Showing $\int_{-1}^1 \left|\frac{1}{x}\cos\left(\frac{1}{x^2}\right)\right|dx$ diverges I want to show that $$\int_{-1}^1 \left|\frac{1}{x}\cos\left(\frac{1}{x^2}\right)\right|dx$$ diverges. I am doing this to show that a specific function's derivative is not in $L^1$ and I believe this integral must go to $\infty$ for this to be true. Thanks.
 A: The integrand's even, so change the lower limit to $0$, double the integral, and move $\tfrac1x$ outside $|\cdot|$. Use $y=\tfrac{1}{x^2}$ to write the integral as$$\int_1^\infty\tfrac{|\cos y|}{y}dy=\int_1^{3\pi/2}\tfrac{|\cos y|}{y}dy+\sum_{n\ge0}\int_{3\pi/2}^{7\pi/2}\tfrac{|\cos y|}{y+2\pi n}dy.$$The divergence of the harmonic series completes the proof, since$$\int_{3\pi/2}^{7\pi/2}\tfrac{|\cos y|}{y+2\pi n}dy\ge\int_{3\pi/2}^{5\pi/2}\tfrac{\cos y}{y+2\pi n}dy\ge\tfrac{1}{5\pi/2+2\pi n}\underbrace{\int_{3\pi/2}^{5\pi/2}\cos ydy}_2\ge\tfrac{4}{5\pi+4\pi n}.$$
A: Letting $u=1/x^2$, we find that
$$\int_{-1}^1\left|{1\over x}\cos(1/x^2)\right|\,dx=2\int_0^1{|\cos(1/x^2)|\over x}\,dx=\int_1^\infty{|\cos u|\over u}\,du$$
Now on each interval $[2\pi k-4\pi/3,2\pi k-2\pi/3]$, we have $|\cos u|\ge1/2$. Since these intervals are all disjoint and all of width $2\pi/3$ (and also since $2\pi k-4\pi/3\gt1$ for $k=1$), it follows that
$$\int_1^\infty{|\cos u|\over u}\,du\gt\sum_{k=1}^\infty{1/2\over2\pi k}\cdot{2\pi\over3}={1\over6}\sum_{k=1}^\infty{1\over k}=\infty$$
