Tubular neighborhoods of orbits of power-logarithmic germs

Abstract

We consider a class of power-logarithmic germs. We call them parabolic Dulac germs, as they appear as Dulac germs (first-return germs) of hyperbolic polycycles. In view of formal or analytic characterization of such a germ f by fractal properties of several of its orbits, we study the tubular $\varepsilon$-neighborhoods of orbits of f with initial points $x_0$. We denote by $A_f(x_0,\varepsilon )$ the length of such a tubular $\varepsilon$-neighborhood. We show that, even if f is an analytic germ, the function $\varepsilon \mapsto A_f(x_0,\varepsilon )$ does not have a full asymptotic expansion in $\varepsilon$ in the scale of powers and (iterated) logarithms. Hence, this partial asymptotic expansion cannot contain necessary information for analytic classification. In order to overcome this problem, we introduce a new notion: the continuous time length of the $\varepsilon$-neighborhood $A^c_f(x_0,\varepsilon )$. We show that this function has a full transasymptotic expansion in $\varepsilon$ in the power, iterated logarithm scale. Moreover, its asymptotic expansion extends the initial, existing part of the asymptotic expansion of the classical length $\varepsilon \mapsto A_f(x_0,\varepsilon )$. Finally, we prove that this initial part of the asymptotic expansion determines the class of formal conjugacy of the Dulac germ f.

Appendix

Proof of Proposition 3.5

Take ${\widehat{F}}\in \widehat{{\mathcal {L}}}_1^I$ and let $F\in {\mathcal {G}}$ be its one generalized integral sum, as in (3.4). It is sufficient to verify that the transfinite algorithm of Poincaré described in [9, Section 3.1] applied to F with respect to any coherent section ${\mathbf {s}}$ gives the asymptotic expansion ${\widehat{F}}$. It is sufficient to prove the following:

• 1. That the terms of ${\widehat{F}}\in \widehat{{\mathcal {L}}}_1^I$ from (3.3) can be regrouped as:

  $$\begin{aligned} {\widehat{F}}(y)=\sum _{i=1}^{\infty } {\widehat{g}}_i\big (\varvec{\ell }(y)\big ) y^{\alpha _i}, \end{aligned}$$

  (10.1)

  where $(\alpha _i)$ is a strictly increasing sequence of real numbers tending to $+\infty$ or a finite sequence and ${\widehat{g}}_i$ are convergent.

• 2. At the same time, that the generalized integral sum F of ${\widehat{F}}$ given in (3.4) satisfies:

  $$\begin{aligned} F(y)-\sum _{i=1}^{n} g_i\big (\varvec{\ell }(y)\big )y^{\alpha _i}=o(y^{\alpha _n}),\ y\rightarrow 0,\ n\in {\mathbb {N}}, \end{aligned}$$

  (10.2)

  where $g_i$ are the sums of the convergent transseries ${\widehat{g}}_i$ and $(\alpha _i)$ are the same as in 1.

First, by Fubini’s theorem and absolute convergence of transseries ${\widehat{G}}_i$, $i=0,1,$ and ${\widehat{h}}$ in (3.3), we have that:

$$\begin{aligned}&{\widehat{G}}_i(y)=\sum _{j=1}^{\infty }{\widehat{g}}_j^i\big (\varvec{\ell }(y)\big ) y^{\beta _j^i},\ i=0,1,\nonumber \\&{\widehat{h}}(y)=a y^{\gamma _1}+\sum _{j=2}^{\infty } {\widehat{h}}_j\big (\varvec{\ell }(y)\big ) y^{\gamma _j},\ a\in {\mathbb {R}}. \end{aligned}$$

(10.3)

Here, $(\beta _j^i)_{j}$ and $(\gamma _j)_j$ are strictly increasing and tending to $+\infty$ or a finite sequence, and ${\widehat{g}}_j^i\in \widehat{{\mathcal {L}}}_0^\infty$, $i=0,1,$ and ${\widehat{h}}_j\in \widehat{{\mathcal {L}}}_0^\infty$, $j\in {\mathbb {N}}$, are convergent with the sums $g_j^i$, $h_j$ respectively. For the sums $G_i\in {\mathcal {G}},\ i=0,1,$ and $h\in {\mathcal {G}}$ of convergent transseries from (3.4), it also holds that:

$$\begin{aligned}&G_i(y)-\sum _{j=1}^{n}g_j^i\big (\varvec{\ell }(y)\big ) y^{\beta _j^i}=o(x^{\beta _n^i}),\ n\in {\mathbb {N}},\ i=0,1,\nonumber \\&h(y)-a y^{\gamma _1}-\sum _{j=2}^{n} h_j\big (\varvec{\ell }(y)\big ) y^{\gamma _j}=o(y^{\gamma _n}),\ y\rightarrow 0,\ n\in {\mathbb {N}}. \end{aligned}$$

(10.4)

It is easy to see, with ${\widehat{h}}$ convergent and h as above, that:

$$\begin{aligned} \varvec{\ell }(e^{-\frac{\gamma }{y}}{\widehat{h}}(y))=\frac{y}{\gamma }+{\widehat{k}}(y), \end{aligned}$$

(10.5)

as well as that

$$\begin{aligned} \varvec{\ell }(e^{-\frac{\gamma }{y}} h(y))=\frac{y}{\gamma }+ k(y), \end{aligned}$$

where ${\widehat{k}}\in \widehat{{\mathcal {L}}}$ is a convergent transseries with the sum $k\in {\mathcal {G}}$. In particular,

$$\begin{aligned} {\widehat{k}}(y)=c y^{\delta _0}+\sum _{i=2}^{\infty }{\widehat{k}}_i(\varvec{\ell }(y)) y^{\delta _i},\ k(y)-cy^{\delta _0}- \sum _{i=2}^{n} k_i(\varvec{\ell }(y)) y^{\delta _i}=o(y^{\delta _n}),\ n\in {\mathbb {N}}, \end{aligned}$$

where $(\delta _i)_i$ are strictly increasing to $+\infty$ or a finite sequence, $\delta _0>1$, $c \in {\mathbb {R}}$, and ${\widehat{k}}_i\in \widehat{{\mathcal {L}}}_0^\infty$ are convergent power asymptotic expansions of $k_i\in {\mathcal {G}}$, $i\in {\mathbb {N}}$.

Now suppose that in (3.3) ${\widehat{f}}(y)=\sum _{k=N}^{\infty } a_k y^k \in \widehat{{\mathcal {L}}}_0^I$, $N\in {\mathbb {Z}}$, with exponent of integration $\alpha$. Then

$$\begin{aligned} \frac{d}{dy}\Big (y^\alpha {\widehat{f}}\big (\varvec{\ell }(y)\big )\Big )=y^{\alpha -1}{\widehat{R}}\big (\varvec{\ell }(y)\big ), \end{aligned}$$

with ${\widehat{R}}\in \widehat{{\mathcal {L}}}_0^\infty$ convergent Laurent series. Let f be its integral sum. Then, by [9, Remark 3.13], f admits ${\widehat{f}}$ as its power asympotic expansion. Moreover, differentiating (3.1) and (3.2) and since ${\widehat{R}}$ is a convergent Laurent series, inductively it follows that $f^{(k)}$ admits the formal derivative ${\widehat{f}}^{(k)}$, $k\in {\mathbb {N}}$, as its power asymptotic expansion. Indeed, inductively, ${\widehat{f}}^{(k)}$ is a finite combination of ${\widehat{f}}$, ${\widehat{R}}$, powers of $\varvec{\ell }$ and the formal derivatives ${\widehat{R}}'$, $\ldots$, ${\widehat{R}}^{(k-1)}$, $k\in {\mathbb {N}}$. The same combination holds for the germ counterparts.

Using (10.5), we have the following Taylor expansions (formal and for germs):

$$\begin{aligned} {\widehat{f}}\big (\varvec{\ell }\big (e^{-\frac{\gamma }{y}}{\widehat{h}}(y)\big )\big )={\widehat{f}}\big (\frac{y}{\gamma }\big ) +{\widehat{f}}'\big (\frac{y}{\gamma }\big ){\widehat{k}}(y)+\frac{1}{2!}{\widehat{f}}'' \big (\frac{y}{\gamma }\big ){\widehat{k}}(y)^2+\cdots , \end{aligned}$$

(10.6)

$$\begin{aligned} f\big (\varvec{\ell }\big (e^{-\frac{\gamma }{y}} h(y)\big )\big ) =f\big (\frac{y}{\gamma }\big )+ f'\big (\frac{y}{\gamma }\big ) k(y) +\frac{1}{2!} f''\big (\frac{y}{\gamma }\big ) k(y)^2+\cdots . \end{aligned}$$

(10.7)

Combining (10.3) with (10.6), as well as on the other hand (10.4) with (10.7), we conclude (10.1) formally for ${\widehat{F}}\in \widehat{{\mathcal {L}}}_1^I$ and analogously (10.2) for its sum F. Here, $g_i\in {\mathcal {G}}$ are exactly the sums of convergent series ${\widehat{g}}_i$, $i\in {\mathbb {N}}$, since they are given as the same finite combinations of convergent series, respectively their sums. $\square$

Proposition 10.1

(Uniqueness of the generalized integral sum) Let ${\widehat{F}}\in \widehat{{\mathcal {L}}}_1^I$. Let

$$\begin{aligned} {\widehat{F}}(y)={\widehat{G}}_1(y)\cdot {\widehat{f}}\big (\varvec{\ell }(e^{-\frac{\gamma }{y}} {\widehat{h}}_1(y))\big )+{\widehat{G}}_0(y) \end{aligned}$$

(10.8)

be a decomposition of the form (3.3), not necessarily unique. Let $\alpha \in {\mathbb {R}}$ be the exponent of integration of ${\widehat{f}}$.

• 1. If $\alpha <0$, then the generalized integral sum $F\in {\mathcal {G}}_{AN}$ corresponding to this decomposition is unique up to an additive term $c G_1(y)\cdot \big (e^{-\frac{\gamma }{y}} h_1(y)\big )^{-\alpha }$, $c\in {\mathbb {R}}$. Here, $h_1\in {\mathcal {G}}_{AN}$ is the sum of ${\widehat{h}}_1$.

• 2. If $\alpha \ge 0$, the generalized integral sum $F\in {\mathcal {G}}_{AN}$ is unique.

Morever, let

$$\begin{aligned} {\widehat{F}}(y)={\widehat{H}}_1(y)\cdot {\widehat{g}}\big (\varvec{\ell }(e^{-\frac{\delta }{y}} {\widehat{h}}_2(y))\big )+{\widehat{H}}_0(y) \end{aligned}$$

(10.9)

be another decomposition (3.3) of the same ${\widehat{F}}\in \widehat{{\mathcal {L}}}_1^I$, with the exponent of integration $\beta \in {\mathbb {R}}$ of ${\widehat{g}}$ not necessarily equal to $\alpha$. Then its sum is again equal to F, up to an additive term $c G_1(y)\cdot \big (e^{-\frac{\gamma }{y}} h_1(y)\big )^{-\alpha }$, $c\in {\mathbb {R}}$.

Proof

If $\alpha =0$, then necessarily $\beta =0$ (the case ${\widehat{F}}$ convergent), and the sum $F\in {\mathcal {G}}_{AN}$ is unique. We therefore suppose in the proof that $\alpha ,\ \beta \ne 0$.

The first statement of the proposition follows directly from Definition 3.3 of the integral sum and Remark 3.4. Therefore, the generalized integral sum corresponding to decomposition (10.8) is unique if $\alpha >0$ and unique up to $c G_1(y)\cdot \big (e^{-\frac{\gamma }{y}} h_1(y)\big )^{-\alpha }$, $c\in {\mathbb {R}},$ if $\alpha <0$. Analogously, the generalized integral sum corresponding to decomposition (10.9) is unique if $\beta >0$ and unique up to $d H_1(y)\cdot \big (e^{-\frac{\delta }{y}} h_2(y)\big )^{-\beta }$, $d\in {\mathbb {R}}$, if $\beta <0$. We show below that if (10.8) and (10.9) are two decompositions of the same ${\widehat{F}}$, then formally in $\widehat{{\mathcal {L}}}_2^\infty$:

$$\begin{aligned}&\frac{{\widehat{H}}_1(\varvec{\ell })}{{\widehat{G}}_1(\varvec{\ell })}\cdot \frac{({\widehat{h}}_1(\varvec{\ell }))^\alpha }{({\widehat{h}}_2(\varvec{\ell }))^\beta }=Cx^{\delta \beta -\gamma \alpha },\ \text {for some constant } C\in {\mathbb {R}}. \end{aligned}$$

(10.10)

It follows that $\gamma \alpha =\delta \beta$. Moreover, since ${\widehat{H}}_1,\ {\widehat{G}}_1,\ {\widehat{h}}_1,\ {\widehat{h}}_2\in \widehat{{\mathcal {L}}}^\infty$ are convergent, their sums are unique, and it follows that:

$$\begin{aligned} \frac{H_1(y)}{G_1(y)}\cdot \frac{(h_1(y))^\alpha }{(h_2(y))^\beta }=C,\ \text {for some constant } C\in {\mathbb {R}}. \end{aligned}$$

(10.11)

Consequently, the additive term in the integral sum of ${\widehat{F}}\in \widehat{{\mathcal {L}}}_1^I$ is the same for all decompositions of ${\widehat{F}}$.

We prove now that the integral sums of both decompositions (10.8) and (10.9) of ${\widehat{F}}$ are equal, up to the above mentioned additive term. Substracting (10.8) and (10.9) and using (10.10), we get:

$$\begin{aligned} {\widehat{f}}\big (\varvec{\ell }(x^\gamma {\widehat{h}}_1(\varvec{\ell }))\big )-C\frac{(x^\delta {\widehat{h}}_2(\varvec{\ell }))^\beta }{(x^\gamma {\widehat{h}}_1(\varvec{\ell }))^\alpha }{\widehat{g}}\big (\varvec{\ell }(x^\delta {\widehat{h}}_2(\varvec{\ell }))\big )=\frac{{\widehat{H}}_0(\varvec{\ell })-{\widehat{G}}_0(\varvec{\ell })}{{\widehat{G}}_1(\varvec{\ell })}. \end{aligned}$$

(10.12)

We multiply (10.12) by $\big (x^\gamma {\widehat{h}}_1(\varvec{\ell })\big )^\alpha \in \widehat{{\mathcal {L}}}_2^\infty$ and differentiate formally in $\widehat{{\mathcal {L}}}_2^\infty$ by $\frac{\mathrm d}{\mathrm dx}$. By (3.1),

$$\frac{\mathrm d}{\mathrm dx}\big (x^\alpha {\widehat{f}}(\varvec{\ell })\big )=x^{\alpha -1}{\widehat{R}}_1(\varvec{\ell }),\ \frac{\mathrm d}{\mathrm dx}\big (x^\beta {\widehat{g}}(\varvec{\ell })\big )=x^{\beta -1}{\widehat{R}}_2(\varvec{\ell }),$$

where ${\widehat{R}}_1,\ {\widehat{R}}_2\in \widehat{{\mathcal {L}}}_0^\infty$ are convergent Laurent series. We get formally in $\widehat{{\mathcal {L}}}_2^\infty$:

$$\begin{aligned}&\big (x^\gamma {\widehat{h}}_1(\varvec{\ell })\big )^{\alpha -1}{\widehat{R}}_1\big (\varvec{\ell }(x^\gamma {\widehat{h}}_1(\varvec{\ell }))\big )\cdot \frac{\mathrm d}{\mathrm dx}\big (x^\gamma {\widehat{h}}_1(\varvec{\ell })\big )\nonumber \\&-C\big (x^\delta {\widehat{h}}_2(\varvec{\ell })\big )^{\beta -1}{\widehat{R}}_2\big (\varvec{\ell }(x^\delta {\widehat{h}}_2(\varvec{\ell }))\big )\cdot \frac{\mathrm d}{\mathrm dx}\big (x^\delta {\widehat{h}}_2(\varvec{\ell })\big )=\frac{\mathrm d}{\mathrm dx}\Big (\frac{{\widehat{H}}_0(\varvec{\ell })-{\widehat{G}}_0(\varvec{\ell })}{{\widehat{G}}_1(\varvec{\ell })}\cdot \big (x^\gamma {\widehat{h}}_1(\varvec{\ell })\big )^\alpha \Big ). \end{aligned}$$

(10.13)

Since ${\widehat{h}}_{1,2},\ {\widehat{G}}_{0,1},\ {\widehat{H}}_0$ are convergent in $\widehat{{\mathcal {L}}}^\infty$ and their derivatives commute with the sums, and since ${\widehat{R}}_{1,2}$ are also convergent, we may remove the hats and get the following analogue of (10.13) in ${\mathcal {G}}_{AN}$:

$$\begin{aligned}&\big (x^\gamma h_1(\varvec{\ell })\big )^{\alpha -1}R_1\big (\varvec{\ell }(x^\gamma h_1(\varvec{\ell }))\big )\cdot \frac{\mathrm d}{\mathrm dx}(x^\gamma h_1(\varvec{\ell }))\nonumber \\&-C\big (x^\delta h_2(\varvec{\ell })\big )^{\beta -1}R_2\big (\varvec{\ell }(x^\delta h_2(\varvec{\ell }))\big )\cdot \frac{\mathrm d}{\mathrm dx}(x^\delta h_2(\varvec{\ell }))=\frac{\mathrm d}{\mathrm dx}\Big (\frac{ H_0(\varvec{\ell })-G_0(\varvec{\ell })}{G_1(\varvec{\ell })}\cdot \big (x^\gamma h_1(\varvec{\ell })\big )^\alpha \Big ). \end{aligned}$$

(10.14)

The goal is to get the equivalent of (10.12) for germs. So, once we have reached the Eq. (10.14) for germs by removing hats due to convergence, we reverse the procedure. We integrate (10.14) in ${\mathcal {G}}_{AN}$ with respect to $\int _d^{x} \,\mathrm dt$ or, equivalently, $\int _d^{x^\gamma h_1(\varvec{\ell })} \mathrm d\big (t^\gamma h_1(\varvec{\ell }(t))\big )$, where $d\ge 0$ and $d=0$ iff $\alpha ,\ \beta >0$. The notation means:

$$\frac{\mathrm d\big (t^\gamma h_1(\varvec{\ell }(t))\big )}{\mathrm dt}=t^{\gamma -1}\big (\gamma h_1(\varvec{\ell }(t))+h_1(\varvec{\ell }(t))\varvec{\ell }^2(t)\big ).$$

By (3.2), for any two integral sums $f,\ g\in {\mathcal {G}}_{AN}$ of ${\widehat{f}}, \ {\widehat{g}}\in \widehat{{\mathcal {L}}}_0^\infty$, there exist $d_1,\ d_2\ge 0$ such that:

$$\begin{aligned}&f\big (\varvec{\ell }(x^\gamma h_1)\big )(x^\gamma h_1)^\alpha =\int _{d_1}^{x^\gamma h_1(\varvec{\ell })} R_1\big (\varvec{\ell }(s) \big )s^{\alpha -1}\,\mathrm ds,\nonumber \\&g(\varvec{\ell }(x^\delta h_2))(x^\delta h_2)^\beta =\int _{d_2}^{x^\delta h_2(\varvec{\ell })}R_2\big (\varvec{\ell }(s)\big )s^{\beta -1}\,\mathrm ds, \end{aligned}$$

(10.15)

where $d_1=0$ and $d_2=0$ if and only if $\alpha ,\ \beta >0$, otherwise $d_1,\ d_2>0$. The integral sums f and g are obviously uniquely defined by (10.15) up to an additive constant of integration. Since by (10.11) $CG_1(x^\delta h_2)^\beta =H_1(x^\gamma h_1)^{\alpha }$, using (10.15), in (10.14) after integration, we get:

$$\begin{aligned} f\big (\varvec{\ell }(x^\gamma h_1)\big )-C\frac{(x^\delta h_2)^\beta }{(x^\gamma h_1)^\alpha }g\big (\varvec{\ell }(x^\delta h_ 2)\big )=\frac{H_0(\varvec{\ell })-G_0(\varvec{\ell })}{G_1(\varvec{\ell })}+ D \big (x^\gamma h_1(\varvec{\ell })\big )^{-\alpha },\ D\in {\mathbb {R}}.\nonumber \\ \end{aligned}$$

(10.16)

Comparing (10.8), (10.9) with (10.16), we conclude that the integral sum of ${\widehat{F}}$ is unique up to $D G_1(\varvec{\ell }) (x^\gamma h_1(\varvec{\ell }))^{-\alpha }$, $D\in {\mathbb {R}}$. Note that this term is not dependent on decomposition, and that $G_1$, $\alpha ,\ \gamma ,\ h_1$ are elements of an arbitrarily chosen decomposition.

It remains only to prove (10.10). We have (formally in $\widehat{{\mathcal {L}}}^\infty$):

$$\begin{aligned} {\widehat{F}}(\varvec{\ell })={\widehat{G}}_1(\varvec{\ell })\cdot {\widehat{f}}\big (\varvec{\ell }(x^\gamma {\widehat{h}}_1(\varvec{\ell })))\big )+{\widehat{G}}_0(\varvec{\ell })={\widehat{H}}_1(\varvec{\ell })\cdot {\widehat{g}}\big (\varvec{\ell }(x^\delta {\widehat{h}}_2(\varvec{\ell })))\big )+{\widehat{H}}_0(\varvec{\ell }).\nonumber \\ \end{aligned}$$

(10.17)

It is not possible that ${\widehat{G}}_1\equiv 0$ and ${\widehat{H}}_1\equiv \!\!\!\!\!\!/ \ 0$. Indeed, in that case, we would have that $\varvec{\ell }\mapsto {\widehat{g}}\big (\varvec{\ell }(x^\delta {\widehat{h}}_2(\varvec{\ell }))\big )$ is convergent in $\widehat{{\mathcal {L}}}^\infty$ as a quotient of convergent transseries:

$$\begin{aligned} {\widehat{g}}\big (\varvec{\ell }(x^\delta h_2(\varvec{\ell }))\big )=\frac{G_0(\varvec{\ell })-H_0(\varvec{\ell })}{H_1(\varvec{\ell })}. \end{aligned}$$

(10.18)

We denote convergent transseries in $\widehat{{\mathcal {L}}}^\infty$ without hats. Note that ${\widehat{g}}$ is divergent close to 0. Take $\varvec{\ell }$ sufficiently small so that the convergent transseries on the right-hand side, as well as ${\widehat{h}}_2(\varvec{\ell })$, evaluated at $\varvec{\ell }$, converge. On the other hand, since $\varvec{\ell }(x^\delta h_2(\varvec{\ell }))=\varvec{\ell }(1+o(1)),\ \varvec{\ell }\rightarrow 0$ , by taking $\varvec{\ell }$ sufficiently small, we may ensure that ${\widehat{g}}$ evaluated at $\varvec{\ell }(x^\delta h_2(\varvec{\ell }))$ diverges. This is a contradiction with the equality (10.18). Therefore, either both ${\widehat{G}}_1$ and ${\widehat{H}}_1$ are zero or both are different from zero. In the first case it trivially follows that $G_0\equiv H_0$ (both are convergent) and the decompositions (10.8) and (10.9) are exactly the same.

Suppose now without loss of generality that ${\widehat{G}}_1\equiv \!\!\!\!\!\!/ \ 0$. Dividing both sides of the equality (10.17) by $G_1$, we get formally in $\widehat{{\mathcal {L}}}^\infty$:

$$\begin{aligned} {\widehat{f}}(\varvec{\ell }(x^\gamma h_1(\varvec{\ell })))-\frac{H_1(\varvec{\ell })}{ G_1(\varvec{\ell })}{\widehat{g}}(\varvec{\ell }(x^\delta h_2(\varvec{\ell })))=\frac{H_0(\varvec{\ell })-G_0(\varvec{\ell })}{G_1(\varvec{\ell })}. \end{aligned}$$

Multiplying by $(x^\gamma h_1(\varvec{\ell }))^\alpha \in \widehat{{\mathcal {L}}}_2^\infty$ and differentiating formally by $\frac{\mathrm d}{\mathrm dx}$, we get (in $\widehat{{\mathcal {L}}}_2^\infty$):

$$\begin{aligned}&\frac{\mathrm d}{\mathrm dx}\Big ((x^\gamma h_1(\varvec{\ell }))^\alpha {\widehat{f}}\big (\varvec{\ell }(x^\gamma h_1(\varvec{\ell }))\big )\Big )-\frac{\mathrm d}{\mathrm dx}\Big ({\widehat{g}}\big (\varvec{\ell }(x^\delta h_2(\varvec{\ell }))\big ) (x^\delta h_2(\varvec{\ell }))^\beta \cdot \frac{H_1(\varvec{\ell })}{ G_1(\varvec{\ell })}\frac{(x^\gamma h_1(\varvec{\ell }))^\alpha }{(x^\delta h_2(\varvec{\ell }))^\beta }\Big )\\&\quad =\frac{\mathrm d}{\mathrm dx}\Big (\frac{H_0(\varvec{\ell })-G_0(\varvec{\ell })}{G_1(\varvec{\ell })}(x^\gamma h_1(\varvec{\ell }))^\alpha \Big ). \end{aligned}$$

Differentiating, dividing by $x^{\alpha \gamma -1}$ and regrouping the convergent transseries we get that

$$\begin{aligned} \varvec{\ell }\mapsto {\widehat{g}}\big (\varvec{\ell }(x^\delta h_2(\varvec{\ell }))\big )h_2(\varvec{\ell })^\beta \cdot \frac{\frac{\mathrm d}{\mathrm dx}\Big (\frac{H_1(\varvec{\ell })}{G_1(\varvec{\ell })}\frac{(x^\gamma h_1(\varvec{\ell }))^\alpha }{(x^\delta h_2(\varvec{\ell }))^\beta }\Big )}{x^{\gamma \alpha -1-\delta \beta }} \end{aligned}$$

is a convergent transseries in $\widehat{{\mathcal {L}}}^\infty$. If the derivative in the parenthesis is different from 0, it follows that $\varvec{\ell }\mapsto {\widehat{g}}(\varvec{\ell }(x^\delta h_2(\varvec{\ell })))$ is a convergent transseries in $\widehat{{\mathcal {L}}}^\infty$. This is a contradiction, as already explained in detail above. Therefore, it necessarily holds that the derivative is 0, that is:

$$\begin{aligned} \frac{H_1(\varvec{\ell })}{G_1(\varvec{\ell })}\frac{(x^\gamma h_1(\varvec{\ell }))^\alpha }{(x^\delta h_2(\varvec{\ell }))^\beta }=C,\ C\in {\mathbb {R}}. \end{aligned}$$

Now (10.10) directly follows. $\square$

Proof of Proposition 5.7

The blocks of the formal inverse ${\widehat{g}}^{-1}$ are by (5.13) convergent transseries. Since every integral section is coherent (respects convergence), it is sufficient to prove that $g^{-1}$ can be expanded in increasing powers in x in the form (5.21), where $f_{\beta _i}$ are the sums of convergent ${\widehat{f}}_{\beta _i}$. Roughly speaking, this is proven by repeating the same steps of construction as described in the proof of Proposition 5.6, but this time on germs in ${\mathcal {G}}_{AN}$.

Let $g_\alpha (x)=x^\alpha P_m(\varvec{\ell }^{-1})$ be the first block in the Dulac expansion. Then

$$\begin{aligned} g=g_\alpha \circ \varphi ,\ g^{-1}=\varphi ^{-1}\circ g_\alpha ^{-1}. \end{aligned}$$

Computing as in the formal case, we get:

$$\begin{aligned}&g_{\alpha }^{-1}(x)=(a\alpha ^{-m})^{-\frac{1}{\alpha }}\cdot x^{\frac{1}{\alpha }}\,\varvec{\ell }^{\frac{m}{\alpha }}\Big (1+F\big (\varvec{\ell }_2,\frac{\varvec{\ell }}{\varvec{\ell }_{2}}\big )\Big ), \end{aligned}$$

(10.19)

$$\begin{aligned}&\varphi (x)=g_\alpha ^{-1}\big (g(x)\big )=x\cdot \big (1+F_2(\varvec{\ell }_2,\frac{\varvec{\ell }}{\varvec{\ell }_2},T(x))\big ). \end{aligned}$$

(10.20)

Here, $F,\ F_2$ are analytic germs in two variables, with Taylor expansions ${\widehat{F}}$ and ${\widehat{F}}_2$ from ${\widehat{g}}_\alpha ^{-1}$ resp. ${\widehat{\varphi }}$. The germ $T\in {\mathcal {G}}_{AN}$ is defined by $g(x)=ax^\alpha \varvec{\ell }^{-m}(1+T(x))$. Since g is a Dulac germ with Dulac expansion ${\widehat{g}}$, it follows that:

$$\begin{aligned} T(x)-\varvec{\ell }^m P_0(\varvec{\ell }^{-1})-\sum _{i=1}^{n} x^{\alpha _i-\alpha }\varvec{\ell }^m P_i(\varvec{\ell }^{-1})=o(x^{\alpha _n-\alpha }),\ \forall n\in {\mathbb {N}}, \end{aligned}$$

(10.21)

with $P_i$ as in (5.17).

Write ${\widehat{\varphi }}(x)=\sum _{i=1}^{\infty } x^{\beta _i}\varvec{\ell }^{m_{\beta _i}}{\widehat{G}}_{\beta _i}(\varvec{\ell }_2,\frac{\varvec{\ell }}{\varvec{\ell }_2})$, $\beta _i>0$ and strictly increasing, $m_{\beta _i}\in {\mathbb {Z}}$, $i\in {\mathbb {N}}$. Putting (10.21) in (10.20), and expanding $F_2$, we get immediately that:

$$\begin{aligned} \varphi (x)=&\sum _{i=1}^{n} x^{\beta _i}\varvec{\ell }^{m_{\beta _i}} G_{\beta _i}(\varvec{\ell }_2,\frac{\varvec{\ell }}{\varvec{\ell }_2})+o(x^{\beta _n}),\ n\in {\mathbb {N}}, \end{aligned}$$

(10.22)

where $G_{\beta _i}$ are analytic germs of two variables with Taylor expansion ${\widehat{G}}_{\beta _i}$, $i\in {\mathbb {N}}$.

In particular, as was the case for ${\widehat{\varphi }}$, the leading term of $\varphi (x)-x$ is of power strictly bigger than 1 in x ($\varphi$ is strictly parabolic).

We now analyze the blocks in the asymptotic expansion of $\varphi ^{-1}$ by increasing powers in x, using the Neumann inverse series. We prove that they are the sums of the corresponding convergent blocks (see (5.19)) of ${\widehat{\varphi }}^{-1}$. By coherence of integral sections, this implies that ${\widehat{\varphi }}^{-1}$ is the sectional asymptotic expansion of $\varphi ^{-1}$ with respect to any integral section.

Recall the Schröder operator ${\widehat{\Phi }}_{\varphi }$ from Lemma 5.4, used for obtaining the formal inverse ${\widehat{\varphi }}^{-1}$ of $\widehat{\varphi }$. We define similarly here the linear operator $\Phi _{\varphi }$ acting on ${\mathcal {G}}_{AN}$, $\Phi _{\varphi }\in L({\mathcal {G}}_{AN})$, by:

$$\begin{aligned} \Phi _{\varphi }\cdot f=f\circ \varphi ,\ f \in {\mathcal {G}}_{AN}. \end{aligned}$$

Denote here $h=\varphi -\mathrm {id}\in {\mathcal {G}}_{AN}$. Furthermore, let us introduce the linear operator $H_{\varphi }:=\Phi _{\varphi }-\mathrm {Id}\in L({\mathcal {G}}_{AN})$,

$$\begin{aligned} H_{\varphi }\cdot f=f\circ \varphi -f,\ f\in {\mathcal {G}}_{AN}. \end{aligned}$$

Let us consider the Neumann series:

$$\begin{aligned} \sum _{k=0}^{\infty }(-1)^{k}H_{\varphi }^{k}\cdot id. \end{aligned}$$

Denote its partial sums by

$$\begin{aligned} S_{n}:=\sum _{k=0}^{n}(-1)^{k}H_{\varphi }^{k}\cdot id\in {\mathcal {G}}_{AN},\ n\in \mathbb {N}. \end{aligned}$$

We prove that the Neumann partial sums $S_n$ approximate $\varphi ^{-1}$, as $n\rightarrow \infty$. More precisely, we prove that, for every $\gamma >0$, there exists $n_{\gamma }\in \mathbb {N}$ such that

$$\begin{aligned} S_{n_{\gamma }}(x)=\varphi ^{-1}(x)+O(x^{\gamma }),\ x\rightarrow 0. \end{aligned}$$

In other words, we prove that:

$$\begin{aligned} S_{n_{\gamma }}\big (\varphi (x)\big )=x+O(x^{\gamma }),\ x\rightarrow 0. \end{aligned}$$

Indeed,

$$\begin{aligned} S_{n_{\gamma }}\big (\varphi (x)\big )&=\big (\Phi _{\varphi }\cdot S_{n_{\gamma }}\big )(x)= \big (\mathrm {Id}+H_{\varphi }\big )\cdot \big (\sum _{k=0}^{n_{\gamma }}(-1)^{k}H_{\varphi }^{k}\cdot \mathrm {id}\big )\nonumber \\&= \sum _{k=0}^{n_{\gamma }}(-1)^{k}H_{\varphi }^{k}\cdot \mathrm {id} +\sum _{k=0}^{n_{\gamma }}(-1)^{k}H_{\varphi }^{k+1} \cdot \mathrm {id}=x+(-1)^{n_{\gamma }}H_{\varphi }^{n_{\gamma }+1}\cdot \mathrm {id}. \end{aligned}$$

(10.23)

Since $\varphi$ is strictly parabolic, there exists some $\delta >0$ such that $H_{\varphi }\cdot \mathrm {id}=o(x^{1+\delta })$. Inductively, there exists $n_{\gamma }\in \mathbb {N}$ such that $H_{\varphi }^{n_{\gamma }+1}\cdot \mathrm {id}=O(x^{\gamma })$. Now (10.23) transforms to:

$$\begin{aligned} S_{n_{\gamma }}\big (\varphi (x)\big )=x+O(x^{\gamma }), \end{aligned}$$

that is

$$\begin{aligned} S_{n_{\gamma }}(x)=\varphi ^{-1}(x)+O\big ((\varphi ^{-1}(x))^{\gamma }\big )=\varphi ^{-1}(x)+O(x^{\gamma }). \end{aligned}$$

(10.24)

By (10.24), we have, for $\gamma \rightarrow \infty$, the following expansion of $\varphi ^{-1}$ in strictly increasing powers of x:

$$\begin{aligned} \varphi ^{-1}=&\,\mathrm {id}- h+\Big ( h\circ \varphi - h\Big )+\Big (\big ( h\circ \varphi -h\big )\circ \varphi -\big ( h\circ \varphi -h\big )\Big )+\cdots +O(x^\gamma ), \end{aligned}$$

(10.25)

as compared with its formal analogue (5.18). In (10.25), for a fixed $\gamma >0$, the number of summands up to the order $O(x^\gamma )$ is finite and equal to $n_\gamma$. We now expand the compositions in summands of $\varphi ^{-1}$ in increasing powers of x, using expansion for $\varphi$ given in (10.22) and the fact that $\varphi$ is strictly parabolic. Since $\varphi$ is strictly parabolic, the order of x in the consecutive brackets of $\varphi ^{-1}$ is strictly increasing. Thus only finitely many terms contribute to a block with a fixed power of x. The blocks in x of $\varphi ^{-1}$ are the sums of the corresponding convergent blocks of ${\widehat{\varphi }}^{-1}$.

Finally, we analyze the blocks with increasing powers of x in the composition $g^{-1}=\varphi ^{-1}\circ g_\alpha ^{-1}$. Using the expansion by blocks of $\varphi ^{-1}$ and (10.19), we show that they are the sums of the corresponding convergent blocks of ${\widehat{g}}^{-1}={\widehat{\varphi }}^{-1}\circ {\widehat{g}}_\alpha ^{-1}$. $\square$

Proposition 10.2

Let ${\widehat{f}}\in \widehat{{\mathcal {L}}}$. Then the formal invariant $\rho$ of ${\widehat{f}}$ is given by⁵

$$\begin{aligned} \rho =\Big [\frac{\varvec{\ell }}{x}\Big ]\frac{1}{{\widehat{g}}(x)}=-[\varvec{\ell }_2^{-1}]\int ^x\frac{ds}{{\widehat{g}}(s)}. \end{aligned}$$

Proof

The second equality is obvious. We prove the first equality in two steps:

• 1. For the formal normal form ${\widehat{f}}_0(x)=x-x^{\alpha }\varvec{\ell }^m+\rho x^{2\alpha -1}\varvec{\ell }^{2m+1}$, it obviously holds that:

  $$\begin{aligned} \Big [\frac{\varvec{\ell }}{x}\Big ]\frac{1}{{\widehat{g}}_0(x)}=\rho . \end{aligned}$$

  This can easily be seen expanding $\frac{1}{{\widehat{g}}_0(x)}.$

• 2. Let ${\widehat{f}}_1={\widehat{\varphi }}\circ {\widehat{f}}\circ {\widehat{\varphi }}^{-1}$, ${\widehat{\varphi }}(x)=x+cx^\beta \varvec{\ell }^r\in \widehat{{\mathcal {L}}},\ c\in {\mathbb {R}},\ \beta \in {\mathbb {R}},\ r\in {\mathbb {Z}},$ such that $(\beta ,r)\succ (1,0)$. We prove that:

  $$\begin{aligned} \Big [\frac{\varvec{\ell }}{x}\Big ]\frac{1}{{\widehat{g}}(x)}=\Big [\frac{\varvec{\ell }}{x}\Big ]\frac{1}{{\widehat{g}}_1(x)}. \end{aligned}$$

  That is, we prove that the coefficient $\big [\frac{\varvec{\ell }}{x}\big ]\frac{1}{{\widehat{g}}(x)}$ does not change by formal changes of variables.

Indeed, we estimate the difference:

$$\begin{aligned}&\Big [\frac{\varvec{\ell }}{x}\Big ]\frac{1}{{\widehat{g}}(x)}-\Big [\frac{\varvec{\ell }}{x}\Big ]\frac{1}{{\widehat{g}}_1(x)}= [\varvec{\ell }_2^{-1}]\int ^x\frac{ds}{{\widehat{g}}_1(s)} -[\varvec{\ell }_2^{-1}]\int ^x\frac{ds}{{\widehat{g}}(s)}\nonumber \\&=[\varvec{\ell }_2^{-1}] \Big (\int ^x\frac{ds}{{\widehat{g}}_1(s)}-\int ^x\frac{ds}{{\widehat{g}}(s)}\Big ). \end{aligned}$$

(10.26)

Compute:

$$\begin{aligned} \int ^x \frac{ds}{{\widehat{g}}_1(s)}&=\int ^x \frac{ds}{s-{\widehat{f}}_1(s)}\nonumber \\&=\int ^{{\widehat{\varphi }}^{-1}(x)} \frac{{\widehat{\varphi }}'(t)}{{\widehat{\varphi }}(t)-\widehat{\varphi }({\widehat{f}}(t))}\, dt \ \Big (\text {with }s={\widehat{\varphi }}(t)\Big )\nonumber \\&=\int ^{{\widehat{\varphi }}^{-1} (x)}\frac{dt}{{\widehat{g}}(t)}\cdot \frac{{\widehat{\varphi }}'(t){\widehat{g}}(t)}{{\widehat{\varphi }}(t)-{\widehat{\varphi }}({\widehat{f}}(t))}=\int ^{{\widehat{\varphi }}^{-1}(x)}\frac{dt}{{\widehat{g}}(t)}\cdot \Big (1+\frac{1}{2}\frac{{\widehat{\varphi }}''(t){\widehat{g}}(t)}{{\widehat{\varphi }}'(t)}+\cdots \Big )\nonumber \\&=\int ^{{\widehat{\varphi }}^{-1}(x)}\frac{dt}{{\widehat{g}}(t)}+\frac{1}{2}\int ^{{\widehat{\varphi }}^{-1}(x)}\frac{d}{dt}(\log {\widehat{\varphi }}'(t))dt+\cdots =\int ^{{\widehat{\varphi }}^{-1}(x)}\frac{dt}{{\widehat{g}}(t)}+o(1). \end{aligned}$$

(10.27)

Here, o(1) denotes infinitesimal terms in the formal series. Furthermore, since ${\widehat{\varphi }}\in \widehat{{\mathcal {L}}}$ and parabolic, it can be seen that

$$\begin{aligned} \int ^{{\widehat{\varphi }}^{-1}(x)}\frac{dt}{{\widehat{g}}(t)}-\int ^x \frac{dt}{{\widehat{g}}(t)}\in \widehat{{\mathcal {L}}}, \end{aligned}$$

(10.28)

that is, the difference does not contain the double logarithm. Indeed, if we denote by ${\widehat{P}}$ the formal antiderivative of $\frac{1}{{\widehat{g}}}$, we get that the difference is equal to ${\widehat{P}}'(x){\widehat{h}}(x)+\frac{1}{2}{\widehat{P}}''(x)({\widehat{h}}(x))^2+\cdots$, where ${\widehat{h}}=\mathrm {id}-{\widehat{\varphi }}^{-1}$. Obviously, ${\widehat{P}}^{(k)}\in \widehat{{\mathcal {L}}}$, for all $k\in {\mathbb {N}}$.

Using (10.27) and (10.28), we conclude in (10.26) that

$$\begin{aligned} \Big [\frac{\varvec{\ell }}{x}\Big ]\frac{1}{{\widehat{g}}(x)}-\Big [\frac{\varvec{\ell }}{x}\Big ]\frac{1}{{\widehat{g}}_1(x)}=0. \end{aligned}$$

$\square$

Proof of Proposition 9.2

By Proposition 10.2, by the change of variables in the integral and by integration by parts, we get:

$$\begin{aligned} \rho&=\Big [\frac{\varvec{\ell }}{x}\Big ]\frac{1}{{\widehat{g}}(x)}=-[\varvec{\ell }_2^{-1}]\int ^x \frac{ds}{{\widehat{g}}(s)}\nonumber \\&=-[\varvec{\ell }_2^{-1}]\int ^{{\widehat{g}}(x)/2} \frac{({\widehat{g}}^{-1})'(2t)}{t}dt \Big (\text {with }s={\widehat{g}}^{-1}(2t)\Big )\nonumber \\&=-[\varvec{\ell }_2^{-1}]\int ^{{\widehat{g}}(x)/2} \frac{{\widehat{g}}^{-1}(2t)}{2t^2}dt. \end{aligned}$$

(10.29)

We show now, in the similar way as in the proof of Proposition 10.2 in the Appendix, that

$$\begin{aligned}{}[\varvec{\ell }_2^{-1}]\int ^{{\widehat{g}}(x)/2} \frac{{\widehat{g}}^{-1}(2t)}{2t^2}dt=[\varvec{\ell }_2^{-1}]\int ^{\frac{1}{2}x^\alpha \varvec{\ell }^m} \frac{{\widehat{g}}^{-1}(2t)}{2t^2}dt. \end{aligned}$$

That is, we put ${\widehat{P}}$ to be the formal antiderivative of $\frac{{\widehat{g}}^{-1}(2x)}{2x^2}$ and prove that the difference

$$\begin{aligned}{\widehat{P}}\Big (\frac{{\widehat{g}}(x)}{2}\Big )&-{\widehat{P}}\Big (\frac{1}{2} x^\alpha \varvec{\ell }^m\Big )\\&={\widehat{P}}'\Big (\frac{{\widehat{g}}(x)}{2}\Big )\Big (\frac{{\widehat{g}}(x)- x^\alpha \varvec{\ell }^m}{2}\Big )+\frac{1}{2}{\widehat{P}}''\Big (\frac{{\widehat{g}}(x)}{2}\Big )\Big (\frac{{\widehat{g}}(x)- x^\alpha \varvec{\ell }^m}{2} \Big )^2+\cdots \end{aligned}$$

belongs to $\widehat{{\mathcal {L}}}$ (we prove that it converges formally and does not contain a double logarithm).

Therefore, by (10.29):

$$\begin{aligned} \rho&=-[\varvec{\ell }_2^{-1}]\int ^{\frac{1}{2}x^\alpha \varvec{\ell }^m} \frac{{\widehat{g}}^{-1}(2t)}{2t^2}dt=\Big [\frac{\varvec{\ell }}{x}\Big ]\frac{{\widehat{g}}^{-1}(x^\alpha \varvec{\ell }^m)}{x^\alpha \varvec{\ell }^m}\frac{d}{dx}\log \big (x^\alpha \varvec{\ell }^m\big )\nonumber \\&=\Big [\frac{\varvec{\ell }}{x}\Big ]\frac{{\widehat{g}}^{-1}(x^\alpha \varvec{\ell }^m)}{x^\alpha \varvec{\ell }^m}\big (\frac{\alpha }{x}+m\frac{\varvec{\ell }}{x}\big )\nonumber \\&=\alpha [\varvec{\ell }]\frac{{\widehat{g}}^{-1}(x^\alpha \varvec{\ell }^m)}{x^\alpha \varvec{\ell }^m}+m[1]\frac{{\widehat{g}}^{-1}(x^\alpha \varvec{\ell }^m)}{x^\alpha \varvec{\ell }^m}\nonumber \\&=\alpha [\varvec{\ell }]\frac{{\widehat{g}}^{-1}(t)}{t}\big |_{t=x^\alpha \varvec{\ell }^m}+m[1]\frac{{\widehat{g}}^{-1}(t)}{t}\big |_{t=x^\alpha \varvec{\ell }^m}. \end{aligned}$$

$\square$

Proposition 10.3

Let f be a (prenormalized) Dulac germ. Then there exists $\delta >0$ such that:

$$\begin{aligned} \widehat{A}^c_{{\widehat{f}}}(\frac{\varepsilon }{2})=-\varepsilon \int ^\varepsilon \frac{{\widehat{g}}^{-1}(t)}{t^2}\,dt-\varepsilon \varvec{\ell }(\varepsilon )^{-1}+o(\varepsilon ^{1+\delta }),\ \varepsilon \rightarrow 0. \end{aligned}$$

(10.30)

Proof

Recall that

$$\begin{aligned} {\widehat{A}}^c_{{\widehat{f}}}(\frac{\varepsilon }{2})={\widehat{g}}^{-1}(\varepsilon )+\varepsilon {\widehat{\Psi }}\big ({\widehat{g}}^{-1}(\varepsilon )\big ). \end{aligned}$$

(10.31)

We have:

$$\begin{aligned} {\widehat{\Psi }}({\widehat{g}}^{-1}(\varepsilon ))=\int ^{{\widehat{g}}^{-1}(\varepsilon )}\frac{ds}{\widehat{\xi } (s)}=\int ^\varepsilon \frac{({\widehat{g}}^{-1})'(t)dt}{{\widehat{\xi }}({\widehat{g}}^{-1}(t)).} \end{aligned}$$

(10.32)

Here, $X=\widehat{\xi }\frac{d}{dx}$ is the formal vector field such that ${\widehat{f}}$ is its time-one map, that is, ${\widehat{f}}=\text {Exp}\Big ({\widehat{\xi }} \frac{d}{dx}\Big ).\mathrm {id}.$ Then ${\widehat{\Psi }}'=\frac{1}{{\widehat{\xi }}}$. Let $P_{-m}$ be the polynomial of degree $-m$, $m\in {\mathbb {N}}_0^-$, such that ${\widehat{g}}(x)=x^\alpha P_{-m}(\varvec{\ell }^{-1})+o(x^{\alpha +\delta })$, for some $\delta >0$. Since $-{\widehat{g}}={\widehat{\xi }}+\frac{1}{2}{\widehat{\xi }}'{\widehat{\xi }}+\cdots$ and $\alpha >1$, we have that

$$\begin{aligned} {\widehat{\xi }}(x)=-x^\alpha P_{-m}(\varvec{\ell }^{-1})+o(x^{\alpha +\delta }),\text { for some } \delta >0, \end{aligned}$$

and

$$\begin{aligned} {\widehat{\xi }}'(x){\widehat{\xi }}(x)={\widehat{g}}'(x){\widehat{g}}(x)+o(x^{2\alpha -1+\delta }),\ \text { for some }\delta >0. \end{aligned}$$

Therefore,

$$\begin{aligned} -{\widehat{g}}(x)={\widehat{\xi }}(x)+{\widehat{g}}\cdot {\widehat{g}}'(x)+o(x^{2\alpha -1+\delta }), \text { for some } \delta >0. \end{aligned}$$

(10.33)

Putting ${\widehat{g}}^{-1}(t)=\alpha ^{-\frac{m}{\alpha }}t^{\frac{1}{\alpha }}\varvec{\ell }^{-\frac{m}{\alpha }}+h.o.t.$ obtained in Sect. 5 in (10.33), we have:

$$\begin{aligned}&{\widehat{\xi }}({\widehat{g}}^{-1}(t))=-{\widehat{g}}({\widehat{g}}^{-1}(t))-{\widehat{g}}({\widehat{g}}^{-1}(t))\cdot {\widehat{g}}'({\widehat{g}}^{-1}(t))+o(t^{2-\frac{1}{\alpha }+\delta }),\\&{\widehat{\xi }}({\widehat{g}}^{-1}(t))=-t-t\cdot {\widehat{g}}'({\widehat{g}}^{-1}(t))+o(t^{2-\frac{1}{\alpha }+\delta }),\ t\rightarrow 0, \end{aligned}$$

for some $\delta >0$. It follows that

$$\begin{aligned} \frac{({\widehat{g}}^{-1})'(t)}{{\widehat{\xi }}({\widehat{g}}^{-1}(t))}&=\frac{({\widehat{g}}^{-1})'(t)}{-t\Big (1+{\widehat{g}}'\big ({\widehat{g}}^{-1}(t)\big )+o(t^{1-\frac{1}{\alpha }+\delta })\Big )}\\&=-\frac{({\widehat{g}}^{-1})'(t)}{t}\Big (1-{\widehat{g}}'\big ({\widehat{g}}^{-1}(t)\big )+o(t^{1-\frac{1}{\alpha }+\delta })\Big )\\&=-\frac{({\widehat{g}}^{-1})'(t)}{t}+\frac{1}{t}+o(t^{-1+\delta }). \end{aligned}$$

Now (10.32) becomes, using integration by parts:

$$\begin{aligned} {\widehat{\Psi }}({\widehat{g}}^{-1}(\varepsilon ))&=-\int ^{\varepsilon }\frac{({\widehat{g}}^{-1})'(t)}{t}dt+\int ^\varepsilon \frac{dt}{t}+\int ^{\varepsilon }o(t^{-1+\delta })\,dt\nonumber \\&=-\frac{{\widehat{g}}^{-1}(\varepsilon )}{\varepsilon }-\int ^{\varepsilon } \frac{{\widehat{g}}^{-1}(t)}{t^2}\,dt-\varvec{\ell }(\varepsilon )^{-1}+o(\varepsilon ^\delta ), \end{aligned}$$

(10.34)

for some $\delta >0$. Putting (10.34) in (10.31), the statement follows. $\square$