Inverse Laplace transform after derivative of transform. I've been using the following theorem in my intro ODE course:

If $F(s) = \mathscr{L}\left\{ f(t) \right\}$ and $n\in\mathbb{N}$, then $\mathscr{L}\left\{t^n f(t) \right\} = (-1)^n\frac{d^n}{ds^n}F(s).$

What is the inverse of this transform? For instance, say I wanted to take $$ \mathscr{L}^{-1}\left\{ \frac{s}{(s^2+16)^2} \right\}.$$ I can see that the result would be $\frac{1}{8}t\sin(4t)$. But, how would one go about manually working this out (and particularly in general)? 
 A: Observe that $$\int \frac s{(s^2+16)^2}ds=-\frac1{s^2+16}+C\iff \frac{d \left(\frac1{s^2+16}\right)}{ds} =- \frac s{(s^2+16)^2}$$
and from this, $$\mathscr{L}^{-1}\left\{ \frac1{(s^2+16)} \right\}=\frac{\sin4t}4$$ 
So, $$\mathscr{L} \left\{ t^1\cdot\frac{ \sin 4t}4 \right\}=(-1)^1\frac{d^1}{ds^1}\left(\frac1{s^2+16}\right)=+\frac{2s}{(s^2+16)^2} $$
A: ILTs like this may be evaluated using the residue theorem: the ILT is, in the this case, the sum of the residues at the poles to the left of the vertical line of integration.  Here, the poles at $s_{\pm}=\pm 4 i$ are double poles.  
The residue of a double pole $s=s_0$ of a function of the form
$$f(s) = \frac{p(s)}{q(s)^2}$$
is given by the expression
$$\text{Res}_{s=s_0} \frac{p(s)}{q(s)^2} = \frac{p'(s_0)}{q'(s_0)^2} - \frac{q''(s_0) p(s_0)}{q'(s_0)^3}$$
In this case, $p(s)=s e^{s t}$ and $q(s)=s^2+16$.  Using the above ingredients, it is not hard to show that the ILT is
$$\begin{align}\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{s}{(s^2+16)^2} e^{s t}\\ &= \left [\frac{1+i 4 t}{-64} - \frac{8 i}{-512 i} \right ] e^{i 4 t} +  \left [\frac{1-i 4 t}{-64} - \frac{-8 i}{512 i} \right ] e^{i 4 t} \\ &= \frac18 t \, \sin{4 t}\end{align}$$
