Step by step
The question
Question: find the Laplace transform of e^(−3t), from the definition.
ℒ{e^(−3t)} =
1 Use the definition
The definition: multiply by e to the minus st, and integrate from 0 to infinity.
ℒ{f(t)} = ∫ from 0 to ∞ of e^(−st) f(t) dt
Here f(t) is e^(−3t).
ℒ{e^(−3t)} = ∫ from 0 to ∞ of e^(−st)e^(−3t) dt
Join the powers: −st − 3t is −(s + 3)t.
= ∫ from 0 to ∞ of e^(−(s + 3)t) dt
e^(a) × e^(b) = e^(a + b)
2 Integrate
Integrate: divide by the number in front of t, which is −(s + 3).
= [−e^(−(s + 3)t)s + 3] from 0 to ∞
When t goes to infinity, e to a big minus power goes to 0.
That needs s > −3.
t → ∞: e^(−(s + 3)t) → 0 when s > −3
Careful!
If s is −3 or less, the power is not negative, and the integral never stops growing.
Careful: only for s > −3
3 Top minus bottom
At infinity: 0.
At 0: e⁰ = 1, so −1/(s + 3).
Top minus bottom.
= 0 − (−1s + 3)
So ℒ = 1/(s + 3), for s > −3.
= 1s + 3
4 Check
Let's check with the table row: e to the at gives 1/(s − a), with a = −3.
The same.
Right!
table: e^(at) → 1s − a, a = −3
Let's look back.
First: use the definition.
Then: integrate.
Then: top minus bottom.
Last: check.
Answer
= 1s + 3