Find the Laplace transform of the function $F(t) = \sin(kt)$.

1. Problem Description

Find the Laplace transform of the function $F(t) = \sin(kt)$.

2. Solution Steps

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

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

Therefore, the Laplace transform of $\sin(kt)$ is

$$L\{\sin(kt)\} = \int_0^\infty e^{-st} \sin(kt) dt$$

We can solve this integral by integration by parts twice. Let $u = \sin(kt)$ and $dv = e^{-st} dt$. Then $du = k \cos(kt) dt$ and $v = -\frac{1}{s} e^{-st}$. Using integration by parts, $\int u dv = uv - \int v du$:

$$L\{\sin(kt)\} = \left[-\frac{1}{s}e^{-st} \sin(kt) \right]_0^\infty + \int_0^\infty \frac{1}{s} e^{-st} k \cos(kt) dt$$

$$= 0 + \frac{k}{s} \int_0^\infty e^{-st} \cos(kt) dt$$

Now, integrate by parts again. Let $u = \cos(kt)$ and $dv = e^{-st} dt$. Then $du = -k \sin(kt) dt$ and $v = -\frac{1}{s} e^{-st}$.

$$\frac{k}{s} \int_0^\infty e^{-st} \cos(kt) dt = \frac{k}{s} \left( \left[ -\frac{1}{s} e^{-st} \cos(kt) \right]_0^\infty - \int_0^\infty \frac{1}{s} e^{-st} k \sin(kt) dt \right)$$

$$= \frac{k}{s} \left( \frac{1}{s} - \frac{k}{s} \int_0^\infty e^{-st} \sin(kt) dt \right) = \frac{k}{s^2} - \frac{k^2}{s^2} \int_0^\infty e^{-st} \sin(kt) dt$$

Thus, we have:

$$L\{\sin(kt)\} = \frac{k}{s^2} - \frac{k^2}{s^2} L\{\sin(kt)\}$$

$$L\{\sin(kt)\} + \frac{k^2}{s^2} L\{\sin(kt)\} = \frac{k}{s^2}$$

$$L\{\sin(kt)\} \left(1 + \frac{k^2}{s^2} \right) = \frac{k}{s^2}$$

$$L\{\sin(kt)\} \left( \frac{s^2 + k^2}{s^2} \right) = \frac{k}{s^2}$$

$$L\{\sin(kt)\} = \frac{k}{s^2 + k^2}$$

3. Final Answer

$L\{\sin(kt)\} = \frac{k}{s^2 + k^2}$