Is $\int_0^\epsilon \frac{\left(\frac{p-t}{p(1-t)}\right)^y-(1-t)^{y(1-p)/p}}{t\log(1-t)}dt$ bounded by a constant for large $y$? For $p\in (0,1)$ and $\epsilon>0$ a small enough constant, consider the function $f:\mathbb{N}\to\mathbb{R}$ given by
$$f(y)=\int_0^\epsilon \frac{\left(\frac{p-t}{p(1-t)}\right)^y-(1-t)^{y(1-p)/p}}{t\log(1-t)}dt.$$
Is it true that $f(y)=O(1)$ for all $p\in(0,1)$? Numerical experiments seem to support this assertion.
A possible first approach would be to compute the Taylor expansion of the numerator around $t=0$. We have
$$ \left(\frac{p-t}{p(1-t)}\right)^y-(1-t)^{y(1-p)/p}=O_y(t^2),$$
where $O_y$ hides terms which depend on $y$. If these terms were independent of $y$ then the result would follow easily, but this is not the case.
 A: Just compute.Fix $p$, for $t$ small enough. Let $a=\frac{p-t}{p(1-t)}=1+(1-1/p)t+(1-1/p)t^2+(1-1/p)t^3+O(t^4)$,
$b=(1-t)^{\frac{p}{1-p}}=1+(1-1/p)t+(1-3/2p+1/2p^2)t^2+(1-1/p+11/6p^2-1/6p^3)t^3+O(t^4)$,
$b-a=\frac{1-p}{2p^2}t^2+O(t^3)>0$.
Now, let $h(y)=b^y-a^y=(-1)\times $numerator of your integral $>0$, view $y$ as in $\mathbb{R}_{>0}$, this function has a upper bound which is independent of $y$, taking value at $ln(\frac{lnb/lna}{ln(b/a)})$, take this upper bound,  substitution and compute the $t$-expension, then you may get the result, which is not hard to you.
A: $\newcommand{\de}{\delta}
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\renewcommand{\th}{\theta}
\newcommand{\R}{\mathbb{R}}
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\newcommand{\PP}{\operatorname{\mathsf P}}
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Let 
\begin{equation}
 h_1(t):=\ln\frac{p-t}{p(1-t)},\quad h_2(t):=\frac{1-p}p\,\ln(1-t),\quad \de:=h_2-h_1. 
\end{equation}
Then $\de(0)=0$ and 
\begin{equation}
 \de'(t)=\frac{(1-p) t}{p (1-t) (p-t)}>0,
\end{equation}
so that $\de>0$ and hence $h_1<h_2$. 
Also, $\de'(t)\sim\frac{(1-p) t}{p^2}$ and hence $\de(t)\sim \frac{(1-p) t^2}{2p^2}$ as $t\to0$. 
Also, $h_2(t)<-\frac{1-p}p\,t$. So,
\begin{multline*}
0< f(y)=\int_0^\ep \frac{e^{h_2(t)y}-e^{h_1(t)y}}{-t\ln(1-t)}dt
 <\int_0^\ep \frac{\de(t)}{-t\ln(1-t)}\,y\, e^{h_2(t)y}\,dt \\ 
 \ll\int_0^\ep y e^{h_2(t)y}\,dt 
 <\int_0^\infty \exp\{-\tfrac{1-p}p\,ty\}y\,dt=\frac p{1-p}=O(1), 
\end{multline*}
as desired. 
