The Borel-Laplace transform of a transeries that contains logarithms

Question

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.

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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.

Answer 1

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.