Inverse laplace transform excercise I want to find the inverse transform of $$\frac{1}{(2s-1)^3}$$
I first applied a shifting theorem to get $$(e^t)\mathcal{L}^{-1}\left( \frac{1}{(2s)^3} \right)$$
I am just wondering is it possible to take the 2 out from here so it becomes 
$$\frac{1}{8}(e^t)\mathcal{L}^{-1}\left( \frac{1}{s^3} \right)$$
Any feed back would be much appreciated
 A: I'm busy, what's this all about? (Tl;dr) You have a small calculation error, but yes, you would be able to do it because $\mathcal{L}^{-1}$ is a linear operator.
Let's take a step back and think about what we are doing, shall we? :)
One of the Laplace transform properties is that
$$\mathcal{L}^{-1} (F(s-a)) = e^{at} \mathcal{L}^{-1}(F(s)).$$
This is great. Let's apply this in our case. We find that the root is $a = \frac{1}{2}$. Oops! This means that what you really have is that
$$e^{\frac{1}{2}t} \mathcal{L}^{-1} \left( F(s) \right).$$
But what is $F(s)$? Let's rewrite it:
$$\frac{1}{(2s-1)^3} = \frac{1}{\left( 2 \left( s - \frac{1}{2} \right) \right)^3} = \frac{1}{8} \frac{1}{\left( s - \frac{1}{2} \right)^3}.$$
Hence the answer is
$$\mathcal{L}^{-1} \left( \frac{1}{(2s-1)^3} \right) = \frac{1}{8} e^{\frac{1}{2}t} \mathcal{L}^{-1} \left( \frac{1}{s^3} \right).$$
A: We could use the inverse Laplace transform integral/Bromwich Integral/Mellin-Fourier integral/Mellin's inverse formula (many names) and then use Residue theory. The pole is at $s = 1/2$ of order three.
The Bromwich contour is a line from $\gamma - i\infty$ to $\gamma + i\infty$ and then a partial circle connecting the line. 
\begin{align}
\frac{1}{2\pi i}\int_{\gamma + -i\infty}^{\gamma + i\infty}\frac{e^{st}}{(2s - 1)^3}ds &= \sum\text{Res}\\
&= \lim_{s\to 1/2}\frac{1}{(3 - 1)!}\frac{d^{3 - 1}}{ds^{3 - 1}}(s - 1/2)^3\frac{e^{st}}{(2s - 1)^3}\\
&= \lim_{s\to 1/2}\frac{1}{2}\frac{d^2}{ds^2}(s - 1/2)^3\frac{e^{st}}{8(s - 1/2)^3}\\
&= \frac{t^2e^{s/2}}{16}
\end{align}
