Question

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. Find ℒ{f(t)}. (Write your answer as a function of s.) f(t) = 9, 0 ≤ t < 5 0, t ≥ 5

Answer 1

By definition of the transform, and the definition of $f$, we have

$\displaystyle\int_0^\infty f(t)e^{-st}\,\mathrm dt=9\int_0^5e^{-st}\,\mathrm dt$

Substitute $u=-st$, so that $\mathrm du=-s\,\mathrm dt$.

$\displaystyle-\frac9s\int_0^{-5s}e^u\,\mathrm du=-\frac9s(e^{-5s}-1)=\boxed{\frac9s-\frac{9e^{-5s}}s}$

Alternatively, if you're familiar with the unit step function, you can write

$f(t)=9(u(t)-u(t-5))$

where the unit step function is defined as

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

Either way, the transform is the same.