b}{(a^2 + b^2)^{n+1}}.$

Here, $n = 1$, $a = 3$, and $b = 1$.

Step 2: Plug in the parameters

Using the formula, we compute: $\int_{0}^{\infty} t e^{-3t} \sin(t) \, dt = \frac{1! \cdot 1}{(3^2 + 1^2)^2}.$

Step 3: Simplify

$\int_{0}^{\infty} t e^{-3t} \sin(t) \, dt = \frac{1}{(9 + 1)^2} = \frac{1}{10^2} = \frac{1}{100}.$

Final Answer:

$\int_{0}^{\infty} t e^{-3t} \sin(t) \, dt = \frac{1}{100}.$

Let me know if you'd like further clarification or details on any specific step.

5 Related Questions:

1. How does the Laplace Transform simplify integral calculations?
2. Can integration by parts be applied to solve this problem? How?
3. What happens if the limits of integration are finite (e.g., 0 to $a$)?
4. How would the result change if the coefficient of $e^{-3t}$ were different?
5. How do we evaluate similar integrals with a $\cos(t)$ function instead of $\sin(t)$?

Tip:

When dealing with infinite integrals of exponential functions multiplied by trigonometric terms, Laplace Transforms are often the fastest and most efficient solution method.