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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.

askquestions4
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  • I would start by writing the integral definition and making a change of variable. That is, $$L(f(t-a)) = \int e^{-st} f(t-a) \ dt$$ and set $y = t-a.$ – Sean Roberson Sep 30 '22 at 18:05
  • Right I tried this, but the issue is that I get stuck with for $t+1 = v$ that $e^s \int_1^{\infty}f(v)e^{-sv}dv$. See that $1$ in the lower bound? Now my laplace transform is ruined. – askquestions4 Sep 30 '22 at 18:13
  • @SeanRoberson I meant – askquestions4 Sep 30 '22 at 18:19
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    Use a clever "add zero" trick. Add and subtract $$\int_0 ^a f(v)e^{-sv} \ dv $$ and you'll have something you can work with (remember how to mess with limits of integration!). – Sean Roberson Sep 30 '22 at 18:21
  • @SeanRoberson Yep I thought of that too, I'm still stuck because them I'm left with $\int_{0}^{1}h(v)e^{-sv}dv$ for which there is no known general formula. – askquestions4 Sep 30 '22 at 18:23
  • $f(t+1)=tf(t)$ is the Gamma function which unfortunately has no Laplace transform. – Kurt G. Sep 30 '22 at 18:42
  • Hmm, then I guess we multiply it by a characteristic function? – askquestions4 Sep 30 '22 at 18:43
  • @KurtG. Shouldn't we be able to "discover" that the laplace transform of a particular function satisfies this functional relationship anyway? – askquestions4 Sep 30 '22 at 22:59
  • $f(t+1)=tf(t)$ or the new $f(t+1)=f(t)/t$ ? If you Laplace transform the simple function $F(x)=e^{2x}$ to $f(t)=1/(t-2)$ you see that $f$ doesn't satisfy these relationships. Meanwhile I am also quite confused what the question is now. As far as I can tell the question started with how to Laplace transform $f$. Time for an edit with clarifications I guess. – Kurt G. Oct 01 '22 at 07:58
  • I'm looking for a formula for $L[f(t+a)],$ the right-hand expression $t f(t)$ or $f(t)/t$ is irrelevant, it's just a placeholder, it could be any expression. The central focus is the $f(t+a)$ term, that's the term I actually care about transforming. – askquestions4 Oct 01 '22 at 14:09

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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.

Leucippus
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