TITLE: limit of an integral of a sequence of functions
QUESTION [9 upvotes]: Suppose that $f$ is continuous on $[0,1]$. ($f'(x)$ may or may not exist).
How can I show that 
$$\lim_{n\rightarrow\infty} \int\limits_0^1 \frac{nf(x)}{1+n^2x^2} dx = \frac{\pi}{2}f(0)\;?$$
My attempt was to recognize that $\int_0^1\frac{n}{1+n^2x^2}dx=\tan^{-1}(nx)$. But I can't simply use integration by parts on the original integral as $f'(x)$ might not exist. Any hints on how to solve this?
Help much appreciated...

REPLY [3 votes]: Take $\epsilon>0$. There exists $\alpha >0$ such that $|f(x)-f(0)|\leq \epsilon/2$ for all $0\leq x\leq \alpha$. Then decompose your integral $L_n$ into three pieces as follows
$$
L_n=\int_0^\alpha\frac{nf(0)}{1+n^2x^2}dx+\int_0^\alpha\frac{n(f(x)-f(0))}{1+n^2x^2}dx+\int_\alpha^1\frac{nf(x)}{1+n^2x^2}dx=I_n+J_n+K_n.
$$
First, note that
$$
I_n=f(0)\arctan(n\alpha)\longrightarrow \frac{\pi}{2}f(0).
$$
Second, we have
$$
|J_n|\leq \int_0^\alpha\frac{n|f(x)-f(0)|}{1+n^2x^2}dx\leq \frac{\epsilon}{2}\int_0^\alpha\frac{n}{1+n^2x^2}dx=\frac{\epsilon}{2}\arctan(n\alpha)\longrightarrow\frac{\pi\epsilon}{4}.
$$
Third, denoting $\|f\|_\infty$ the maximum of $|f|$ on $[0,1]$, we get
$$
|K_n|\leq \|f\|_\infty \int_\alpha^1\frac{n}{1+n^2x^2}dx\leq \|f\|_\infty \int_\alpha^1\frac{n}{1+n^2\alpha^2}dx \leq \frac{\|f\|_\infty}{n\alpha^2}\longrightarrow 0.
$$
So $L_n-\frac{\pi}{2}f(0)$ is bounded by a sequence which converges to $\frac{\pi\epsilon}{4}<\epsilon$. It follows that there exists $N$ such that 
$$
|L_n-\frac{\pi}{2}f(0)|\leq \epsilon\qquad\forall n\geq N.
$$
This proves the desired result, namely that $L_n$ converges to $\frac{\pi}{2}f(0)$.