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

The laplace transform is
`F(s) = (s(1+e^(-s\pi)))/(s^2+1)`

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=
`(-e^(st))/(s)` and du = -sin(t), then, in one step of the integration we have that

`I = \left.(-\cos(t) e^(-st))/(s)\right|_(0)^\pi- \int_(0)^\pi (\sin(t) e^(-st))/(s) dt`

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

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

Therefore,

`I = \left.(-\cos(t) e^(-st))/(s)\right|_(0)^\pi - (\left.(-\sin(t) e^(-st))/(s^2)\right|_(0)^\pi - (1)/(s^2) I`

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

`I((s^2+1)/(s^2)) =\left.(-\cos(t) e^(-st))/(s)\right|_(0)^\pi+\left.(\sin(t) e^(-st))/(s^2)\right|_(0)^\pi`

Then,

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

Answer 2

`F(s) = (s(e^(\pi s)+1))/(s^2 +1)`

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) = (s(e^(\pi s)+1))/(s^2 +1)`.