解答:
在$[0,\infty)-\{\frac{1}{k}:k\in \mathbb{N}^{+}\}$上定义函数
\[f(x)=\sum_{k=1}^{\infty}f_k(x)=\sum_{k=1}^{\infty}\frac{x^2\sin\left(\frac{1}{x-\frac{1}{k}}\right)}{k^2}\]
在$\{\frac{1}{k}:k\in \mathbb{N}^{+}\}$上定义函数值为$0$
(1) 在$\forall 0<x_0\notin\{\frac{1}{k}:k\in \mathbb{N}^{+}\}$,存在$\delta>0$使得\[2\delta<\min_{k\in \mathbb{N}^+}|x_0-\frac{1}{k}|\]那么任意$x\in [x_0-\delta,x_0+\delta]$
\[|f_k(x)|\leq \frac{(x_0+\delta)^2}{k^2}\]
\[|f'_k(x)|=\frac{1}{k^2}\left|\cos\left(\frac{1}{x-\frac{1}{k}}\right)\frac{-x^2}{\left(x-\frac{1}{k}\right)^2}+2x\sin\left(\frac{1}{x-\frac{1}{k}}\right)\right|\leq\left( \left( \frac{x}{x-\frac{1}{k}}\right )^2+2x\right)\frac{1}{k^2}\leq \left (\frac{(x_0+\delta)^2}{\delta^2}+2(x_0+\delta)\right)\frac{1}{k^2}\]
说明$\sum\limits_{k=1}^{\infty}f_k(x),\sum\limits_{k=1}^{\infty}f'_k(x)$均在$[x_0-\delta,x_0+\delta]$上一致收敛,而$f_k(x),f'_k(x)$均在$[x_0-\delta,x_0+\delta]$上连续,故$f(x)$在$x_0$处可导
(2)当$x_0=\frac{1}{n},n\in \mathbb{N}^{+}$时,存在$\delta>0$,使得\[2\delta<\min_{k\in \mathbb{N}^{+},k\neq n}|x_0-\frac{1}{k}|\]
那么任意$x\in [x_0-\delta,x_0+\delta]$,\[\sum_{k\in\mathbb{N}^{+},k\neq n}\left|f_k(x)\right|\leq \sum_{k\in\mathbb{N}^{+},k\neq n}\frac{(x_0+\delta)^2}{k^2}\]
从而$f(x)-f_n(x)=\sum\limits_{k\neq n,k\in \mathbb{N}^{+}}f_k(x)$在$[x_0-\delta,x_0+\delta]$内一致收敛。再由$k\neq n$时,$f_k(x)$在$[x_0-\delta,x_0+\delta]$内连续,从而$f(x)-f_n(x)$在$x_0$处连续。
但是$x_0$为$f_n(x)$的第二类间断点,从而$x_0$为$f(x)=(f(x)-f_n(x))+f_n(x)$的第二类间断点
(3)注意到$f(0)=0$,\[\left|\frac{f(x)-f(0)}{x-0}\right|\leq \sum_{k=1}^{\infty}\left|\frac{x\sin\left(\frac{1}{x-\frac{1}{k}}\right)}{k^2}\right|\leq x\sum_{k=1}^{\infty}\frac{1}{k^2}\]
因此$f'(0)=0$
题目:$\int_{x=1}^{+\infty}f(x)dx$收敛,但是$\lim\limits_{x\to +\infty}f(x)$不存在
例1:
记\[f(x)=\frac{x}{1+x^6\sin^2(x)},S_n= \sum_{i=1}^n a_i =\sum_{i=1}^n\int_{(i-1)\pi}^{i\pi} f(x)dx\]只要证明$S_n$有界,对$a_k$放缩:
\[a_k=\int_{(k-1)\pi}^{\pi}f(x)dx\leq \int_{(k-1)\pi}^{k\pi} \frac{k\pi}{1+(k-1)^6{\pi}^6\sin^2(x)}dx=\int_{0}^{\pi} \frac{k\pi}{1+(k-1)^6{\pi}^6\sin^2(x)}dx=\int_{0}^{\frac{\pi}{2}} \frac{2k\pi}{1+(k-1)^6{\pi}^6\sin^2(x)}dx\]
再利用$\sin(x)\geq\frac{2}{\pi}x$有\[a_k\leq \int_{0}^{\frac{\pi}{2}}\frac{2k\pi}{1+4(k-1)^6{\pi}^4x^2}dx=\frac{k}{\pi (k-1)^3}\int_0^{(k-1)^3\pi^3}\frac{dy}{1+y^2}=\frac{k}{\pi (k-1)^3}\arctan((k-1)^3\pi^3)\sim\frac{1}{2k^2}(k\to \infty)\]
\[f(x)=\left\{\begin{matrix} n^4(x-n) &x\in[n,n+\frac{1}{2n^3}] \\ n^4(n+\frac{1}{n^3}-x) & x\in (n+\frac{1}{2n^3},n+\frac{1}{n^3}]\\ 0&x\in (n+\frac{1}{n^3},n+1) \end{matrix}\right.\]
显然$f(x)$在$[1,+\infty)$连续,且\[\int_1^{\infty}f(x)dx=\sum_{n=1}^{\infty}\int_{n}^{n+1}f(x)dx=\sum_{n=1}^{\infty}\frac{1}{4n^2}<+\infty\]
但是$$f(n+\frac{1}{2n^3})=\frac{n}{2}\to \infty(n\to \infty)$$