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

Question

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

I was trying use 12 but i couldn't arrive to a solution

Answer 1

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,

$$t f(t) = \sin{4 t}$$

Answer 2

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: $$ \mathcal{L}^{-1}\left(\arctan\frac{4}{s}\right) = \sum_{k=0}^{+\infty}\frac{(-1)^k 4^{4k+1}}{(2k+1)!}\,t^{2k}=\color{red}{\frac{\sin(4t)}{t}}.$$

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}