<s>[INST] Prove $\int^{\infty}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ diverges First of all, I noticed that $\sin^2(\pi(x+\frac{1}{x}))$ "approaches" $\sin^2(\pi x)$ as $x \to \infty$. In other words, it seems that the area under one period of $\sin^2(\pi(x+\frac{1}{x}))$ should converge to the value of the area under one period of $\sin^2(\pi x)$ as $x \to \infty$.
The following is my best attempt at formalizing this idea:

Let $f(x) = \sin^2(\pi(x+\frac{1}{x}))$ and  $\xi_n = \frac{n+\sqrt{n^2-4}}{2}$, where $n \geq 2$. Notice that $f(\xi_n)=0$.
Now, consider the limit of $\xi_{n+1} - \xi_{n}$ as $n \to \infty$:
$$\begin{align} \lim_{n\to\infty} \xi_{n+1}-\xi_n &= \lim_{n\to\infty}\frac{(n+1)+\sqrt{(n+1)^2-4}}{2}-\frac{n+\sqrt{n^2-4}}{2} \\ &= \frac{1}{2}\lim_{n\to\infty}1+\sqrt{(n+1)^2-4}-\sqrt{n^2-4} \\ &= \frac{1}{2}\lim_{n\to\infty}\frac{2+\frac{2}{n}+2\sqrt{1+\frac{2}{n}-\frac{3}{n^2}}}{\frac{1}{n}+\sqrt{1+\frac{2}{n}-\frac{3}{n^2}}+\sqrt{1-\frac{4}{n^2}}} \\ &= 1 \end{align}$$ What the result above tells us is that as x
increases, the period of $\sin^2(\pi(x+\frac{1}{x}))$ approaches the
period of $\sin^2(\pi x)$. Furthermore, $x+\frac{1}{x} \approx x$ as
$x\to \infty$, which implies that $\sin^2(\pi(x+\frac{1}{x}))$
"approaches" $\sin^2(\pi x)$ as $x \to \infty$. Using both of these
facts, we can claim that $\int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx$ approaches $\int^{1}_{0} \sin^2(\pi x) dx$ as $n \to \infty$.
We can also compute the value of $\int^{1}_{0} \sin^2(\pi x) dx$:
$$\begin{align} \int^{1}_{0} \sin^2(\pi x) dx &= \frac{1}{2}\int^{1}_{0}1-\cos(2\pi x)dx \\ &=\frac{1}{2}\{(1-\frac{1}{2\pi}\sin(2\pi))-(0-\frac{1}{2\pi}\sin(0))\} \\ &= \frac{1}{2}\end{align}$$ Thus, we can state that:
$$\lim_{n\to\infty} \int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx = \frac{1}{2}$$
Notice that we can write $\int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx$ as... $$\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx - \int^{\xi_{n}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$$ If we consider $\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ as a sequence, then $\lim_{n\to\infty} \int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx = \frac{1}{2}$
tells us that $\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ is
not a Cauchy sequence, and hence $\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ is divergent as $n \to \infty$.
Thus, we can finally claim that $\int^{\infty}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ is divergent.

Now, the part of my argument which I find to be weak is:

What the result above tells us is that as $x$
increases, the period of $\sin^2(\pi(x+\frac{1}{x}))$ approaches the
period of $\sin^2(\pi x)$. Furthermore, $x+\frac{1}{x} \approx x$ as
$x\to \infty$, which implies that $\sin^2(\pi(x+\frac{1}{x}))$
"approaches" $\sin^2(\pi x)$ as $x \to \infty$. Using both of these
facts, we can claim that $\int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx$ approaches $\int^{1}_{0} \sin^2(\pi x) dx$ as $x \to \infty$.

How can I make this part more rigorous/stronger? I don't really know how to achieve any more rigor with my current level of mathematics.
 [/INST] What's wrong with this less formal approach?
Replace the original integrand with sin^2 x , since the factor of pi goes away with a simple change of variable, and since the x term dominates the 1/x term over the interval. Then every occurrence of the new integrand as x increases is non-negative and identical in magnitude, so the integral must diverge.
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