If $f_n (x)=\frac{n \sin x}{x (1+n^2 x^2)}$ then evaluate limit of integration $f_n(x)$ over $0 \to 1$ as $n \to \infty$

Question

In this problem,
I tried to dominated convergence theorem but I couldn't get any dominated function. How to find limit of this integration?
Any hints or comments are welcomed.

Answer 1

I am assuming that you have $$ f_n(x)=\frac{\sin(x)}{x}\frac{n}{1+n^2x^2} $$ and want $$ \lim_{n\to\infty}\int_0^1f_n(x)\,\mathrm{d}x $$ First note that $$ \begin{align} \lim_{n\to\infty}\int_0^1\frac{n}{1+n^2x^2}\,\mathrm{d}x &=\lim_{n\to\infty}\int_0^n\frac1{1+x^2}\,\mathrm{d}x\\ &=\frac\pi2 \end{align} $$ For $n\gt\frac1x$, $\frac{n}{1+n^2x^2}$ tends monotonically to $0$. Thus, using Dominated convergence on intervals avoiding $0$, we can deduce that $$ \lim_{n\to\infty}\int_0^1g(x)\frac{n}{1+n^2x^2}\,\mathrm{d}x=\frac\pi2\lim_{x\to0}g(x) $$ for any $g$ continuous on $[0,1]$.

For this question, $g(x)=\frac{\sin(x)}{x}$.

Answer 2

We have for $f_n(x)=\frac{n}{1+n^2x^2}$ $\lim_{n\to \infty}\frac{n}{1+n^2x^2}=0$ for $x\ne 0$ and $\lim_{n\to \infty}\frac{n}{1+n^2x^2}=\infty$ for $x=0$.

Heuristically, then we anticipate that for an adequately smooth function $\phi$ we have

$$\begin{align} \lim_{n\to \infty}\int_0^1 f_n(x)\phi(x)dx&=\phi(0)\lim_{n\to \infty}\int_0^1 f_n(x)\,dx\\\\ =\frac{\pi}{2}\phi(0) \end{align}$$

To make this more rigorous, let's assume that $\phi$ is continuous with $|\phi(x)|\le M$ on $[0,1]$. Now, we examine the integral

$$\begin{align} \left|\int_0^1 f_n(x)\,(\phi(x)-\phi(0))\,dx\right| & \le \int_0^1 f_n(x)\,\left|\phi(x)-\phi(0)\right|\,dx \tag 1\\\\ &=\int_0^{\delta} f_n(x)\,|\phi(x)-\phi(0)|\,dx\\\\ &+\int_{\delta}^1 f_n(x)\,|\phi(x)-\phi(0)|\,dx \tag 2 \end{align}$$

for $\delta>0$, where we simply split the integral in going from $(1)$ to $(2)$.

Given $\epsilon>0$, we choose $\delta(\epsilon)$ so that $|\phi(x)-\phi(0)|<\epsilon/\pi$ for $|x|<\delta(\epsilon)$. Thus, we have for the first integral on the right-hand side on $(2)$

$$\begin{align} \int_0^{\delta} f_n(x)\,|\phi(x)-\phi(0)|\,dx & \le \frac{\epsilon}{\pi}\int_0^{\delta} f_n(x)\,dx\\\\ &=\frac{\epsilon}{\pi}\,\arctan(n\delta)\\\\ &\le\frac{\epsilon}{2} \end{align}$$

We have for the second integral on the right-hand side of $(2)$

$$\begin{align} \int_{\delta}^{1} f_n(x)\,|\phi(x)-\phi(0)|\,dx & \le 2M \int_{\delta}^{1} f_n(x)\,dx\\\\ &\le 2M \frac{n}{1+n^2\delta^2}\\\\ &\le\frac{2M}{n\delta^2}\\\\ &\le \frac{\epsilon}{2} \end{align}$$

whenever $n>N(\epsilon)=\frac{4M}{\delta^2\,\epsilon}$.

Putting it all together, we have that for any given $\epsilon>0$, there exists a $\delta(\epsilon)>0$, and an $N(\epsilon)=\frac{4M}{\delta^2(\epsilon)\,\epsilon}$ such that

$$\left|\int_0^1 f_n(x)\,(\phi(x)-\phi(0))\,dx\right|<\epsilon$$

whenever $n>N$, which yields the desired result. We simply take $\phi(x)=\sin x/x$, for $x \ne 0$ and $\phi(0)=1$.