Inverse Laplace with $\ln$

Question

How can I compute the inverse Laplace of

1) $\ln\left(\dfrac{s+1}{s-1}\right)$

2) $\ln\left(\dfrac{s-1}{s}\right)$. Can someone please hep me to do these two problems

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\large\tt\mbox{With}\quad \gamma > 1:}$ \begin{align}&\color{#66f}{\large% \int_{\gamma - \infty\ic}^{\gamma + \infty\ic} \ln\pars{s + 1 \over s - 1}\,\expo{st}\,{\dd s \over 2\pi\ic}} =\int_{\gamma - \infty\ic}^{\gamma + \infty\ic} \int_{-1}^{1}{\dd x \over x + s}\,\expo{st}\,{\dd s \over 2\pi\ic} \\[3mm]&=\int_{-1}^{1}\int_{\gamma - \infty\ic}^{\gamma + \infty\ic} {\expo{st} \over s + x}\,{\dd s \over 2\pi\ic}\,\dd x =\int_{-1}^{1}\expo{-xt}\,\dd x ={\expo{-t} - \expo{t} \over -t} = \color{#66f}{\large{2\sinh\pars{t} \over t}} \end{align}

The other one can be evaluated in a similar way.

Answer 2

Given \begin{align} \mathcal{L}\{f(t); s\} = \int_{0}^{\infty} e^{-st} f(t) \ dt \end{align} then the inverse Laplace transforms are given by the following \begin{align} \mathcal{L}\{ \frac{2 \sinh(at)}{t}; s\} = \ln\left(\frac{s+a}{s-a}\right) \end{align} and \begin{align} \mathcal{L}\{ \frac{1 - e^{at}}{t}; s\} = \ln\left(\frac{s-a}{s}\right). \end{align}

The process to find the values given above is to use the relation \begin{align} \mathcal{L}\{ \frac{f(t)}{t}; s\} = \int_{s}^{\infty} g(u) \ du \end{align} where \begin{align} g(s) = \int_{0}^{\infty} e^{- st} f(t) \ dt. \end{align}

Answer 3

Podes usar algunas propiedades de las transformadas.

[You can use some properties of the transforms]

Ej; [Example]

$$((-1)^n) (t^n) f(t)=F^n(s)$$

Donde f(t) es la antitransformada de $F^n(s)$.

[Where $f(t)$ is in the inverse transform of $F^{n}(s)$.