Is The Inverse Laplace Transform of $e^{st}\operatorname{Log}\left(\frac{s+1}{s}\right)$ doable using inversion formula? I'm trying to solve inverse laplace transform using inversion formula and given by this integral:
$$\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i \infty} e^{st}\operatorname{Log}\left(\frac{s+1}{s}\right)\,\Bbb ds.$$
Here is my contour, since the branch points of $\operatorname{Log} \left(\frac{s+1}{s}\right)$ are $0$ and $-1$

First, i want to show integral on $L_u\cup L_d$ is $0$ by bounding the integral with ML and then take the limit when $R$ goes to $\infty$.
By letting $L_u,\, L_d: s= \xi\pm iR,0\leq \xi\leq \gamma$, where $\gamma$ is the real number that the vertical line of the given contour passed by.
Since the $L$ is $\lvert e^{t(\xi\pm iR)} \rvert$, then i have ML inequality as below:
$$\lvert F(s)e^{st} \rvert \leq M_R \lvert e^{t(\xi\pm iR)} \rvert = M_R e^{\xi t} \leq M_R e^{at}$$
Next, i need to find $M_R$ and take the limit.
$$\begin{align}
\left|F(s)\right| &= \left|\operatorname{Log}\left(\frac{s+1}{s}\right)\right|\\
&= \left|\operatorname{Log}\left(\frac{\xi\pm iR+1}{\xi\pm iR}\right)\right| = M_R
\end{align}$$
And by taking the limit of the last expression when $R$ goes to infinity yields $0$. Meaning the integrals along those lines are $0$.
So, from here, am i doing this right? I'm not sure my work is correct. Maybe there are some mistakes there. Help me please!
Edit:
Working with my $L_u$ with $ML$ inequality, i have $L=\gamma$. Assuming $-\pi<\operatorname{arg}{s}\leq \pi$ and parametrizing $s=-\xi+iR$, $\xi\in [-\gamma,0]$:
$\begin{align}
\left|\int_{L_u}\right| 
&\leq \left|e^{st} \log\left(1 + \frac 1s\right)\right|\\
&\leq \left|e^{-\xi t}\right| \left|e^{iRt}\right|\left|\ln\left|1+ \frac{1}{-\xi+iR}\right| + i\pi\right|\\
&\leq 1\cdot 1 \cdot \ln\left(1+\frac 1R\right) + \pi\\
&\approx \frac 1R + \pi
\end{align}$
Combining the $ML$ i have
$$\frac{\gamma}{R} +\gamma\pi$$
Which does NOT approach to $0$. Why? Please spot my mistake. I can't think about how to make it goes to $0$ since yesterday. Hope you kind to help me.
New Edit: Now my main question is not about ML. I managed to set the big and small arc goes to $0$. My main question now is "is it possible to evaluate this by using inversion formula (without differentiate both sides) and use log form instead?"
What i mean by differentiate both sides is:
$$\begin{align}
\mathcal{L}^{-1} &= \operatorname{Log}\left(\frac{s+1}{s}\right)\\
(\mathcal{L}^{-1})' &= \frac{1}{1+s} - \frac{1}{s}
\end{align}$$
I don't want this kind of solution. What i really want is evaluating $\operatorname{Log}\left(\frac{s+1}{s}\right)$ by inversion formula. That's it.
because I keep getting integrals on the contour lines above and below the branch cut cancel each other and make everything 0, which of course I don't want it.
Attempt:

 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\LARGE\left. a\right)}$
One idea, in this particular case, is to get rid of the $\ds{\ln}$-function by introducing an integral representation of it. Namely,
\begin{align}
&\bbox[5px,#ffd]{\left.{1 \over 2\pi\ic}
\int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty} \expo{st}\ln\pars{s + 1 \over s}
\,\dd s\,\right\vert_{\,\gamma\ >\ 0}}
\\[5mm] = &\
\int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty} \expo{st}\
\overbrace{\int_{0}^{1}{\dd x \over x + s}}
^{\ds{\ln\pars{s + 1 \over s}}}\
\,{\dd s \over 2\pi\ic}
\\[5mm] = &\
\int_{0}^{1}\int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty} 
{\expo{st} \over s + x}\,{\dd s \over 2\pi\ic}\,\dd x
\\[5mm] = &\
\int_{0}^{1}\bracks{t > 0}\expo{-xt}\,\dd x =
\bbx{\bracks{t > 0}\,{1 - \expo{-t} \over t}} \\ &
\end{align}

$\ds{\LARGE\left. b\right): {\large Contour Integration}}$'
Note that $\ds{{s + 1 \over s} < 0}$ whenever
$\ds{s \in \pars{-1,0}}$: It indicates that the $\mbox{$\ds{\ln}$-branch-cut}$ lies along $\ds{\bracks{-1,0}}$. Also,
$\ds{{s \pm \ic\epsilon + 1 \over s \pm \ic\epsilon}
\,\,\,\stackrel{{\rm as}\ \epsilon\ \to\ 0^{+}}{\sim}
\,\,\, \pars{1 + s \over s} \mp {\epsilon \over s^{2}}\,\ic}$ such that the $\ds{\ln}$-$\ds{arg}$ is $\pars{-\pi}$ when $\ds{s}$ is above the real axis and $\ds{\pi}$ when it's below the real axis.
\begin{align}
&\bbox[5px,#ffd]{\left.{1 \over 2\pi\ic}
\int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty} \expo{st}\ln\pars{s + 1 \over s}
\,\dd s\,\right\vert_{\,\gamma\ >\ 0}}
\\[5mm] = &\
-\int_{-\infty}^{-1}\ln\pars{s + 1 \over s}\expo{ts}
{\dd s \over 2\pi\ic}\label{1}\tag{1}
\\[2mm] - &\
\int_{-1}^{0}\bracks{\ln\pars{-s - 1 \over s} - \ic\pi}\expo{ts}
{\dd s \over 2\pi\ic}
\\[2mm] - &\
\int_{0}^{-1}\bracks{\ln\pars{-s - 1 \over s} + \ic\pi}\expo{ts}
{\dd s \over 2\pi\ic}
\\[2mm] - &
\int_{-1}^{-\infty}\ln\pars{s + 1 \over s}\expo{ts}
{\dd s \over 2\pi\ic}\label{2}\tag{2}
\end{align}
I included (\ref{1}) and (\ref{2}) "by completeness" but, indeed, it's not necessary because the $\ds{\ln}$-branch-cut lies along $\ds{\bracks{-1,0}}$
albeit it guarantees the vanishing out of the integration along the"big-arc".
Then,
\begin{align}
&\bbox[5px,#ffd]{\left.{1 \over 2\pi\ic}
\int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty} \expo{st}\ln\pars{s + 1 \over s}
\,\dd s\,\right\vert_{\,\gamma\ >\ 0}}
\\[5mm] = &\
-\int_{0}^{1}\bracks{\ln\pars{-s + 1 \over s} - \ic\pi}\expo{-ts}
{\dd s \over 2\pi\ic}
\\[2mm] &\
\,\, + \int_{0}^{1}\bracks{\ln\pars{-s + 1 \over s} + \ic\pi}\expo{-ts}
{\dd s \over 2\pi\ic}
\\[5mm] = &\
\int_{0}^{1}\expo{-ts}\,\dd s =
\bbx{1 - \expo{-t} \over t} \\ &
\end{align}
A: Let's pick a sufficiently large $\gamma$ such that $|s|>1$, so that we can plug in
$$
\log\left(1+\frac1s\right)=\sum_{k=1}^\infty{(-1)^{k+1}\over ks^k}
$$
and get
$$
\begin{aligned}
f(t)
&={1\over2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\sum_{k=1}^\infty{(-1)^{k+1}e^{st}\over s^k}\mathrm ds \\
&=\sum_{k=1}^\infty{(-1)^{k+1}\over k}{1\over2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}{e^{st}\over s^k}\mathrm ds
\end{aligned}
$$
Provided that
$$
\mathcal L\{t^{k-1}\}(s)={(k-1)!\over s^k}
$$
we can deduce
$$
\begin{aligned}
f(t)
&=\sum_{k=1}^\infty{(-t)^{k-1}\over k!} \\
&=-\frac1t\left[\sum_{k=0}^\infty{(-t)^k\over k!}-1\right] \\
&={1-e^{-t}\over t}
\end{aligned}
$$
where the final step follows from the fact that
$$
e^z=\sum_{n=0}^\infty{z^k\over n!}
$$
