How to calculate Lebesgue integral in this type?

Question

For Lebesgue integral in this type, like$$\int_\pi^\infty \left({1 \over {x \sin^{1/3}x} }\right)^2$$ can anyone give me some general idea? I don't know use which inequalities to start with the this one.

Answer 1

I assume that you only want to show that the integral is finite, not calculate the precise value.

I would do the following: $$ \int_\pi^\infty \frac{1}{x^2 \cdot \sin^{2/3}(x)} /, dx =\sum_{n=1}^\infty \int_{n\pi}^{(n+1)\pi} \frac{1}{x^2 \cdot \sin^{2/3}(x) }\, dx \asymp \sum_{n=1}^{\infty} n^{-2} \int_0^\pi \sin^{-2/3}(x)\,dx <\infty. $$

Here, I used $\pi$-periodicity of $\sin^2$. Finally, the final integral should be finite, since $\sin (x) \sim x$ near $0$ and $\int x^{-2/3}\, dx<\infty$. The blowup at $\pi$ can be handled d similarly. I leave the details to you.

A similar approach will work for similar integrals.