In my friends’ uni notes I have found this formula: $$\tag{1} L\{r^t\}(s) = \frac{1-e^{-s}}{s(1-re^{-s})}$$, however, I can’t find a proof of this formula anywhere.
Edit: the actual formula that's in the notes uses assumption $(4)$ but it doesn't state that later which became the source of my confusion. If we include that assumption in the formula directly it becomes $$ L\{r^{\lfloor t \rfloor} \}(s) = \frac{1-e^{-s}}{s(1-re^{-s})} $$ Proof to which I have included in the answer to this question
Afaik the formula should be like this: $$\tag{2} L\{r^t\}(s) = \frac{1}{s-\ln{r}}$$ The context in which this formula occurs are second order difference equations (e.g. $$\tag{3} y(t+2) -8y(t+1) + 12y(t) = 0 \\ y(0) = 1, y(1) = 1, \forall_{t<0}\ y(t)=0 $$. My friend says that they don’t remember seeing a proof of this formula and the notes they have are from 2-3 years before they started the course. They also don’t seem to know of any textbooks that the lecturers are refencing. I think the extra assumption that is used here is that the $y(t)$ function is a step function instead of a regular one (i.e. $$\tag{4} \forall_{t\in\mathbb{Z},\ 0\leq\epsilon<1}\ y(t+\epsilon) = y(t)$$
Could I ask someone to point me to some resources where I could find more information on the topic (ideally the proof of this formula and other techniques related to solving second order difference equations). Also I tried to solve the equation $(3)$ in a different way using Laplace transform (but without using the formula $(1)$) but I couldn’t get anywhere with this. I tried to use this formula $$\tag{5} L\{u_k(t)y(t-k)\}=e^{-ks}Y(s)$$ to get (by taking Laplace of both sides) $$\tag{6} e^{2s}Y(s) -8e^{s}Y(s) +12Y(s) = 0$$ but at this point it’s not clear to me how should I proceed. My guess is that I should use the initial conditions somehow but I don’t know how.
Ok I have figured it out. It was indeed necessary to use property $(4)$. Instead of writing it like that I will say that I’m taking the Laplace transform of $f(t) = r^{\lfloor t \rfloor}$. For any interval from $t$ to $t+1$, where $t \in \mathbb{Z}$ the function $f(t)$ takes on the value $r^t$. If we take the Laplace transform of $f(t)$ we get: $$ L\{f(t)\}(s) = \int_0^\infty e^{-st}f(t)dt$$ If we break it up into intervals of length $1$ we get: $$ \begin{align}L\{f(t)\}(s) &= \int_0^1 e^{-st}f(t)dt + \int_1^2 e^{-st}f(t)dt + \int_2^3 e^{-st}f(t)dt + … \\ &= \int_0^1 e^{-st}r^0dt + \int_1^2 e^{-st}r^1dt + \int_2^3 e^{-st}r^2dt + … \\ &= -\frac{r^0}{s} e^{-st}|^1_0 -\frac{r^1}{s} e^{-st}|^2_1 -\frac{r^2}{s} e^{-st}|^3_2 - … \\ &= \frac{r^0}{s} (1 - e^{-s}) + \frac{r^1}{s} (e^{-s} - e^{-2s}) + \frac{r^2}{s} (e^{-2s} - e^{-3s}) + … \end{align}$$ The last bit should be clear that it’s a geometric series with ratio $r e^{-s}$ and first term $\frac{1 - e^{-s}}{s}$ so that sum is equal to $$ \frac{1 - e^{-s}}{s(1-re^{-s})}$$ which concludes the proof of $(1)$