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8bitgrafix
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hi guys, this is literally my first post here on physicsforums so i apologize in advance that my latex formatting sucks.
prove the alternative form of the 2nd translation theorem of the laplace transform:
L[f(t)u(t-a)]=e^{-sa}L[f(t+a)]
where u(t-a) is the unit step function and L[f(t)]=\int^\infty_0e^{-st}f(t)dt is the definition of the laplace transform of f(x)
L[f(t)u(t-a)]=\int^\infty_0 e^{-st}f(t)u(t-a)\,dt
=\int^a_0 e^{-st}f(t)u(t-a)\,dt + \int^\infty_a e^{-st}f(t)u(t-a)\,dt
the first integral is 0 since the unit step function 0 for any t < a, and u(t-a)=1 for t >= a so the 2nd integral becomes, \int^\infty_a e^{-st}f(t)\,dt
then since the laplace transform is an integral from 0 to \infty we need to make the substitutions u = t-a, \ t = u+a,\ du = dt to change the lower limit of integration from a to 0. then the integral becomes, \int^\infty_0 e^{-s(u+a)}f(u+a)\,du
then you pull the e^{-sa} outside the integral to give e^{-sa}\int^\infty_0 e^{-su}f(u+a)\,du and this is where I'm stuck. I don't know what to do here since the definition of the laplace transform is L[f(u)]=\int^\infty_0e^{-su}f(u)du and i can't make any more substitutions without changing the limits of integration.
Homework Statement
prove the alternative form of the 2nd translation theorem of the laplace transform:
L[f(t)u(t-a)]=e^{-sa}L[f(t+a)]
where u(t-a) is the unit step function and L[f(t)]=\int^\infty_0e^{-st}f(t)dt is the definition of the laplace transform of f(x)
Homework Equations
The Attempt at a Solution
L[f(t)u(t-a)]=\int^\infty_0 e^{-st}f(t)u(t-a)\,dt
=\int^a_0 e^{-st}f(t)u(t-a)\,dt + \int^\infty_a e^{-st}f(t)u(t-a)\,dt
the first integral is 0 since the unit step function 0 for any t < a, and u(t-a)=1 for t >= a so the 2nd integral becomes, \int^\infty_a e^{-st}f(t)\,dt
then since the laplace transform is an integral from 0 to \infty we need to make the substitutions u = t-a, \ t = u+a,\ du = dt to change the lower limit of integration from a to 0. then the integral becomes, \int^\infty_0 e^{-s(u+a)}f(u+a)\,du
then you pull the e^{-sa} outside the integral to give e^{-sa}\int^\infty_0 e^{-su}f(u+a)\,du and this is where I'm stuck. I don't know what to do here since the definition of the laplace transform is L[f(u)]=\int^\infty_0e^{-su}f(u)du and i can't make any more substitutions without changing the limits of integration.