Answer to Question #260220 in Real Analysis for abc

Question

If 𝑓 is a continuous function on [0,1], show that lim𝑛→∞ ∫ 𝑛𝑓(𝑥) 1+𝑢2𝑥 2 1 0 𝑑𝑥 = 𝜋 2 𝑓(0). 

Answer 1

Let

$I=\displaystyle{\int^{1}_{0}}\frac{nf(x)}{1+u^2x^2}dx=I_1+I_2$

where

$I_1=\displaystyle{\int^{n^{-1/3}}_{0}}\frac{nf(x)}{1+u^2x^2}dx$

and

$I_2=\displaystyle{\int^{1}_{n^{-1/3}}}\frac{nf(x)}{1+u^2x^2}dx$

$|f(x)|≤M$ because f is continuous. We have:

$I_2\le\displaystyle{\int^{1}_{n^{-1/3}}}|\frac{nf(x)}{1+u^2x^2}|dx\le \displaystyle{\int^{1}_{n^{-1/3}}}\frac{nM}{1+u^2x^2}dx \le \displaystyle{\int^{1}_{n^{-1/3}}}\frac{nM}{1+u^2(n^{-1/3})^2}dx\\$

$=\frac{nM}{1+n^2(n^{-1/3})^2}(1-n^{-1/3})=o(1)$

Put

$I_3=\displaystyle{\int^{n^{-1/3}}_{0}}\frac{nf(0)}{1+n^2x^2}dx$

We have:

$|I_1-I_3|=|\displaystyle{\int^{n^{-1/3}}_{0}}\frac{n(f(x)-f(0))}{1+n^2x^2}dx|\le$

$\le \displaystyle{\sup_{n\in [0,n^{-1/3}]}}|f(x)-f(0)|\cdot \displaystyle{\int^{n^{-1/3}}_{0}}\frac{ndx}{1+n^2x^2}dx=o(1)\cdot \displaystyle{\int^{n^{-1/3}}_{0}}\frac{d(nx)}{1+n^2x^2}dx=$

$=o(1)\cdot arctan(n\cdot u^{-1/3})=o(1)\cdot O(1)$

Hence

$I=I_1+I_2=I_3+(I_1-I_3)+I_2=I_3+o(1)+o(1)$

where

$I_3=f(0)\displaystyle{\int^{n^{-1/3}}_{0}}\frac{d(nx)}{1+u^2x^2}dx=f(0)\cdot arctan(n\cdot u^{-1/3})=f(0)\frac{\pi}{2}+o(1)$

So,

$I\to f(0)\frac{\pi}{2}$ as $n\to \infin$