Question

Find the inverse Laplace of
`\displaystyle{L^(-1)\left \{ (1)/(s^3) + (1)/(s)\right \}}`

Answer 1

Compute the Laplace transform of
`f(t)=t^n`, where
`n` is a non-negative integer.

`\displaystyle L\left\{t^n\right\} = I_n = \int_0^\infty t^n e^(-st) \, dt`

Integrate by parts with

`u=t^n \implies du = nt^(n-1) \, dt`

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

`\displaystyle I_n = \frac ns \int_0^\infty t^(n-1) e^(-st) \, dt = \frac ns I_(n-1)`

By substitution,

`\displaystyle I_n = \frac ns I_(n-1) = (n(n-1))/(s^2) I_(n-2) = \cdots = (n(n-1)\cdots(n-(k-1)))/(s^k) I_(n-k)`

so that for
`k=n`, we end up with

`\displaystyle I_n = (n(n-1)\cdots1)/(s^n) I_0 = (n!)/(s^n) \int_0^\infty e^(-st) \, dt = (n!)/(s^(n+1))`

We then have the inverse transform relation

`\displaystyle L^(-1)\left\{ \frac1{s^n} \right\} = (t^(n-1))/((n-1)!)`

and so

`L^(-1)\left\{ \frac1{s^3} + \frac1s \right\} = (t^2)/(2!) + (0!)/(t^0) = \boxed{\frac{t^2}2 + 1}`