Inversion of Laplace transform $F(s)=\log(\frac{s+1}{s})$ (Bromwich integral) I am looking for the inversion of Laplace transform $F(s)=\log(\frac{s+1}{s})$.
I started by using the general formula of the Bromwich integral:
$\displaystyle \lim_{R\to\infty} \int_{a-iR}^{a+iR} \frac{1}{2\pi i}\log\left(\frac{s+1}{s}\right) e^{st}ds $
Then, I used that: $\displaystyle \log\left(\frac{s+1}{s}\right)=\sum_{n=1}^{\infty} \frac{ (-1)^{n+1} }{n} (1/s)^n $ for $|s|>1$.
Since  $|s|>1$ the Bromwich line should be to the right of $1$.
So:
$\displaystyle \sum_{n=1}^{\infty} \lim_{R\to\infty} \int_{a-iR}^{a+iR} \frac{1}{2\pi i}\frac{ (-1)^{n+1} }{n} (1/s)^n  e^{st}ds $
Can someone help me to continue this?
 A: EDIT
I am not happy with the original explanation of how the integrals over the arcs vanish as $R \to \infty$.  The point is that, while we can split the integral up to treat one branch point at a time, it is only when we combine the integrals that we get cancellation of the contribution from the arcs.  I am editing the solution below to reflect that.

I don't think your expansion is the right way to go.  Rather, you want to define a Bromwich contour as a closed loop. Unfortunately, you have branch points at $s=-1$ and $s=0$, so the contour would need to circle around those so as to avoid them.  You then use Cauchy's theorem to find the ILT in terms of a real-valued integral that you may evaluate.
In this case, I would split the log into two pieces, i.e., $\log{(1+1/s)} - \log{(1+s)}-\log{s}$.  Now we wish to compute
$$\int_{c-i R}^{c+iR} ds \, \log{(1+s)} \, e^{s t}$$
where we will take the limit as $R \to \infty$, and we only require that $c \gt 0$, or generally, to the right of the pole having maximum real part.
Consider then
$$\oint_C dz \, \log{(1+z)} \, e^{z t}$$
where $C$ is the Bromwich contour including the vertical line $[c-i R,c+i R]$, plus an arc of radius $R$ traversed counterclockwise from $c+i R$ to $c-i R$. However, there is a branch cut along the negative real line up to $z=-1$, so the contour also includes a traversal up the negative real line to $z=-1$ from the arc, a circle about $z=-1$, and a traversal back up the negative real axis to the arc.  Thus, the above contour integral may be written as several pieces:
$$\int_{c-i R}^{c+iR} ds \, \log{(1+s)} \, e^{s t} + i R \int_{\pi/2}^{\pi} d\theta\, e^{i \theta} \, \log{(1+R e^{i \theta})}\, e^{R t \cos{\theta}} \, e^{i R t \sin{\theta}} \\ + e^{i \pi} \int_R^{1 + \epsilon} dx \, \log{(1+x e^{i \pi})} + i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \,\log{(1+\epsilon e^{i \phi})} e^{\epsilon e^{i \phi}t}\\ + e^{-i \pi} \int_{1 + \epsilon}^R dx \, \log{(1+x e^{-i \pi})} + i R \int_{-\pi}^{-\pi/2} d\theta\, e^{i \theta} \, \log{(1+R e^{i \theta})}\, e^{R t \cos{\theta}} \, e^{i R t \sin{\theta}}$$
Now, as $\epsilon \to 0$, the fourth integral vanishes.  Note that, by Cauchy's theorem, the contour integral is $0$; thus, we have
$$\int_{c-i R}^{c+iR} ds \, \log{(1+s)} \, e^{s t} = \int_1^{R} dx \left [\log{(1+x e^{-i \pi})}-\log{(1+x e^{i \pi})} \right ] \, e^{-x t} \\ + i 2 R \int_0^{\pi/2} d\theta \, \Im{\left [e^{i(\theta+R t \cos{\theta})} \log{\left ( 1+R e^{i \theta} \right )} \right ]} e^{-R t \sin{\theta}}$$
Note the way I wrote this; it is suggestive of the fact that, at the branch point $x=0$, there is a change of phase by $2 \pi$.  In fact, you may have noticed that the argument of the log terms are negative, and so we must add $\pm i \pi$ accordingly; we know which sign because of the choice of phase we made along the way.  Thus,
$$\int_1^{R} dx \left [\log{(1+x e^{-i \pi})}-\log{(1+x e^{i \pi})} \right ] \, e^{-x t} = \int_1^{R} dx \left [-i \pi +\log{(x-1)}- i \pi - \log{(x-1)} \right ] \, e^{-x t} \\ = -i 2 \pi \int_1^{R} dx\, e^{-x t}$$
Thus
$$\frac{1}{i 2 \pi}\int_{c-i R}^{c+iR} ds \, \log{(1+s)} \, e^{s t} = -\frac{e^{-t} - e^{-R t}}{t}\\ + i 2 R \int_0^{\pi/2} d\theta \, \Im{\left [e^{i(\theta+R t \cos{\theta})} \log{\left ( 1+R e^{i \theta} \right )} \right ]} e^{-R t \sin{\theta}}$$
We will deal with that ugly piece later, when we are able to consider the limit as $R \to \infty$.  
Using a similar reasoning as above, but having a branch point at $z=0$ rather than $z=-1$, you may show that
$$\frac{1}{i 2 \pi}\int_{c-i R}^{c+iR} ds \, \log{s} \, e^{s t} = -\frac{1-e^{-R t}}{t}\\ + i 2 R \int_0^{\pi/2} d\theta \, \Im{\left [e^{i(\theta+R t \cos{\theta})} \log{\left ( i R e^{i \theta} \right )} \right ]} e^{-R t \sin{\theta}}$$
Now we may consider the ugly pieces, which we can now subtract; the result is
$$i 2 R \int_0^{\pi/2} d\theta \, \Im{\left [e^{i(\theta+R t \cos{\theta})} \log{\left (1+ \frac{i}{R} e^{-i \theta} \right )} \right ]} e^{-R t \sin{\theta}}$$
which, as $R$ gets large, becomes asymptotically, using $\log{(1+y)} \sim y$ for small $y$:
$$i 2 \int_0^{\pi/2} d\theta \, \cos{(R t \cos{\theta})} \, e^{-R t \sin{\theta}}$$
whose magnitude is bounded by
$$2 \int_0^{\pi/2} d\theta \, e^{-R t \sin{\theta}} \le 2 \int_0^{\pi/2} d\theta \, e^{-2 R t \theta/\pi} \le \frac{\pi}{R t}$$
Thus, the integrals over the arcs vanish as $R \to \infty$, and in this limit, we may finally write the sought-after ILT as
$$\frac{1}{i 2 \pi}\int_{c-i \infty}^{c+i\infty} ds \, \log{\left (1+\frac{1}{s}\right)} \, e^{s t} = \frac{1-e^{-t}}{t}$$
NOTE
I could have (in fact, should have) done this using a simple exterior contour minus an interior contour (a "dogbone") around the branch points at $z=0$ and $z=-1$, which would have helped to avoid all of the cancelling of the divergences above. 
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\LARGE\left.a\right)}$
\begin{align}
&\mbox{By definition,}\quad\ln\pars{s + 1 \over s} = \int_{1}^{\pars{s + 1}/s}{\dd z \over z}\qquad
\left\{\substack{%
where\ the\ integration\ path\ doesn't\ pass\ through\ the\ origin
\\[1mm]
or\ cross\ the\ negative\ real\ axis.}\right.
\\[5mm] = &
\int_{0}^{1}{\dd z \over z + s}\ \mbox{such that}\
\int_{\pars{-z}^{+} - \infty\ic}^{\pars{-z}^{+} + \infty\ic}
{\expo{ts} \over s + z}\,{\dd s \over 2\pi\ic} = \expo{-zt}
\\[5mm] &\
\implies \int_{0}^{1}\expo{-tz}\,\dd z = \bbx{1 - \expo{-t} \over t}\\ &
\end{align}
