Maximal increasing behaviour of $\sqrt{-\log x}\operatorname{Li}_{1/2}(x)$ This is an extension of a problem in mathematical biology. It appears that

*

*For any $\varepsilon>0$, there exists an interval $I\subset(0,1)$ on which $\sqrt{-\log x}^{1+\varepsilon}\operatorname{Li}_{1/2}(x)$ is strictly decreasing;


*The largest value $\nu$ such that $\sqrt{-\log x}\operatorname{Li}_\nu(x)$ is strictly increasing on $(0,1)$ is $\nu=1/2$.
Is there a proof for the first claim?

In the limiting case, it is not difficult to show that $\sqrt{-\log x}\operatorname{Li}_{1/2}(x)$ is strictly increasing on $(0,1)$, as the condition that its derivative is positive is equivalent to $$\sum_{k\ge1}\frac d{dk}(x^k\sqrt k)<0$$ and applying Euler-Maclaurin completes the proof. When the indices $\varepsilon,\nu$ are arbitrary we don't have such a simple expression.
Edit: The second claim can be proved by writing the function as a series (from the Hankel contour) $$\sqrt{-\log x}\operatorname{Li}_\nu(x)=\frac{\Gamma(1-\nu)}{\log^{1/2-\nu}y}+\sum_{k\ge0}\frac{(-1)^k\zeta(\nu-k)}{k!}\log^{k+1/2}y$$ where $y=1/x\in(1,\infty)$. For all $\nu>1/2$, we can see that the logarithmic terms cause $$\lim_{x\to1^-}\sqrt{-\log x}\operatorname{Li}_\nu(x)=0$$ and since the function is positive on $(0,1)$, it cannot be strictly increasing over the entire interval. We have shown above that $\nu=1/2$ works, so that settles the question.
(Cross-posted on MSE but no answers yet)
 A: $\newcommand{\ep}{\varepsilon}\newcommand{\Ga}{\Gamma}\newcommand{\de}{\delta}$Here is a proof of your first claim:
Integral representation:
We have the integral representation
\begin{equation*}
    \sqrt\pi\,\text{Li}_{1/2}(x)=\int_0^\infty \frac{t^{-1/2}\,dt}{e^t/x-1}. 
\end{equation*}
Here and in what follows, $\ep\in(0,\infty)$, $x\in(0,1)$, and $t\in(0,\infty)$.
So,
\begin{equation*}
    f(x):=\sqrt\pi\,\big(\sqrt{-\ln x}\big)^{1+\ep}\text{Li}_{1/2}(x)
=\int_0^\infty dt\,t^{-1/2} g_t(x), \label{1}\tag{1}
\end{equation*}
where
\begin{equation*}
    g_t(x):=\frac{\big(\sqrt{-\ln x}\big)^{1+\ep}}{e^t/x-1}. 
\end{equation*}
So,
\begin{equation*}
    f'(x)
=\int_0^\infty dt\,t^{-1/2} g_t'(x) 
\end{equation*}
and hence
\begin{equation*}
    -2\big(-\ln x\big)^{(1-\ep)/2}f'(x)
=\int_0^\infty dt\,t^{-1/2} h_x(t), \label{2}\tag{2}
\end{equation*}
where
\begin{equation*}
    h_x(t):=\frac{(1+\ep) (e^t-x)+2 e^t \ln x}{(e^t-x)^2}. 
\end{equation*}
Main difficulty: It is that, for each $\ep\in(0,1]$ and each $x$ in a left neighborhood of $1$, $h_x(t)$ changes its sign (from $-$ to $+$) as $t$ goes from $0$ to $\infty$. Moreover, it does not seem easy to appropriately bound $h_x(t)$ from below.
If such a lower bound on $h_x(t)$ were easily available, we could just note that $h_x(t)\underset{x\uparrow1}\longrightarrow\frac{1+\ep}{e^t-1}>0$ and then use Fatou's lemma to conclude that $f'<0$ in a left neighborhood of $1$, which would imply that
for each $\ep\in(0,\infty)$ there is a nonempty interval $I\subset(0,1)$ such that $\big(\sqrt{-\ln x}\big)^{1+\ep}\text{Li}_{1/2}(x)$ is strictly decreasing in $x\in I$.
To overcome this "sign-changing" difficulty, we seem to have to work hard, as follows.
Replacing the integrand by a simpler, appropriately approximating integrand:
We have $|\ln x-(x-1)|\le(1-x)^2/2$ and hence
\begin{equation*}
    |h_x(t)-H_x(t)|\le\de_x(t):=\frac{e^t(1-x)^2}{(e^t-x)^2}\le e^{-t}, 
\end{equation*}
where
\begin{equation*}
    H_x(t):=\frac{(1+\ep) (e^t-x)-2 e^t(1-x)}{(e^t-x)^2}. \label{2.5}\tag{2.5}
\end{equation*}
Also, for each real $t>0$ we have $\de_x(t)\to0$ as $x\to1$.
Therefore and by dominated convergence,
\begin{equation*}
    \int_0^\infty dt\,t^{-1/2}|h_x(t)-H_x(t)|\to0
\end{equation*}
as $x\uparrow1$.
So, it suffices to show that
\begin{equation*}
    K(x):=\int_0^\infty dt\,t^{-1/2} H_x(t)\underset{x\uparrow1}\longrightarrow\infty \label{3}\tag{3}
\end{equation*}
-- then, in view of \eqref{1} and \eqref{2}, it will follow that
for each $\ep\in(0,\infty)$ there is a nonempty interval $I\subset(0,1)$ such that $\big(\sqrt{-\ln x}\big)^{1+\ep}\text{Li}_{1/2}(x)$ is strictly decreasing in $x\in I$.
Integrating part of the integrand by parts, to get a further improved integrand:
Integrating by parts, we get
\begin{equation*}
    \int_0^\infty dt\,t^{-1/2} \frac{(1+\ep)(e^t-x)}{(e^t-x)^2}
    =\int_0^\infty dt\,t^{-1/2} \frac{2(1+\ep)te^t}{(e^t-x)^2}
\end{equation*}
and hence, in view of \eqref{3} and \eqref{2.5},
\begin{align*}
    \tfrac12 K(x)&=\int_0^\infty dt\,t^{-1/2} \frac{(1+\ep)te^t- e^t(1-x)}{(e^t-x)^2} \\ 
&   =\int_0^\infty dt\,t^{-1/2}e^{-t} \frac{(1+\ep)t-(1-x)}{(1-xe^{-t})^2}. \label{3.5}\tag{3.5}
\end{align*}
Minorizing the further improved (sign-changing) integrand by a yet simpler one, with an elementary integral:
Note that the latter integrand has the same sign as $t-t_x$, where
\begin{equation*}
    t_x:=\frac{1-x}{1+\ep},
\end{equation*}
and, by the convexity of the exponential function, $e^{-t}-(1-k_x t)$ also has the same sign as $t-t_x$, where
\begin{equation*}
    k_x:=\frac{1-e^{-t_x}}{t_x}. 
\end{equation*}
Also, $v\frac1{(1-xv)^2}$ is increasing in $v\in(0,1)$, for each $x\in(0,1)$.
It follows that
\begin{equation*}
    \tfrac12 K(x)>L(x)
    :=\int_0^{\sqrt{1-x}} dt\,t^{-1/2}(1-k_x t) \frac{(1+\ep)t-(1-x)}{(1-x(1-k_x t))^2} \label{4}\tag{4}
\end{equation*}
(since $\sqrt{1-x}>t_x$, the integrand in \eqref{3.5} is $>0$ for $t>\sqrt{1-x}$.)
Completing the proof:
The integral $L(x)$ is elementary, even if rather cumbersome. It is also elementary, even if tedious, to show that $L(x)\to\infty$ as $x\uparrow1$.
Thus, \eqref{3} follows from \eqref{4}, and we are done.

Details of the calculation of $L(x)$ and its left limit $L(1-)$ are presented in the image of a Mathematica notebook below (click on the image to magnify it):

