I have a function of the form as follows $$ a \int_0^1 \frac{1}{1-(p(x)/(p(x)+a))^N}dx=A $$ In the function, $p(x)$ is a known function between $[p_0,1]\subset [0,1]$. $A$ is a positive constant.
Thus, the equation actually describes relationship between $a$ and $N\in \mathbb{N}$. I have already shown that $a>0$ is increasing in $N$ by implicit function theorem. My question is, how can I find the limit of $a$ as $N\rightarrow \infty$?
It seems obvious that if let $\lim_N a$ be finite, then what's in the integral is always equal to one as $N$ goes to infinity. Then we get $\lim_N a=A$, which is also finite. But I don't know how to show the result in a rigorous way? Or, I was wrong about it.
Thanks a lot in advance.
I will show that $A > a > A-\frac1{N}\int_0^1 p(x)dx $, so $\lim_{N \to \infty} a =A $.
First of all, $\int_0^1 \frac{1}{1-(p(x)/(p(x)+a))^N}dx > 1$, so $A > a$.
Then, using Bernoulli's inequality,
$\begin{array}\\ (p(x)+a)/p(x) &=1+(a/p(x))\\ \text{so}\\ ((p(x)+a)/p(x))^N &=(1+(a/p(x)))^N\\ &>1 + Na/p(x)\\ \text{or}\\ (p(x)/(p(x)+a))^N &<1/(1 + Na/p(x))\\ &=p(x)/(p(x) + Na)\\ \text{Therefore}\\ 1-(p(x)/(p(x)+a))^N &>1-(p(x)+Na)/p(x)\\ &=Na/(p(x) + Na)\\ \text{so that}\\ A &=a \int_0^1 \frac{1}{1-(p(x)/(p(x)+a))^N}dx\\ &<a\int_0^1 \frac{dx}{(Na)/(p(x) + Na)}\\ &=\int_0^1 \frac{(p(x) + Na)dx}{N}\\ &=\frac1{N}\int_0^1 p(x)dx+a\\ \end{array} $
Therefore $A > a > A-\frac1{N}\int_0^1 p(x)dx $, so $\lim_{N \to \infty} a =A $.