# Definition of the Laplace Transform

## Related Calculator: Laplace Transform Calculator

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))=int_0^oo e^(-st)f(t)dt, where s is a complex parameter.

Let's go through a couple of examples.

Example 1. Calculate L(1).

L(1)=int_0^ooe^(-st)*1dt=int_0^ooe^(-st)dt=lim_(a->oo)(int_0^a e^(-st)dt)=lim_(a->oo)(-1/se^(-st)|_0^a)=lim_(a->oo)(-1/se^(-sa)+1/se^(-s*0))=lim_(a->oo)(-1/se^(-sa)+1/s)

Note that if s<0, lim_(a->oo)(-1/se^(-sa)+1/s)=oo and the integral is divergent.

In case s>0, lim_(a->oo)(-1/se^(-sa)+1/s)=1/s and the integral is convergent.

So, L(1)=1/s, provided that s>0.

Example 2. Calculate L(e^(at)).

L(e^(at))=int_0^ooe^(-st)*e^(at)dt=int_0^ooe^((a-s)t)dt=lim_(b->oo)(int_0^be^((a-s)t)dt)=lim_(b->oo)(1/(a-s)e^((a-s)t)|_0^b)=

=lim_(b->oo)(1/(a-s)e^((a-s)b)-1/(a-s)e^((a-s)*0))=lim_(b->oo)(1/(a-s)e^((a-s)b)-1/(a-s))

Note that if a-s>0, we have thatlim_(b->oo)(1/(a-s)e^((a-s)b)-1/(a-s))=oo and the integral is divergent.

In case that a-s<0, lim_(b->oo)(1/(a-s)e^((a-s)b)-1/(a-s))=-1/(a-s)=1/(s-a) and the integral is convergent.

So, L(e^(at))=1/(s-a), 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 last two 'non-standard' examples.

Example 3. Calculate the Laplace transform of the derivative: L(f'(t)).

L(f'(t))=int_0^ooe^(-st)f'(t)dt=lim_(a->oo)(int_0^a e^(-st)f'(t)dt)

Using integration by parts:

lim_(a->oo)(int_0^a e^(-st)f'(t)dt)=lim_(a->oo)(int_0^a e^(-st)df(t))=lim_(a->oo)(e^(-st)f(t)|_0^a-int_0^a(-se^(-st)f(t))dt)=

=lim_(a->oo)(e^(-sa)f(a)-e^(-s*0)f(0)+s\int_0^a(e^(-st)f(t))dt)=

=lim_(a->oo)(e^(-sa)f(a)-f(0))+s\int_0^oo(e^(-st)f(t))dt=lim_(a->oo)(e^(-sa)f(a))-f(0)+sL(f(t))=

=lim_(a->oo)(e^(-sa)f(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)|<Ae^(\alphat) for all t>=B.

This condition implies that lim_(a->oo)(e^(-sa)f(a))=0, provided that s>\alpha.

So, L(f'(t))=sL(f(t))-f(0), provided that s>\alpha.

Example 4. Calculate the Laplace transform of the integral: L(int_0^tf(\tau)d\tau).

L(int_0^tf(\tau)d\tau)=int_0^oo(e^(-st)int_0^tf(\tau)d\tau)dt=int_0^oo(int_0^te^(-st)(f(\tau)d\tau))dt

On this stage, we need to change the order of integration. t is changing from 0 to oo, while \tau is changing from 0 to t. So, if we change the order of integration, we have that \tau is changing from 0 to oo and t is changing from \tau to oo:

int_0^oo(int_0^te^(-st)(f(\tau)d\tau))dt=int_0^oo(int_\tau^oo(e^(-st)f(\tau))dt)d\tau=int_0^oo(f(\tau)int_\tau^oo(e^(-st))dt)d\tau=int_0^oo(f(\tau)*(-1/s)e^(-st)|_\tau^oo)d\tau=int_0^oo(f(\tau)1/se^(-s\tau))d\tau=1/sL(f(t))=1/sF(s)

So, L(int_0^tf(\tau)d\tau)=(F(s))/s.

For a list of common Laplace transforms, see Table of Laplace Transforms.