Bromwich integral de $1/s^k$ con k real (no entero) y $1<k$

Question

¿Existe una forma sencilla de calcular la transformada inversa de Laplace de $1/s^k$ con k no entero utilizando la integral de Bromwich (básicamente sin utilizar la conocida transformada de laplace de $t^n$ )?

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\int_{0}^{-\infty}{\expo{st} \over \pars{-s}^{k}\expo{\ic\pi k}} \,{\dd s \over 2\pi\ic} + \int_{-\infty}^{0}{\expo{st} \over \pars{-s}^{k}\expo{-\ic\pi k}} \,{\dd s \over 2\pi\ic} \\[3mm]&= -\expo{-\ic\pi k}\int_{0}^{\infty}{\expo{-st} \over s^{k}}\,{\dd s \over 2\pi\ic} + \expo{\ic\pi k}\int_{0}^{\infty}{\expo{-st} \over s^{k}}\,{\dd s \over 2\pi\ic} ={\sin\pars{\pi k} \over \pi}\int_{0}^{\infty}s^{-k}\expo{-st}\,\dd s \\[3mm]&= {\sin\pars{\pi k} \over \pi}\,t^{k - 1}\int_{0}^{\infty}s^{-k}\expo{-s}\,\dd s ={\sin\pars{\pi k} \over \pi}\,t^{k - 1}\Gamma\pars{1 - k} ={\sin\pars{\pi k} \over \pi}\,t^{k - 1}\,{\pi \over \Gamma\pars{k}\sin\pars{\pi k}} \\[3mm]&=\color{#000}{\large{t^{k - 1} \over \Gamma\pars{k}}} \end{align}