Definition of the Laplace Transform

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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 that`lim_(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.