Get asymptotic behaviour of the equation $y' = C e^{-(\log x)^s}$, as $x \rightarrow \infty$ I have the following differential equation
$$y''+ s \frac{(\log x)^{s-1}}{x}y'=0$$
where $s>0$. It  is very easy to see that
$$y' = C_1 e^{-(\log x)^s},$$
which cannot be integrated in closed form, as far as I know. What I am interested in is finding how $y$ behaves as $x \rightarrow \infty$, that is, it's asymptotic behaviour.
A first idea would be to write the integrand as a series in $(\log x )^s$ and integrate term by term, but that doesn't really help since as $x$ is arbitrarily large, every term is larger than the previous one (in absolute terms), and so it doesn't make sense to terminate the series somewhere.
How can I do that in this case?
EDIT:
One idea might be to make the ansatz
$$y(x) = \sum_{n=0}^\infty y_n(x) s^n,$$
and solve for the $y_n(x)$ term by term. The zeroth order term is
$$y_0(x) = C_1 e^{-d} x+C_2,$$
while the first order term is
$$y_1(x) = C_1 d e^{-d} \int^x \log (\log u) du.$$
In general, it is easy to see that
$$y_n(x) \propto \int^x[ \log (\log u)]^n du.$$
If we are able to compute the limit of the ratio
$$ \lim_{x \rightarrow \infty} \frac{y_{n+1}(x)}{y_n(x)} \propto \lim_{x \rightarrow \infty}\frac{\int^x[ \log (\log u)]^{n+1} du
}{\int^x[ \log (\log u)]^n du}
$$
and show that it vanishes (I don't know what it is equal to, but it'd be lovely if it vanished ), then indeed we have the asymptotic behaviour. If it diverges, then we haven't made much progress. I think it probably diverges, given that it is easy to see that the limit is
$$ \lim_{x \rightarrow \infty} \frac{y'_{n+1}(x)}{y'_n(x)} \propto \lim_{x \rightarrow \infty}\frac{[ \log (\log x)]^{n+1}
}{[ \log (\log x)]^n}=\lim_{x \rightarrow \infty} \log (\log x) = \infty.
$$
This of course assumes that the original limit is indeterminate which is likely but not known yet.
NOTE: I set all integration constants for the terms $y_n(x)$ with $n \geq 1$ equal to zero, and keep them only for the zeroth term.
 A: Assuming  the conditions
$$y(1) = 0,\quad y'(0) = 1,\tag1$$
one can get
$$y=I(s,x)=\int\limits_1^x e^{\large-\ln^{\LARGE s}y}\text dy.\tag2$$
Integral $(2)$ contains parameter $s,$ which defines its behavior.
In particular,
$$\lim\limits_{\large s\to 0}\,\text I(s,x) = \int\limits_1^x\,\text dx = x-1,\tag{3.0}$$
$$\text I(1,x) = \int\limits_1^x\dfrac{\text dy}y = \ln x,\tag{3.1}$$
$$\text I(2,x) = \int\limits_1^x e^{\large-\ln^{\Large 2}y}\,\text dy
= \int\limits_1^x e^{\large-\ln^{\Large 2}y+\ln y}\;\dfrac{\text dy}y
= \int\limits_0^{\large\ln x} e^{\large z-z^2}\,\text dz
= \sqrt[4]e\int\limits_0^{\large\ln x} e^{\large-\frac14(2z-1)^2}\,\text dz,$$
$$\text I(2,x) = \dfrac{\sqrt\pi\sqrt[4]e}{2}\left(\operatorname{erf}\left(\dfrac12-\ln2\right)-\operatorname{erf}\left(\dfrac12-\ln(x+1)\right)\right).\tag{3.2}$$

In the common case,
$$\text I(s,x) = \int\limits_1^x e^{\large-\ln^{\LARGE s}y}\,\text dy
= \int\limits_1^x e^{\large-\ln^{\LARGE s}y+\ln y}\;\dfrac{\text dy}y
= \int\limits_0^{\large\ln x} e^{\large z-z^{\LARGE s}}\,\text dz,\tag4$$
$$\text I(s,x) = \int\limits_0^{\large\ln x} e^{\large-z^{\LARGE s}}\,\text de^z
\quad\overset{\text{IBP}}{=\!=}\quad e^{\large z-z^{\LARGE s}}\bigg|_0^{\ln x}
+ s\int\limits_0^{\large\ln x} z^{\Large s-\large1} e^{\large z-z^{\LARGE s}}\,\text dz$$
$$ = xe^{\large-\ln^{\LARGE s}x}-1 + s\sum\limits_{k=0}^\infty\dfrac1{k!} \int\limits_0^{\large\ln x} z^{k+s-1} e^{-z^{\LARGE s}}\,\text dz$$
$$= e^{\large-\ln^{\LARGE s-\Large1}x}-1 +\sum\limits_{k=0}^\infty\dfrac1{k!}\Gamma\left(1+\dfrac ks,z^{\large s}\right)\bigg|_0^{\ln x},$$
$$\text I(s,x)= e^{\large-\ln^{\LARGE s-\Large1}x}-1 +\sum\limits_{k=0}^\infty\dfrac1{k!} \Gamma\left(1+\dfrac ks,\ln^{\large s}x\right).\tag5$$
If $\;s>1,\;$ then the integral $(2)$ has the horizontal asymptote with a value
$$\text I(s,\infty)= \sum\limits_{k=0}^\infty\dfrac1{k!} \Gamma\left(1+\dfrac ks \right) -1.\tag6$$
In particular, Wolfram Alpha obtained
$\text I(\frac76,\infty)=$

$\text I(\frac54,\infty)=$

$\text I(\frac43,\infty)=$

$\text I(\frac32,\infty)=$

$\text I(\frac53,\infty)=$

$\text I(2,\infty)=$

$\text I(\frac52,\infty)=$

$\text I(3,\infty)=$

$\text I(4,\infty)=$

$\text I(6,\infty)=$

$\text I(8,\infty)=$

Note that the minimum of the limit by $\;s\;$ is not linked with calculations error.
P.S. The conditions $(1)$ provide suitable calculations and visualiation of the results.
Good luck!
