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.



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 $y$?
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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. 


