Show $\int_{0}^{\infty} \sin^2{[\pi(x+\frac{1}{x})]}\,\mathrm{d}x$ diverges.
Having trouble doing this. Can't seem to integrate. Not sure how to setup a comparison. Any ideas?
2026-03-26 01:07:57.1774487277
Show $\int_{0}^{\infty} \sin^2{[\pi(x+\frac{1}{x})]}\,\mathrm{d}x$ diverges.
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We have that $\sin^2$ is a Lipschitz-continuous function, so for any $x\geq 1$ $$ \left|\sin^2\left(\pi\left(x+\tfrac{1}{x}\right)\right)-\sin^2(\pi x)\right|\leq \frac{\pi}{x}\tag{1}$$ holds. In particular we have $$ \int_{1}^{N}\sin^2\left(\pi\left(x+\tfrac{1}{x}\right)\right)\,dx = \int_{1}^{N}\sin^2(\pi x)\,dx + O(\log N) \tag{2}$$ but for any $N\in\mathbb{N}^+$ the RHS is $\frac{N-1}{2}+O(\log N)$, which is unbounded as $N\to +\infty$.