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.

The contour $\Gamma_theta$

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.

  • $\begingroup$ Your question suggests you may be familiar with the work of Dudko-Sauzin concerning Fatou coordinates: arxiv.org/abs/1307.8095 Several years ago I asked them a question similar to yours. Their answer was the transseries for inverses to Fatou coordinates - these should typically contain infinitely many terms involving iterated logarithms. I do not know if this has been worked out explicitly in an accessible reference. $\endgroup$ – Adam Epstein Nov 15 '17 at 13:24
  • $\begingroup$ Yes, I've seen it, It does not help. The log term is treated in the beginning of section 2. There $v$ is assumed to be of the form $z+\rho \log z +\phi(z)$ and $\phi$ does not have log terms. The inverse of $v$ is a transseries with infinitely many log terms. So even though the treat the non vanishing resiter case, it is kind of trivial in the end (in the sense that the classical Borel transform suffices). $\endgroup$ – tst Nov 17 '17 at 4:19

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 $$ \mathcal L[l_n*l_k](t) = \mathcal L[l_n](t)\cdot\mathcal L[l_m](t). $$


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.

  • 1
    $\begingroup$ That is interesting. I have 2 questions though. How certain are you that the theory is correct? After all, it is a non-standard theory. Also how do you know that there is no constant? How do you know that the arbitrary constants cancel out? $\endgroup$ – rom Nov 23 '17 at 23:32
  • $\begingroup$ I added an explanation for your first question. You were right about the second, there is a non-vanishing constant. Thanks for that! $\endgroup$ – tst Nov 24 '17 at 19:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.