Laplace transform of non-unit-multiplied delay, $f(t +a)$? This wiki article says https://en.wikipedia.org/wiki/Laplace_transform
that (where $L$ is the laplace transform) $L( f(t-a)u(t-a) ) = e^{-as}F(s).$
However, I don't have a unit-step-function multiplied in my expression, I have
$f(t + 1) = f(t)/t$
and I'm getting unexpected errors that leads me to believe I did something incorrectly. So I'm going back over me work and I think I found where I went wrong, this leaves me with a question:
what is the laplace transform of just plain old "$f(t - a)$" and NOT $f(t-a)u(t-a)$?
The other option is we multiply both sides by the delayed unit step function
$f(t+1) \cdot u(t+a) = f(t)/t \cdot u(t + a),$
but then taking the Laplace transform of both sides gives me another functional equation in terms of $F(s-a)$ that doesn't appear any easier than the start.
 A: It would seem that
\begin{align}
f(t+a) &\doteqdot \int_{0}^{\infty} e^{-s t} \, f(t+a) \, dt \\
&\doteqdot \int_{a}^{\infty} e^{-s (u-a)} \, f(u) \, du \hspace{5mm} u = t+a \\
&\doteqdot e^{a s} \, \left( \int_{0}^{\infty} e^{-s u} \, f(u) \, du - \int_{0}^{a} e^{- s u} \, f(u) \, du \right) \\
&\doteqdot e^{a s} \, \mathcal{L}\{f(t)\} - e^{a s} \, \mathcal{L}_{f}\{ f(t) \}
\end{align}
where $\mathcal{L}\{f(t)\}$ is the regular Laplace transform and $\mathcal{L}_{f}\{ f(t) \}$ is the finite Laplace transform. The definitions are:
\begin{align}
\mathcal{L}\{f(t); t \to s, t \in (0, \infty)\} &= \int_{0}^{\infty} e^{-s t} \, f(t) \, dt \\
\mathcal{L}_{f}\{f(t); t \to s, t \in (0, x)\} &= \int_{0}^{x} e^{-s t} \, f(t) \, dt.
\end{align}
Another way would be to consider the expansion
$$ f(t + a) = \sum_{n=0}^{\infty} \frac{b_{n}}{n!} \, (t + a)^{n} = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{b_{n}}{n!} \, \binom{n}{k} \, a^{n-k} \, t^{k} $$
This suggests the Laplace transform would take the form:
\begin{align}
f(t+a) &\doteqdot \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{b_{n}}{n!} \, \binom{n}{k} \, a^{n-k} \, \mathcal{L}\{t^k \} \\
&\doteqdot \frac{1}{s} \, \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{a^{n-k} \, b_{n}}{(n-k)! \, s^k} \\
&\doteqdot \frac{1}{s} \, \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{a^{n} \, b_{n+k}}{n! \, s^{k}}.
\end{align}
From here it would depend on the coefficients $b_{n}$ to reduce the series to some known form.
