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^ooe^(-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 integral is divergent.

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

So, `L(1)=1/s` provided 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 then `lim_(b->oo)(1/(a-s)e^((a-s)b)-1/(a-s))=oo` and integral is divergent.

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

So, `L(e^(at))=1/(s-a)`, provided s>a.

Note, that we put restriction on s in order to compute Laplace Transform. In general, all Laplace transforms have restrictions on s.

Let's go through last two "non-standard" examples.

Example 3. Calculate Laplace Transform of 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 restriction: assume that exist 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 `s>\alpha`.

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

Example 4. Calculate Laplace Transform of 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 change order of integration. t is changing from 0 to `oo` , while `\tau` is changing from 0 to t. So, if we change order of integration then `\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 list of common Laplace Transforms see Table of Laplace Transforms.