Laplace transform of the following: $f(t)=\int\limits_{0}^{t} \frac{\cosh (\tau) - 1}{\tau} d\tau$

Question

I thought it would be a simple one, however, that integral of $f(t)$ cannot be expressed in terms of standard functions...and I'm pretty much confused.

What should I do?

Answer 1

We have $$L(\cosh t-1)=\frac{s}{s^2-1}-\frac{1}{s}$$ so $$L(\frac{\cosh t-1}{t})=\int_{s}^{\infty}(\frac{u}{u^2-1}-\frac{1}{u})du$$ and $$Lf(t)=\frac{1}{s}\int_{s}^{\infty}(\frac{u}{u^2-1}-\frac{1}{u})du=\frac{1}{s}\left[Ln\frac{\sqrt{u^2-1}}{u}\right]_{s}^{\infty}=\frac{1}{s}Ln\frac{s}{\sqrt{s^2-1}}$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \tilde{\fermi}\pars{s}}\equiv\int_{0}^{\infty}\fermi\pars{t}\expo{-st}\,\dd t =\int_{0}^{\infty}\expo{-st}\int_{0}^{t}{\cosh\pars{\tau} - 1 \over \tau} \,\dd\tau\,\dd t \\[5mm]&=\int_{0}^{\infty} {\cosh\pars{\tau} - 1 \over \tau}\int_{\tau}^{\infty}\expo{-st}\,\dd t\,\dd\tau ={1 \over s}\int_{0}^{\infty} {\cosh\pars{\tau} - 1 \over \tau}\expo{-s\tau}\,\dd\tau \\[5mm]&={1 \over 2s}\int_{0}^{\infty} {\expo{\pars{1 - s}\tau} + \expo{-\pars{1 + s}\tau} - 2\expo{-s\tau} \over \tau}\,\dd\tau \\[5mm]&=-\,{1 \over 2s}\int_{0}^{\infty}\ln\pars{\tau}\bracks{% {\expo{\pars{1 - s}\tau}\pars{1 - s} -\expo{-\pars{1 + s}\tau}\pars{1 + s}} + 2\expo{-s\tau}s}\,\dd\tau \\[5mm]&=-\,{1 \over 2s}\,\lim_{\mu\ \to\ 0}\partiald{}{\mu} \int_{0}^{\infty}\bracks{-\tau^{\mu}\expo{-\pars{s - 1}\tau}\pars{s - 1} -\tau^{\mu}\expo{-\pars{s + 1}\tau}\pars{s + 1} +2\tau^{\mu}\expo{-s\tau}s}\,\dd\tau \\[5mm]&=-\,{1 \over 2s}\,\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{% -\,{1 \over \pars{s - 1}^{\mu}} - {1 \over \pars{s + 1}^{\mu}} +{2 \over s^{\mu}}}\Gamma\pars{\mu + 1} \\[5mm]&=\color{#66f}{\large -\,{1 \over 2s}\,\ln\pars{1 - {1 \over s^{2}}}} \end{align}