Question

Use Definition 7.1.1, DEFINITION 7.1.1 Laplace Transform LetUse Definition 7.1.1, DEFINITION 7.1.1 Laplace Transform Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = [infinity] e−stf(t) dt 0 is said to be the Laplace transform of f, provided that the integral converges. to find ℒ{f(t)}. (Write your answer as a function of s.) WebAssign Plotf be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = [infinity] e−stf(t) dt 0 is said to be the Laplace transform of f, provided that the integral converges. to find ℒ{f(t)}. (Write your answer as a function of s.) f(t) = cos(t), 0 ≤ t < π 0, t ≥ π

Answer 1

Answer:

The laplace transform is $F(s) = \frac{s(1+e^{-s\pi})}{s^2+1}$

Step-by-step explanation:

Let us asume that f(t) =0 for t<0. So, by definition, the laplace transform is given by:

$I = \int_{0}^\pi e^{-st}\cos(t) dt$

To solve this integral, we will use integration by parts. Let u= cos(t)  and dv = $e^{-st}$, so v=$\frac{-e^{st}}{s}$ and du = -sin(t), then, in one step of the integration we have that

$I = \left.\frac{-\cos(t) e^{-st}}{s}\right|_{0}^\pi- \int_{0}^\pi \frac{\sin(t) e^{-st}}{s} dt$

Let $I_2 = \int_{0}^\pi \frac{\sin(t) e^{-st}}{s} dt$. We will integrate I_2 again by parts. Choose u = sin(t) and dv = $\frac{e^{-st}}{s}$. So

$I_2 = \left.\frac{-\sin(t) e^{-st}}{s^2}\right|_{0}^\pi + \int_{0}^\pi \frac{\cos(t) e^{-st}}{s^2}dt$

Therefore,

$I = \left.\frac{-\cos(t) e^{-st}}{s}\right|_{0}^\pi - (\left.\frac{-\sin(t) e^{-st}}{s^2}\right|_{0}^\pi - \frac{1}{s^2} I$

which is an equation for the variabl I. Solving for I we have that

$I(\frac{s^2+1}{s^2}) =\left.\frac{-\cos(t) e^{-st}}{s}\right|_{0}^\pi+\left.\frac{\sin(t) e^{-st}}{s^2}\right|_{0}^\pi$

Then,

$I = \left.\frac{-s\cos(t) e^{-st}}{s^2+1}\right|_{0}^\pi+\left.\frac{\sin(t) e^{-st}}{s^2+1}\right|_{0}^\pi$.

Note that since the sine function is 0 at 0 and pi, we must only care on the first term. Then

$I = \left.\frac{-s\cos(t) e^{-st}}{s^2+1}\right|_{0}^\pi = \frac{s}{s^2+1}(1-(-1)e^{-s\pi}} = \frac{s(1+e^{-s\pi})}{s^2+1}$

Answer 2

Answer:

$F(s) = \frac{s(e^{\pi s}+1)}{s^2 +1}$

Step-by-step explanation:

Using the formula for Laplace the transformations if $F(s)$  is the converted function then

$F(s) = \int\limits_{0}^{\infty} e^{-st} \cos(t) dt = \int\limits_{0}^{\pi} e^{-st} \cos(t) dt$

To solve that integral you need to use integration by parts, when you do integration by parts you get that

$F(s) = \frac{s(e^{\pi s}+1)}{s^2 +1}$.