I am not very good at memorizing things, so I opted for the original(?) method of Laplace transformation: $$\int_0^\infty e^{-sxf(x)} dx$$
In the example and solution section, I found a puzzling answer where $f(x)$ suddenly come down:
Find the Laplace transformation of $f(x)= e^{ax}$
Solution:
$$F(s)L[e^{ax}]=\int_0^\infty e^{-sx}e^{ax}dx=\lim\limits_{R \to \infty} \int_0^R e^{(a-s)x}dx=\lim\limits_{R \to \infty}[\frac {e^{(a-s)}x} {a-s}]=\lim\limits_{R \to \infty}[\frac {e^{(a-s)^R} -1} {a-s}]=\frac {1}{s-a}(s>a)$$
I have two questions:
1.The other examples just used the $L$ symbol while solving the example, while this one used F(s) as well, was this intentional or is it a typo of sort? If it is intentional why did they use it?
2.As stated in the title, why did the function come down? Did they skip a dozen steps or does this have something to do with euler's number odd traits?
Thanks in advance!
It's just a typo. You'll usually write either $L[f]$ (or a variant) or $F(s)$. Further, the transform is defined as
$$ L[f] = \int_0 ^\infty e^{-sx} f(x) \ dx $$
where $f(x)$ is multiplied by the exponential. In general, integral transforms are operators $I$ for which
$$ I[f] = \int_\mathbb{R} K(s, x) f(x) \ dx $$
for some kernel $K$. Here the kernel is $K(s, x) = e^{-sx}$.