The Laplace transform of a function f(t), defined for all t≥0, is the function F(s), defined as follows:
F(s)=L(f(t))=∫0∞e−stf(t)dt, where s is a complex parameter.
Let's go through a couple of examples.
Example 1. Calculate L(1).
L(1)=∫0∞e−st⋅1dt=∫0∞e−stdt=lima→∞(∫0ae−stdt)=lima→∞(−s1e−st∣0a)=lima→∞(−s1e−sa+s1e−s⋅0)=lima→∞(−s1e−sa+s1)
Note that if s<0, lima→∞(−s1e−sa+s1)=∞ and the integral is divergent.
In case s>0, lima→∞(−s1e−sa+s1)=s1 and the integral is convergent.
So, L(1)=s1, provided that s>0.
Example 2. Calculate L(eat).
L(eat)=∫0∞e−st⋅eatdt=∫0∞e(a−s)tdt=limb→∞(∫0be(a−s)tdt)=limb→∞(a−s1e(a−s)t∣0b)=
=limb→∞(a−s1e(a−s)b−a−s1e(a−s)⋅0)=limb→∞(a−s1e(a−s)b−a−s1)
Note that if a−s>0, we have thatlimb→∞(a−s1e(a−s)b−a−s1)=∞ and the integral is divergent.
In case that a−s<0, limb→∞(a−s1e(a−s)b−a−s1)=−a−s1=s−a1 and the integral is convergent.
So, L(eat)=s−a1, provided that s>a.
Note that we put a restriction on s in order to compute the Laplace transform. In general, all Laplace transforms have restrictions on s.
Let's go through the last two 'non-standard' examples.
Example 3. Calculate the Laplace transform of the derivative: L(f′(t)).
L(f′(t))=∫0∞e−stf′(t)dt=lima→∞(∫0ae−stf′(t)dt)
Using integration by parts:
lima→∞(∫0ae−stf′(t)dt)=lima→∞(∫0ae−stdf(t))=lima→∞(e−stf(t)∣0a−∫0a(−se−stf(t))dt)=
=lima→∞(e−saf(a)−e−s⋅0f(0)+s∫0a(e−stf(t))dt)=
=lima→∞(e−saf(a)−f(0))+s∫0∞(e−stf(t))dt=lima→∞(e−saf(a))−f(0)+sL(f(t))=
=lima→∞(e−saf(a))+sF(s)−f(0)=sF(s)−f(0)
On this stage, we need to add a restriction: assume that there are constants A and B, such that ∣f(t)∣<Aeαt for all t≥B.
This condition implies that lima→∞(e−saf(a))=0, provided that s>α.
So, L(f′(t))=sL(f(t))−f(0), provided that s>α.
Example 4. Calculate the Laplace transform of the integral: L(∫0tf(τ)dτ).
L(∫0tf(τ)dτ)=∫0∞(e−st∫0tf(τ)dτ)dt=∫0∞(∫0te−st(f(τ)dτ))dt
On this stage, we need to change the order of integration. t is changing from 0 to ∞, while τ is changing from 0 to t. So, if we change the order of integration, we have that τ is changing from 0 to ∞ and t is changing from τ to ∞:
∫0∞(∫0te−st(f(τ)dτ))dt=∫0∞(∫τ∞(e−stf(τ))dt)dτ=∫0∞(f(τ)∫τ∞(e−st)dt)dτ=∫0∞(f(τ)⋅(−s1)e−st∣τ∞)dτ=∫0∞(f(τ)s1e−sτ)dτ=s1L(f(t))=s1F(s)
So, L(∫0tf(τ)dτ)=sF(s).
For a list of common Laplace transforms, see the table of Laplace transforms.