To find the Laplace transform of the function $\frac{1 - \cos t}{t}$, we can use the following method:

Step 1: Identify the function

The given function is $\frac{1 - \cos t}{t}$.

Step 2: Use the known Laplace transform

We know that the Laplace transform of the function $\text{Si}(t) = \int_0^t \frac{\sin u}{u} \, du$ is $\mathcal{L}\{\text{Si}(t)\} = \frac{\tan^{-1}(s)}{s}$.

Step 3: Express the given function in terms of a known transform

The function $\frac{1 - \cos t}{t}$ can be recognized as the derivative of the sine integral function $\text{Si}(t)$: $\frac{d}{dt} \text{Si}(t) = \frac{\sin t}{t} \implies \int_0^t \frac{\sin u}{u} \, du = \text{Si}(t)$ Thus, $\frac{1 - \cos t}{t} = \int_0^t \frac{\sin u}{u} \, du = \text{Si}(t)$ Therefore, the Laplace transform of $\frac{1 - \cos t}{t}$ is the same as the Laplace transform of $\text{Si}(t)$.

Step 4: Find the Laplace transform

$\mathcal{L}\left\{\frac{1 - \cos t}{t}\right\} = \frac{\tan^{-1}(s)}{s}$

So, the Laplace transform of $\frac{1 - \cos t}{t}$ is $\frac{\tan^{-1}(s)}{s}$.

Would you like further details on this process, or have any questions?

Here are 5 related questions:

1. What is the Laplace transform of $\frac{\sin t}{t}$?
2. How do you find the inverse Laplace transform of $\frac{\tan^{-1}(s)}{s}$?
3. What are the applications of the Laplace transform in solving differential equations?
4. How do you derive the Laplace transform of $\sin(at)$?
5. What is the Laplace transform of a Heaviside step function?

Tip: The Laplace transform is particularly useful for solving linear ordinary differential equations, as it converts differential equations into algebraic equations that are easier to manipulate.