Question

Section 8.1 Introduction to the Laplace Transforms

Problem 1.
Find the Laplace transforms of the following functions by evaluating the integral

`F(s) = {∫}^( \infty ) _(0) {e}^( - st) f(t)dt`

`(a)t`

`(b) {te}^( - t)`

`(c) {sinh} \: bt`

`(d) {e}^(2t) - {3e}^(t)`

`(e) {t}^(2)`

​

Answer 1

For the integrals in (a), (b), and (e), you'll end up integrating by parts.

(a)

`\displaystyle \int_0^\infty t e^(-st) \, dt`

Let

`u = t \implies du = dt`

`dv = e^(-st) \, dt \implies v = -\frac1s e^(-st)`

Then

`\displaystyle \int_0^\infty t e^(-st) \, dt = uv\bigg|_(t=0)^(t\to\infty) - \int_0^\infty v\, du`

`\displaystyle \int_0^\infty t e^(-st) \, dt = \left(-\frac1s te^(-st)\right)\bigg|_0^\infty + \frac1s \int_0^\infty e^(-st) \, dt`

`\displaystyle \int_0^\infty t e^(-st) \, dt = -\frac1s \left(\lim_(t\to\infty) te^(-st) - 0\right) + \frac1s \int_0^\infty e^(-st) \, dt`

`\displaystyle \int_0^\infty t e^(-st) \, dt = \frac1s \int_0^\infty e^(-st) \, dt`

`\displaystyle \int_0^\infty t e^(-st) \, dt = -\frac1{s^2} e^(-st) \bigg|_0^\infty e^(-st) \, dt`

`\displaystyle \int_0^\infty t e^(-st) \, dt = -\frac1{s^2} \left(\lim_(t\to\infty)e^(-st) - 1\right) = \boxed{\frac1{s^2}}`

(b)

`\displaystyle \int_0^\infty t e^(-t) e^(-st) \, dt = \int_0^\infty t e^(-(s+1)t) \, dt`

Let

`u = t \implies du = dt`

`dv = e^(-(s+1)t) \, dt \implies v = -\frac1{s+1} e^(-(s+1))t`

Then

`\displaystyle \int_0^\infty t e^(-(s+1)t) \, dt = -\frac1{s+1} te^(-(s+1)t) \bigg|_0^\infty + \frac1{s+1} \int_0^\infty e^(-(s+1)t) \, dt`

`\displaystyle \int_0^\infty t e^(-(s+1)t) \, dt = -\frac1{s+1} \left(\lim_(t\to\infty)te^(-(s+1)t) - 0\right) + \frac1{s+1} \int_0^\infty e^(-(s+1)t) \, dt`

`\displaystyle \int_0^\infty t e^(-(s+1)t) \, dt = \frac1{s+1} \int_0^\infty e^(-(s+1)t) \, dt`

`\displaystyle \int_0^\infty t e^(-(s+1)t) \, dt = -\frac1{(s+1)^2} e^(-(s+1)t) \bigg|_0^\infty`

`\displaystyle \int_0^\infty t e^(-(s+1)t) \, dt = -\frac1{(s+1)^2} \left(\lim_(t\to\infty)e^(-(s+1)t) - 1\right) = \boxed{\frac1{(s+1)^2}}`

(e)

`\displaystyle \int_0^\infty t^2 e^(-st) \, dt`

Let

`u = t^2 \implies du = 2t \, dt`

`dv = e^(-st) \, dt \implies v = -\frac1s e^(-st)`

Then

`\displaystyle \int_0^\infty t^2 e^(-st) \, dt = -\frac1s t^2 e^(-st) \bigg|_0^\infty + \frac2s \int_0^\infty t e^(-st) \, dt`

`\displaystyle \int_0^\infty t^2 e^(-st) \, dt = -\frac1s \left(\lim_(t\to\infty) t^2 e^(-st) - 0\right) + \frac2s \int_0^\infty t e^(-st) \, dt`

`\displaystyle \int_0^\infty t^2 e^(-st) \, dt = \frac2s \int_0^\infty t e^(-st) \, dt`

The remaining integral is the transform we found in (a), so

`\displaystyle \int_0^\infty t^2 e^(-st) \, dt = \frac2s * \frac1{s^2} = \boxed{\frac2{s^3}}`

Computing the integrals in (c) and (d) is much more immediate.

(c)

`\displaystyle \int_0^\infty \sinh(bt) e^(-st) \, dt = \int_0^\infty \frac{e^(bt)-e^(-bt)}2 * e^(-st) \, dt`

`\displaystyle \int_0^\infty \sinh(bt) e^(-st) \, dt = \frac12 \int_0^\infty \left(e^((b-s)t) - e^((b+s)t)\right) \, dt`

`\displaystyle \int_0^\infty \sinh(bt) e^(-st) \, dt = \frac12 \left(\frac1{b-s} e^((b-s)t) - \frac1{b+s} e^((b+s)t)\right) \bigg|_0^\infty`

`\displaystyle \int_0^\infty \sinh(bt) e^(-st) \, dt = \frac12 \left[\lim_(t\to\infty)\left(\frac1{b-s} e^((b-s)t) - \frac1{b+s} e^((b+s)t)\right) - \left(\frac1{b-s} - \frac1{b+s}\right)\right]`

`\displaystyle \int_0^\infty \sinh(bt) e^(-st) \, dt = \frac12 \left(\frac1{b+s} - \frac1{b-s}\right) = \boxed{(s)/(s^2-b^2)}`

(d)

`\displaystyle \int_0^\infty (e^(2t) - 3e^t) e^(-st) \, dt = \int_0^\infty \left(e^((2-s)t) - 3e^((1-s)t)\right) \, dt`

`\displaystyle \int_0^\infty (e^(2t) - 3e^t) e^(-st) \, dt = \left( \frac1{2-s} e^((2-s)t) - \frac3{1-s} e^((1-s)t) \right) \bigg|_0^\infty`

`\displaystyle \int_0^\infty (e^(2t) - 3e^t) e^(-st) \, dt = \lim_(t\to\infty) \left( \frac1{2-s} e^((2-s)t) - \frac3{1-s} e^((1-s)t) \right) - \left( \frac1{2-s} - \frac3{1-s} \right)`

`\displaystyle \int_0^\infty (e^(2t) - 3e^t) e^(-st) \, dt = \frac3{1-s} - \frac1{2-s} = \boxed{-(2s-5)/(s^2-3s+2)}`