inverse laplace transformation of $\arctan(\frac{4}{s})$ inverse laplace transformation of $\arctan(\frac{4}{s})$ using 

I was trying use 12 but i couldn't arrive to a solution 
 A: Here's a solution that avoids convergence problems of the series representation.  Note that
$$F(s) = \arctan{\left ( \frac{4}{s} \right )} \implies F'(s) = -\frac{4}{s^2+16} $$
Then, by 12 and 5,
$$t f(t) = \sin{4 t}$$
A: Since:
$$\arctan\frac{4}{s}=\sum_{k=0}^{+\infty}\frac{(-1)^k}{2k+1}\left(\frac{4}{s}\right)^{2k+1} $$
and $\mathcal{L}^{-1}\left(\frac{1}{s^{2k+1}}\right)=\frac{t^{2k}}{(2k)!}$ by $(2)$, we have:
$$ \mathcal{L}^{-1}\left(\arctan\frac{4}{s}\right) = \sum_{k=0}^{+\infty}\frac{(-1)^k 4^{4k+1}}{(2k+1)!}\,t^{2k}=\color{red}{\frac{\sin(4t)}{t}}.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}&\color{#66f}{\large%
\int_{\gamma - \infty\ic}^{\gamma - \infty\ic}
\arctan\pars{4 \over s}\expo{st}\,{\dd s \over 2\pi\ic}}
=\int_{\gamma - \infty\ic}^{\gamma - \infty\ic}\expo{st}\int_{0}^{1}{4s\,\dd x \over 16x^{2} + s^{2}}\,{\dd s \over 2\pi\ic}
\\[5mm]&=4\int_{0}^{1}\int_{\gamma - \infty\ic}^{\gamma - \infty\ic}
\expo{st}{s\over s^{2} + 16x^{2}}\,{\dd s \over 2\pi\ic}\,\dd x
=4\int_{0}^{1}\bracks{\expo{-4\ic xt}\,{-4\ic xt \over -8\ic xt}
+\expo{4\ic xt}\,{4\ic xt \over 8\ic xt}}\,\dd x
\\[5mm]&=4\int_{0}^{1}\cos\pars{4xt}\,\dd x
=\color{#66f}{\large{\sin\pars{4t} \over t}}
\end{align}
