I am having trouble with the derivation of the Laplace transform of the heaviside step function and a time-shifted function.
$$u_c(t) = \Big \lbrace_{1, \; t \geq c}^{0, \;t \leq c}$$
Deriving: $$ L \lbrace u_c(t)f(t-c) \rbrace = \int _{0}^{\infty} e^{-st} u_c(t)f(t-c) dt$$
Since the step function is 0 until t = c, the integral bounds change and the function is changed to 1: $$ L \lbrace u_c(t)f(t-c) \rbrace = \int _{c}^{\infty} e^{-st} f(t-c) dt$$
Change of variable: $$ \tau = t -c; \; d \tau = dt$$
The integral bounds change as well, at t=c:
$$ L \lbrace u_c(t)f(t-c) \rbrace = \int _{0}^{\infty} e^{-s(\tau + c)} f(\tau) d \tau$$ $$ L \lbrace u_c(t)f(t-c) \rbrace = \int _{0}^{\infty} e^{-s\tau -sc} f(\tau) d \tau$$ $$\boxed{ L \lbrace u_c(t)f(t-c) \rbrace = e^{-sc} \underbrace{\int _{0}^{\infty} e^{-s\tau} f(\tau) d \tau}_{= L \lbrace f(\tau) \rbrace} = e^{-sc}F(s) }$$
Now approaching it again, but substituting back in for tau: $$ L \lbrace u_c(t)f(t-c) \rbrace = e^{-sc} \underbrace{\int _{0}^{\infty} e^{-s(t-c)} f(t-c) d t}_{= L \lbrace f(t-c) \rbrace}$$
What seems confusing to me is by substituting back in t for tau, it seems as if you need to do the Laplace transform on the timeshifted function as opposed to the non-timeshifted function.
Without substitution: $$L \lbrace u_c(t)f(t-c) \rbrace = e^{-sc} L \lbrace f(t) \rbrace$$
With substitution: $$L \lbrace u_c(t)f(t-c) \rbrace = e^{-sc} L \lbrace f(t-c) \rbrace$$
Can someone explain where I have made the error that makes these two approaches diverge?
EDIT: Answering in case someone stumbles on this in the future. As eyeballfrog said, if the integral is evaluated from c to infinity, then the two approaches are equivalent.
$$\boxed{ L \lbrace u_c(t)f(t-c) \rbrace = e^{-sc} \underbrace{\int _{c}^{\infty} e^{-s(t-c)} f(t-c) d t}_{= L \lbrace f(t-c) \rbrace} }$$
With an example $$u_c(t) = \Big \lbrace_{1, \; t \geq 1}^{0, \;t \leq 1}$$
$$f(t) = t - (t-1)u_c$$
$$L \lbrace f(t) \rbrace = L \lbrace t \rbrace + L \lbrace u_c(t-1) \rbrace$$
$$L \lbrace t \rbrace = \frac{1}{s^2}$$ $$L \lbrace u_c (t-1) \rbrace = \int _{1}^{\infty} e^{-s(t-1)}(t-1) \; dt = \frac{e^{-s}}{s^2}$$