# Properties of the Laplace Transform

## Related Calculators: Laplace Transform Calculator , Inverse Laplace Transform Calculator

Laplace transform has some interesting properties.

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

Proof is straightforward through definition:

L(a(f(t)+bg(t)))=int_0^ooe^(-st)(af(t)+bg(t))dt=int_0^oo(af(t)e^(-st)+bg(t)e^(-st))dt.

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 property will still hold.

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

From 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 following:

L(5e^(3t)-2sin(7t)+cos(t))=5L(e^(3t))-2L(sin(7t))+L(cos(t))=

=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 positive constant.

Proof is straightforward:

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

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

So,

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)

Proof.

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 limits of integration are constant):

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

Or

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. Calculate L(tsin(t))

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

L(tsin(t))=-d/(ds)1/(s^2+1)=1/(s^2+1)^2*(d(s^2+1))/(ds)=(2s)/(s^2+1)^2

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

Proof.

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 Laplace transform of both sides gives:

L(f(t))=L(tg(t))

Or

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

Integrating this equality with respect to s and taking 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 then lim_(s->oo)G(s)=0 .

So,

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

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

This completes the proof.

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

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

L((sin(t))/t)=int_s^oo 1/(p^2+1)dp=arctan(p)|_s^oo=arctan(oo)-arctan(s)=pi/2-arctan(s)=a r c c o t(s)