Let $H(x)=x^2\sin(\pi/x^2)$ if $x\in]0,1]$ and $H(0)=0$. Show that $H'(x)=h(x)$ exists for all $x\in[0,1]$.
Also, show that $H'$ is not Riemann integrable in $[0,1]$. This way, $h$ has an anti derivative in $[0,1]$ but it isn't Riemann integrable in said interval.
Finally, show that $\lim_{a\to 0+}\int_{a}^{1}h(x)dx$ exists.
I think I was able to prove $H'(x)=h(x)$ exists for all $x\in[0,1]$.
My attempt: If $x=0$ $$\lim_{a\to 0}\frac{H(0+a)-H(0)}{a}$$ $$=\lim_{a\to 0} a\sin(\pi/a^2)=0,$$ using $|\sin(\beta)|\leq 1$ and squeeze theorem. So, $H'(0)=0$
Now, for $x\in(0,1]$ we have that $$\frac{d}{dx}\left(x^2\sin(\pi/x^2)\right)$$ $$=2x\sin(\pi/x^2)-\frac{2\pi\cos(\pi/x^2)}{x}.$$ Is my attempt correct? Does it prove what I was asked?
Also, how can I do the rest of the exercise? I honestly don't know how to prove that $H'$ isn't integrable in $[0,1]$ or that $\lim_{a\to 0+}\int_{a}^{1}h(x)dx$ exists.
Isn't the function $$h(x)=H'(x)=2x\sin(\pi/x^2)-\frac{2\pi\cos(\pi/x^2)}{x}$$ if $x\in(0,1],$ $$H'(0)=0$$ bounded? Thanks in advance.
The first part is correct, to show that $h$ is not Riemann integrable notice that $h$ is unbounded, let $M>0$ be a real number and let $N$ be a natural number such that $N$ is odd (this implies that $N^2$ is odd) and $M<N$, then:
\begin{align*} h(1/N) &= 2(1/N)\sin\left(\frac{\pi}{(1/N)^2}\right) - \frac{2\pi\cos\left(\frac{\pi}{(1/N)^2}\right)}{1/N}\\ \\ &= \frac{2}{N}\sin(N^2\pi) - 2N\pi\cos(N^2\pi)\\ \\ &= 0 + 2N\pi > M \end{align*}
this means that $h$ is not bounded, which implies that $h$ is not Riemann integrable. Now notice that, when $x\in [a,1]$ and $a\in (0 , 1)$, the restriction of $h$ in $[a,1]$ is continuous since is the combination of other continuous functions in $[a,1]$, this implies that the restriction is Riemann integrable in $[a,1]$ for all $a\in (0,1)$, in special you can use the fundamental theorem of Calculus in $[a,1]$ and you get that:
$$\lim_{a\to 0+} \int^{1}_{a} h = \lim_{a\to 0+} \big(H(1) - H(a)\big) =\lim_{a\to 0+} -a^2\sin(\pi/a^2) = 0$$