How to Prove the Laplace Transform of f(t/b) Equals bF(bs) for b ≠ 0?

[magnifik]

show that L{f(t/b)} = bF(bs), b is not equal to 0

i know that
L{f(t)} = \inte^-stf(t) dt = F(s)
so
L{f(t/b)} = \inte^-stf(t/b) dt

any tips on how to start? thx

 

[Dick]

Do a u substitution to change the integration variable. Substitute u=t/b.

 

[magnifik]

do i also have to plug in (t/b) into e^-st so that it becomes e^-s(t/b) ?

 

[Dick]

do i also have to plug in (t/b) into e^-st so that it becomes e^-s(t/b) ?

You need to write e^(-st) in terms of u.

 

[magnifik]

so..
u = t/b
t = bu
e^(-st)
e^(-sbu)

 

[Dick]

so..
u = t/b
t = bu
e^(-st)
e^(-sbu)

Sure. Now don't forget to add a factor to change the dt to a du.