Properties of the Laplace Transform

Related Calculators: Laplace Transform Calculator , Inverse Laplace Transform Calculator

The Laplace transform has a number of interesting properties.

Property 1. Linearity of the Laplace transform: `L(af(t)+bg(t))=aL(f(t))+L(g(t))`, where `a` and `b` are some constants.

The proof is straightforward through the definition:


Now, using linearity, we can write that `int_0^oo(af(t)e^(-st)+bg(t)e^(-st))dt=a int_0^ooe^(-st)f(t)dt+b int e^(-st)g(t)dt=aL(f(t))+bL(g(t))`.

This completes the proof.

Note that there can be more than 2 terms, and the property will still hold.

Example 1 . Calculate `L(5e^(3t)-2sin(7t)+cos(t))`.

From the table of Laplace transforms, it is known that `L(e^(3t))=1/(s-3)`, `L(sin(7t))=7/(s^2+7^2)`, and `L(cos(t))=s/(s^2+1^2)`; so, using Property 1, we obtain the following:


`=5/(s-3)-2 7/(s^2+49)+s/(s^2+1)=5/(s-3)-14/(s^2+49)+s/(s^2+1)`.

Property 2. Scaling by constant: `L(f(at))=1/sF(s/a)`, where `a` is a positive constant.

The proof is straightforward:

`L(f(at))=int_0^oo e^(-st)f(at)dt`.

Using the change of variable technique, we make the substituion `u=at`; then, `du=adt`, which yields `dt=(du)/a`. Since `t` is changing from `0` to `oo`, we have that `u` is changing from `a*0=0` to `a*oo=oo`.


`int_0^oo e^(-st)f(at)dt=int_0^oo e^(-su/a)f(u)(du)/a=1/a int_0^oo e^(-s/au)f(u)du=1/aF(s/a)`.

This completes the proof.

Property 3. `L(tf(t))=-d/(ds)F(s)`, where `L(f(t))=F(s)`.


From the definition, it follows that

`F(s)=int_0^oo e^(-st)f(t)dt`.

Differentiating this equality with respect to `s` gives (note that the limits of integration are constant):

`d/(ds)F(s)=int_0^oo d/(ds)e^(-st)f(t)dt`,


`d/(ds)F(s)=int_0^oo -te^(-st)f(t)dt`,

which finally gives

`-d/(ds)F(s)=int_0^oo te^(-st)f(t)dt=L(tf(t))`.

This completes the proof.

Example 2. Calculate `L(tsin(t))`.

Since `L(sin(t))=1/(s^2+1)`, using property 3, we obtain that


Property 4. `L(f(t)/t)=int_s^ooF(u)du`, where `F(s)=L(f(t))`.


This property follows from property 3.

Let `g(t)=f(t)/t` and `L(g(t))=G(s)`. We need to prove that `G(s)=int_s^oo F(u)du`.

So, `f(t)=tg(t)`.

Taking the Laplace transform of both sides gives:



`F(s)=-d/(ds)G(s)` (using property 3).

Integrating this equality with respect to `s` and taking the limits `s` and `oo` yields:

`int_s^oo -d/(ds)G(s)ds=int_s^ooF(u)du`.

`-(lim_(a->oo)G(a)-G(s))=int_s^oo F(u)du`.

Since `G(s)=int_0^oo e^(-st)g(t)dt`, we have that `lim_(s->oo)G(s)=0`.


`-(0-G(s))=int_s^oo F(u)du`.

`G(s)=int_s^oo F(u)du`.

This completes the proof.

Example 3. Calculate `L((sin(t))/t)`.

Since `L(sin(t))=1/(s^2+1)`, using property 4, we obtain that

`L((sin(t))/t)=int_s^oo 1/(p^2+1)dp=lim_(m->oo)int_s^m 1/(p^2+1)dp=lim_(m->oo)atan(p)|_s^m=`