Let , $f[0,1]\to \mathbb R$ be defined by $\displaystyle f(x)=\begin{cases}\frac{1}{x}\sin \frac{1}{x} &\text{ if }x\not =0\\0 &\text{ if } x=0\end{cases}$. Show that $f$ is NOT Lebesgue Integrable.
I want to show $\displaystyle \int_0^1|f|=\infty$.
$\displaystyle \int_0^1|f|=\int_1^{\infty}\left|\frac{\sin x}{x}\right|\,dx=\lim_n\int_1^{n\pi}\frac{|\sin x|}{x}\,dx$
$$=\int_1^{\pi}\frac{|\sin x|}{x}\,dx+\lim_n\sum_{k=2}^n\int_{(k-1)\pi}^{k\pi}\frac{|\sin x|}{x}\,dx.$$ Now put , $x=(k-a)\pi +z$. Also let $\displaystyle A=\int_1^{\pi}\frac{|\sin x|}{x}\,dx$. Then , $$\int_0^1|f|=A+\lim_n\sum_2^n\int_0^{\pi}\frac{|\sin\{(k-a)\pi+z\}|}{(k-1)\pi+z}\,dz$$ $$\ge A+\lim_n\sum_2^n\int_0^{\pi}\frac{|\sin\{(k-a)\pi+z\}|}{k\pi}\,dz$$ $$=A+\lim_n\sum_2^n\frac{1}{k\pi}\int_0^{\pi}|\sin z|\,dz$$
$$A+\frac{1}{\pi}\lim_n\sum_2^n\frac{2}{k}=\infty.$$Is my proof correct ?
Any other way to this ?
$$\text {For } n\in N \text { let } x(n)=\frac {1}{\pi (2 n+1/4)},\quad y(n)=\frac {1}{\pi (2 n+1/2)}.$$ We have $0<x(n+1)<y(n)<x(n)$.
For $x\in [y(n),x(n)]$ we have $\sin x\geq 1/\sqrt 2$ and $1/x\geq 1/x(n)>n+1$ and therefore $f(x)>(n+1)/\sqrt 2.$
We have $x(n)-y(n)>1/16\pi (n+1)^2.$
Therefore $$\int_0^1 |f(x)|\;dx\geq\sum_{n\in N}\int_{y(n)}^{x(n)} f(x) dx\geq$$ $$\geq \sum_{n\in N}[x(n)-y(n]\;[(n+1)/\sqrt 2]\geq$$ $$\sum_{n\in N}[1/16 \pi (n+1)^2][(n+1)/\sqrt 2]=$$ $$=\sum_{n\in N}\frac {1}{16\pi (n+1)\sqrt 2}=\infty.$$