calculus - Show that $int_0^1 |frac{1}{x}sinfrac{1}{x}| dx$ diverges

I read that the improper Riemann integral $$\int_0^1 \Bigg|\frac{1}{x}\sin\frac{1}{x}\Bigg|\ dx$$ diverges.

I have tried comparison criteria for $\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx$, but I cannot find a function $f$ with a divergent integral such that $0\leq f(x)\leq|\frac{1}{x}\sin\frac{1}{x}|$. I also notice, by using a change of variable, that $\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx=\int_1^\infty |\frac{1}{x}\sin x|dx$, but I have not found a use of this equality to prove the divergence of the integral. I have also tried to think about some use of complex analysis, but found none useful to prove the desired result. Coud anybody give a proof of this divergence? $\infty$ thanks!

Answer

I also notice, by using a change of variable, that $\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx=\int_1^\infty |\frac{1}{x}\sin x|dx$, but I have not found a use of this equality to prove the divergence of the integral.

Subdivide $(1,\infty)$ into an infinite number of intervals of the form $\Big(k\pi,~(k+1)\pi\Big)$, and notice that for each of them, their graphic is tangent to a hyperbola, and that the area of each slice is greater than that of a triangle of height $\dfrac1{\pi~\bigg(k+\dfrac12\bigg)}$, due to the convexity/concavity of the sine function. This shows that the value of our integral is greater than a constant multiple of the harmonic series, which is known to be divergent.