Laplace of cos t

Steps to Solve

1. Write the definition of the Laplace transform

The Laplace transform of a function $f(t)$ is defined as:

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

where $s$ is a complex number.

2. Substitute the function into the definition

Substituting $f(t) = \cos(t)$, we get:

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

3. Use integration by parts

To compute this integral, we'll use integration by parts. Let:

• $u = \cos(t)$ and $dv = e^{-st} dt$

Then, we find:

• $du = -\sin(t) dt$ and $v = \frac{e^{-st}}{-s}$

4. Apply integration by parts formula

Using the integration by parts formula $\int u , dv = uv - \int v , du$, we compute:

$$ \mathcal{L}{\cos(t)} = \left[ \cos(t) \cdot \frac{e^{-st}}{-s} \right]_0^\infty - \int_0^\infty \left( \frac{e^{-st}}{-s} \right)(-\sin(t)) , dt $$

5. Evaluate the boundary terms

As $t \to \infty$, the term $e^{-st}$ will dominate. If $s > 0$, this limit goes to $0$. At $t = 0$:

$$ \left[ \cos(t) \cdot \frac{e^{-st}}{-s} \right]_0^\infty = 0 - \frac{1}{-s} = \frac{1}{s} $$

6. Find the remaining integral

Now we need to evaluate the remaining integral:

$$ \int_0^\infty e^{-st} \sin(t) , dt = \frac{1}{s^2 + 1} $$

The Laplace transform formula gives us:

$$ \mathcal{L}{\cos(t)} = \frac{1}{s} + \frac{1}{s^2 + 1} $$

7. Simplify the result

The final expression simplifies to:

$$ \mathcal{L}{\cos(t)} = \frac{s}{s^2 + 1} $$