The result below assumes $t>0$ (as usual for Laplace transforms).
In the Bromwich contour $\gamma$ has to be chosen large enough that it is to the right of all singularities (poles and branch points) so $\gamma = 0^+$ is perfectly valid. The singularities of $\log s/(1+s)$ are a pole at $s=-1$ and two branch points at $s=0$ and $s=\infty$ which we connect via a branch cut along the negative real line.
Then we can deform the contour further to a path which starts at $-\infty -i 0^+$. Runs along the negative real line just below the branch cut. Ends at $0-i 0^+$ in a little semi-circle and then runs back from $0+i 0^+$ to $-\infty +i0^+$ just above the branch cut.
The Bromwich integral thus is given by
$$f(t)=\frac1{2\pi i} \int_{-\infty}^0\!dx\, \left(
\frac{\log (x-i0^+)}{1+x-i0^+ } - \frac{\log (x+i0^+)}{1+x+i0^+ } \right) e^{x t} $$
as the small circle around the branch point at $0$ does not contribute ($|z|\log z \to 0$ for $|z|\to0$).
In the remaining integral, we use $\log(x \pm i 0^+) = \log |x| \pm i \pi$ valid for $x<0$:
$$\begin{align} f(t) &= \frac1{2\pi i} \int_{-\infty}^0\!dx\, \left(
\frac{\log |x|-i\pi}{1+x-i0^+ } - \frac{\log |x|+i\pi}{1+x+i0^+} \right) e^{x t}\\
&= \frac1{2\pi i} \overbrace{\int_{-\infty}^0\!dx\, \log |x| \underbrace{\left(
\frac1{1+x-i0^+ } - \frac1{1+x+i0^+}\right)}_{2\pi i \delta(x+1)} e^{x t}}^{=0}\\
&\quad -\frac12 \int_{-\infty}^0\!dx \underbrace{\left(
\frac1{1+x-i0^+ } + \frac1{1+x+i0^+}\right)}_{2\mathcal{P}\,(1+x)^{-1}} e^{x t} \\
&= -\int_{-\infty}^0\!dx \,\mathcal{P} \frac{e^{x t}}{1+x}
=- e^{-t} \int_{-\infty}^t\!ds \,\mathcal{P} \frac{e^{s}}{s}\\
&=- e^{-t} \mathop{\rm Ei}(t)
\end{align}$$
with $s=(1+x)t$ and Ei the exponential integral.
The way to attack a problem like this is via Cauchy's theorem on a properly distorted Bromwich contour. Here, we want our contour to avoid the branch point at $z=0$. This, we consider
$$\oint_C dz \, z^{-a-1} e^{-z^a} e^{z t} $$
where $a \in (0,1)$ and $C$ is the following contour:
We will define $\text{Arg}{z} \in (-\pi,\pi]$, so the branch is the negative real axis. There are $6$ pieces to this contour, $C_k$, $k \in \{1,2,3,4,5,6\}$, as follows.
$C_1$ is the contour along the line $z \in [c-i R,c+i R]$ for some large value of $R$.
$C_2$ is the contour along a circular arc of radius $R$ from the top of $C_1$ to just above the negative real axis.
$C_3$ is the contour along a line just above the negative real axis between $[-R, -\epsilon]$ for some small $\epsilon$.
$C_4$ is the contour along a circular arc of radius $\epsilon$ about the origin.
$C_5$ is the contour along a line just below the negative real axis between $[-\epsilon,-R]$.
$C_6$ is the contour along the circular arc of radius $R$ from just below the negative real axis to the bottom of $C_1$.
When $t \gt 0$, the integral over the contours $C_2$ and $C_6$ vanish in the limit as $R \to \infty$.
The contour integral is thus equal to, in this limit,
$$\int_{c-i \infty}^{c+i \infty} ds \, s^{-a-1} e^{-s^a} e^{s t} + e^{-i \pi a} \int_{\infty}^{\epsilon} dx \, x^{-a-1} e^{-e^{i \pi a} x^a} e^{-x t} \\ + i \epsilon^{-a} \int_{\pi}^{-\pi} d\phi \, e^{-i a \phi} e^{-\epsilon^a e^{i a \phi}} e^{\epsilon t e^{i \phi}} + e^{i a \pi} \int_{\epsilon}^{\infty} dx \, x^{-a-1} e^{-e^{-i \pi a} x^a} e^{-x t}$$
Note that there is an apparent singularity at $\epsilon = 0$; however, the divergences cancel in the limit as $\epsilon \to 0$.
In this limit, the third integral has the following leading behavior:
$$-i \frac{2}{a} \epsilon^{-a} \sin{\pi a} +i 2 \pi$$
Rescaling and combining the second and fourth integrals, we get for the contour integral:
$$\int_{c-i \infty}^{c+i \infty} ds \, s^{-a-1} e^{-s^a} e^{s t} -i 2 t^a \operatorname{Im}{\left [e^{-i \pi a}\int_{\epsilon t}^{\infty} \frac{du}{u^{1+a}} e^{-u} e^{-e^{i \pi a} t^{-a} u^a} \right]}-i \frac{2}{a} \epsilon^{-a} \sin{\pi a} +i 2 \pi $$
We may Taylor expand the second exponential in the integrand because it is subdominant to the first exponential (at least for the first $n$ terms, where $n$ is that largest integer such that $\lfloor n a \rfloor = 0$). We need only expand to the first two terms to treat the limit as $\epsilon \to 0$. Note that
$$\int_{\epsilon t}^{\infty} \frac{du}{u^{1+a}} e^{-u} = \frac{t^{-a}}{a} \epsilon^{-a} + \Gamma(-a) + O \left ( \epsilon^a \right ) $$
The second term produces
$$-e^{i \pi a} t^{-a}\int_{\epsilon t}^{\infty} \frac{du}{u} e^{-u} $$
Because the exponentials outside the integral cancel, the imaginary part of the term is zero. Thus, we now take the limit as $\epsilon \to 0$. Because the contour integral is zero by Cauchy's theorem, we get for the ILT,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-a-1} e^{-s^a} e^{s t} =
\frac{t^a}{\pi} \operatorname{Im}{\left [e^{-i \pi a}\int_{0}^{\infty} \frac{du}{u^{1+a}} e^{-u} \left ( e^{-e^{i \pi a} t^{-a} u^a} - 1 + e^{i \pi a} t^{-a} u^a \right ) \right]} + \frac{t^a}{\Gamma(1+a)} - 1$$
ADDENDUM
Even though the above result is suitable for numerical calculation, we can illustrate the above result with an analytical example. Consider the case $a=1/2$. Subbing $u=x^2$ and taking the imaginary part of the integral, we end up with
$$\operatorname{Im}{\left [e^{-i \pi a}\int_{0}^{\infty} \frac{du}{u^{1+a}} e^{-u} \left ( e^{-e^{i \pi a} t^{-a} u^a} - 1 + e^{i \pi a} t^{-a} u^a \right ) \right]} = 2 \int_{-\infty}^{\infty} dx \, e^{-x^2} \frac{\sin^2{\beta x}}{x^2}$$
where $\beta = 1/(2 \sqrt{t})$.
The latter integral may be evaluated using Parseval's theorem, because the individual factors of the integrand are inverse Fourier transforms of simple functions. For example,
$$\int_{-\infty}^{\infty} dx \, e^{-x^2} e^{i k x} = \sqrt{\pi} e^{-k^2/4} $$
$$\int_{-\infty}^{\infty} dx \, \frac{\sin^2{\beta x}}{x^2} e^{i k x} =\begin{cases} \pi \beta \left ( 1-\frac{|k|}{2 \beta} \right ) & |k| \lt 2 \beta \\ 0 & |k| \gt 2 \beta \end{cases}$$
The integral is then equal to
$$2 \frac1{2 \pi} \sqrt{\pi} \pi \beta \int_{-2 \beta}^{2 \beta} dk \, \left ( 1-\frac{|k|}{2 \beta} \right ) e^{-k^2/4} = \sqrt{\pi} \beta \int_0^{2 \beta} dk \, \left ( 1-\frac{k}{2 \beta} \right ) e^{-k^2/4}$$
The evaluation is fairly straightforward using the definition of the error function. The result is, for the integral,
$$2 \pi \beta \operatorname{erf}{\beta} - 2 \sqrt{\pi} \left (1-e^{-\beta^2}\right ) $$
Now plugging this back into the main result above and using $\beta = 1/(2 \sqrt{t})$, we get that
$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-3/2} e^{-s^{1/2}} e^{s t} &= \operatorname{erf}{\left ( \frac1{2 \sqrt{t}} \right )} + \frac{2}{\sqrt{\pi}} \sqrt{t} e^{-\frac1{4 t}} - 1 \\ &= \frac{2}{\sqrt{\pi}} \sqrt{t} e^{-\frac1{4 t}} - \operatorname{erfc}{\left ( \frac1{2 \sqrt{t}} \right )}\end{align}$$
Best Answer
Hint.
As $\hat f(s) = \frac{1}{1+e^s}$ the residues for $\hat f(s)$ are located at $1+e^s = 0$ or over the imaginary axis at $(2k+1)\pi i,\ \ k\in \mathbb{Z}$ and $Res[\hat f((2k+1)\pi i)] = -1$ then
$$ \hat f(s) =-\lim_{n\to \infty}\sum_{k=0}^{k=n}\frac{2s}{s^2+(2k+1)^2\pi^2} $$
also
$$ f(t) = -2\lim_{n\to\infty}\sum_{k=0}^{k=n}\cos\left((2k+1)\pi t\right) $$
Follows the graphics for $\int_0^T f(t)dt$ with $n = 20$