Showing an integral is unbounded Let $$F(x)=x^2\sin\left(\frac{1}{x^2}\right)\quad\forall x\in(0,1]$$ and $$G(x)=F'(x)=2x\sin\left(\frac{1}{x^2}\right)-2\frac{1}{x}\cos\left(\frac{1}{x^2}\right)\quad\forall (0,1].$$ I want to show that $$\lim_{t\searrow0}\int_t^1\left|G(x)\right|\,\mathrm{d}x=+\infty.$$
I've had a hard time proving this, given that the values of $|G(x)|$ oscillate between $0$ and an ever increasing number at an ever increasing frequency as $x$ approaches zero, making it hard to bound the integral from below to establish that it is unbounded.
Any hints would be greatly appreciated.
 A: The $2x\sin\left(\dfrac{1}{x^2}\right)$ part is bounded, and can therefore be ignored when investigating the convergence or divergence of the integral.
So let's look at the integral
$$\int\limits_{a_{k+1}}^{a_k} \left\lvert \frac{2}{x}\cos \left(\frac{1}{x^2}\right)\right\rvert\, dx$$
where the $a_k$ are the successive zeros of $\cos (1/x^2)$, $a_k = 1/\sqrt{(k-\frac12)\pi}$.
We can estimate it from below by shrinking the interval, so that $\cos (1/x^2)$ stays away from $0$, say
$$I_k = \int\limits_{b_k}^{c_k} \left\lvert\frac{2}{x}\cos\left(\frac{1}{x^2}\right) \right\rvert\,dx$$
with
$$b_k = \frac{1}{\sqrt{(k+\frac13)\pi}},\quad c_k = \frac{1}{\sqrt{(k-\frac13)\pi}},$$
so that $\lvert\cos (1/x^2)\rvert \geqslant \frac12$ for $b_k\leqslant x \leqslant c_k$. That yields
$$I_k \geqslant \frac{1}{c_k}\left(c_k - b_k\right).$$
For $k$ sufficiently large, you have constants independent of $k$ such that
$$\frac{1}{c_k} \geqslant A\sqrt{k},\quad c_k - b_k \geqslant \frac{B}{k\sqrt{k}},$$
so $\sum_{k\geqslant K} I_k$ is bounded below by a harmonic series for some $K$ sufficiently large, hence
$$\lim_{t\to 0} \int_t^1 \lvert G(x)\rvert\,dx = +\infty.$$
