The Borel-Laplace transform of a transeries that contains logarithms I am interested in Ecalle's generalization of the Borel-Laplace summation. I would like to see an explicit treatment of a summation of a transeries that include logarithmic terms.
The only example I have seen is the construction of Fatouc coordinates in the case of non-vanishing resiter, however this series has a single logarithm term that can just be subtracted.
If this is not possible, I would like to know if a closed form for the following integral can be found
$$ I(z) = \int_{\Gamma_\theta} \frac{\log s}{2 \pi i s} e^{-z s} ds. $$
The contour of integration is shown in the following figure.


A bit of context
We define the Laplace transform as 
$$L^{\theta}[f](z) = \int_{e^{i\theta}\mathbb{R}^+} f(s) e^{-z s} ds. $$
The classical Borel transform is defined as the formal inverse of Laplace transform. Since for any $\theta$ we have $L^\theta[s^n](z)=n!z^{-n-1}$ we define the Borel transform as
$$B[z^{-n-1}](s) = \frac{s^n}{n!}.$$
Beacause the Borel transform introduces a factorial, it may happen that the Borel transform of a formal series can be a germ. If this germ can be extended towards infinity, we may be able to take its Laplace transform. This procedure is called Borel-Laplace summation. This procedure however can deal only with negative integer powers of $z$.
Ecalle generalized this, by defining the Laplace transform of a function $F$ by defining 
$$\mathcal{L}^{\theta}[F](z) = \int_{\Gamma_\theta} F(s) e^{-z s} ds, $$
with $\Gamma_\theta$ being the contour in the above figure.
If $f$ is entire, then we define $F(s)=f(s)\frac{\log s}{2\pi i}$ and we have
$$ L^{\theta}[f](z) = \mathcal{L}^{\theta}[F](z). $$
In this setting the laplace transform of $\frac{\log s}{2 \pi i s}$ should the logarithm or something very close.
 A: I figured out an answer to this. My answer assumes that the theory works, so it cannot be used to validate it, but I feel comfortable with this assumption.
As above, I will use the following definition of the Laplace transform
$$\mathcal{L}^{\theta}[F](z) = \int_{\Gamma_\theta} F(s) e^{-z s} ds.$$
However, an important point that I ignored before is what really is this $F$.
Assume that $F$ is a function with the origin as a possible ramification point. Then if $\phi$ is entire, $\mathcal{L}^{\theta}[\phi](z) = 0$, which of course implies that $\mathcal{L}^{\theta}[F](z) = \mathcal{L}^{\theta}[F+\phi](z)$.
This indicates that the proper space on which the transform is to be applied is some kind of space of functions with ramification points quotient by analytic functions. I will not go into details here, but it is important to keep in mind that $F = F + \phi$ for any analytic $\phi$.
I realized that the solution to my question is to look at the differential equation $$\dot x(t)=1/t.$$
This of course can be solved by a simple integration, so we have
$$ x(t) = \log (t) + c. $$
Now we can solve the same equation after taking the Borel transform. If $B[x](z) = \hat x(z)$, then $$B[\dot x(t)] = -z\, \hat x(z).$$
For the classical Borel transform we have $B[1/t] = 1$, however we need to use the generalized Borel transform so we define
$$ \mathcal B [1/t](z) = \frac{\log(z)}{2\pi i}. $$
Now if we recall the above discussion about the nature of the space where we apply the transform, we see that it is more accurate if we write
$$ \mathcal B [1/t](z) = \frac{\log(z)}{2\pi i} + \phi(z) $$
with $\phi$ being any analytic function.
Finally the simple differential equation above can be written as
$$ -z\, \hat x(z) = \frac{\log(z)}{2\pi i} + \phi(z). $$
Which can be solved to 
$$ \hat x(z) = -\frac{\log(z)}{2\pi i z} +\frac{\phi(0)}{z} + \frac{\phi(z) - \phi(0)}{z} . $$
The last term is analytic, so it vanishes with the Laplace transform. The constant $\phi(0)$ plays the role of the integration constant, so finally for any $\theta$ we can write
$$ \mathcal L^\theta\left[ \frac{\log(z)}{2\pi i z} \right](t) = -\log(t). $$
Let's define
$$l_n(z) = \frac{\log(z)}{2\pi i z^n}. $$
Using integration by parts we get
$$ \mathcal L[l_{n+1}](t) = -\frac{t}{n}\mathcal L[l_n](t) + \frac{(-1)^n t^n}{n\,n!}.$$
The solution to this recurrence equation is
$$ \mathcal L[l_n](t) = \frac{(-1)^{n}t^{n-1}}{(n-1)!}\log(t)+(-1)^{n-1} t^{n-1}\sum_{k=1}^{n-1}\frac{(n-1-k)!}{(n-k)!(n-1)!}. $$
Similarly we can resolve the convolution after the Laplace transform by writing
$$ \mathcal L[l_n*l_k](t) = \mathcal L[l_n](t)\cdot\mathcal L[l_m](t). $$
ADDED LATER:
Let's define $$ x(t) = - \int_\Gamma e^{-st} \frac{\log(s)}{2\pi i s}ds. $$
Then it's derivative is:
$$ \dot{x}(t)= \int_\Gamma e^{-st} \frac{\log(s)}{2\pi i}ds.$$
This integral (by collapsing the path $\Gamma$ to a line) is equal to $ \int_0^\infty e^{-st} ds=\frac{1}{t}$, so we finally get
$$ \dot{x}(t) =  \frac{1}{t}.$$
So $x(t) = \log(t) + c$.
But unfortunately I was wrong above and the constant does not vanish. We have $c = x(1)$ and the relations for $l_n$'s have to change.
