Inverse Laplace transform via the table formulas In my inverse Laplace table there is this inversion "formula":
$(1) \frac{1}{s-a} \rightarrow e^{at}$
I understand that
$\mathcal{L}^{-1}[\frac{1}{s+4}] = \frac{1}{2}\sin(2t)$
But why can I not do the following:
$\mathcal{L}^{-1}[\frac{1}{s+4}] = \mathcal{L}^{-1}[\frac{1}{s-(-4)}]$
Using $(1): \mathcal{L}^{-1}[\frac{1}{s-(-4)}]=e^{-4t}$?
 A: Let's try to understand what is going on with the inverse Laplace transform. The inverse Laplace transform is
$$
\mathcal{L}^{-1}\{F(s)\} = \frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty}F(s)e^{st}ds =\sum\operatorname{Res}\{F(s);s_j\}
$$
where the residues are the poles of $F(s)$. With your first equation, we have a pole when $s = a$; that is, division by zero. The pole is of order one since we have $1/(s-a)^1$. Now, let's find the inverse Laplace transform
$$
\mathcal{L}^{-1}\{1/(s-a)\}=\frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{st}}{s-a}ds =\lim_{s\to a}(s-a)\frac{e^{st}}{s-a}=e^{at}
$$
Now you seem to have some confusion with your next ILT. Again, let's tackle this by definition.
$$
\mathcal{L}^{-1}\{1/(s+4)\}=\frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{st}}{s+4}ds =\lim_{s\to -4}(s+4)\frac{e^{st}}{s+4}=e^{-4t}
$$
In order to obtain $1/2\sin(2t)$, we need $F(s) = \frac{1}{s^2+4}$. Let's prove this to ourselves. In the case, $s^2+4=(s-2i)(s+2i)$ so we have two simple poles of order one at $s=\pm 2i$.
\begin{align}
\mathcal{L}^{-1}\{1/(s^2+4)\}&=\frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{st}}{s^2+4}ds\\ 
&=\lim_{s\to -2i}(s+2i)\frac{e^{st}}{s^2+4}+\lim_{s\to 2i}(s-2i)\frac{e^{st}}{s^2+4}\tag{1}\\
&=\frac{e^{-2it}}{-4i}+\frac{e^{2it}}{4i}\tag{2}
\end{align}
In equation $(1)$, you see we summed the two residues. Now to reduce equation $(2)$, we just need to recall a trig identity. That is,
$$
\sin(t) = \frac{e^t-e^{-t}}{2i}
$$
so equation $(2)$ is really $1/2\sin(2t)$.
