I am trying to find the Laplace transform of $e^{kt}$.
So I have that
$$\mathcal{L} \{ e^{kt} \} = \int_{0}^\infty e^{kt} e^{-st} \ dt,$$
where $s$ is a complex number.
$$\begin{align} \int_{0}^\infty e^{kt} e^{-st} \ dt &= \int_{0}^\infty e^{t(k - s)} \ dt \\ &= \lim_{x \to \infty} \int_{0}^x e^{t(k - s)} \ dt \\ &= \dfrac{1}{k - s} \lim_{x \to \infty} \int_{0}^{x(k - s)} e^u \ du \\ &= \dfrac{1}{k - s} \lim_{x \to \infty} [e^u]^{x(k - s)}_0 \\ &= \dfrac{1}{k - s}\lim_{x \to \infty} \left[ e^{x(k - s)} - 1 \right] \\ &= \lim_{x \to \infty} e^{x(k - s)} - \dfrac{1}{k - s} \\ &= \lim_{x \to \infty} e^{xk}e^{-sx} - \dfrac{1}{k - s} \\ &= \lim_{x \to \infty} e^{xk}e^{-x(\Re(s) + \Im(s))} - \dfrac{1}{k - s} \\ &= \lim_{x \to \infty} e^{xk}e^{-x \Re(s)} \cos(-x \Im(s)) + i \sin(-x \Im(s)) - \dfrac{1}{k - s} \end{align}$$
The author claims that the solution is
$$\int_{0}^\infty e^{kt} e^{-st} \ dt = \dfrac{1}{s - k}$$
I don't see how this is the case. It seems to me that, whatever $\Re(s)$ happens to be, the solution will depend then on what $k$ is?
I would greatly appreciate it if people could please take the time to clarify this.
Note that $\int_0^\infty e^{- (s-k)t} dt $ only converges for $\operatorname{re} (s-k) >0$.
When this is true, we have $\lim_{t\to \infty} e^{- (s-k)t} = 0$.
Then $\int_0^\infty e^{- (s-k)t} dt = ({1 \over k-s} e^{- (s-k)t})\mid_{t=0}^{t=\infty} = { 1\over s-k}$.
Note: (You may want to ignore this if it seems like babble.) There is one rather technical point that used to bug me until I took a complex analysis course, which is the whole $\operatorname{re} s > R$ thing. In general the Laplace transform defines an analytic function $\hat{f}(s)=({\cal L f})(s)$ for some region of convergence, usually of the form $\operatorname{re} s > R$. In many cases, such as above, there is a function, call it $\hat{\phi}$, that is defined on a much larger (connected) set (so, in the above case the function $s \mapsto {1 \over s-k}$ is defined on $\mathbb{C} \setminus \{k\}$) and agrees with $\hat{f}$ for $\operatorname{re} s > R$. In this case we refer to $\hat{\phi}$ (the analytic continuation) as the Laplace transform of $s$ and more or less ignore the $\operatorname{re} s > R$ restriction subsequently.