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.
 A: Here's another approach that uses your intuition that $x+1/x\sim x$ for large $x$ by doing a change of variable. Denoting the integrand by $f(x)$, we have
$$
\int_0^{\infty} f(x)\,dx = \int_0^1 f(x)\,dx +\int_1^{\infty}f(x)\,dx
$$Set $y=x+1/x$, or $x = \frac{y\mp  \sqrt{y^2-4}}{2}$, with $-$ in the first region and $+$ in the second region. The limits of integration become  $0\to\infty,$ $1\to 2,$ and $\infty\to\infty.$ Then we have $dx = \frac{1}{2}\left(1\mp \frac{y}{\sqrt{y^2-4}}\right)dy$, giving
$$
\frac{1}{2}\int_{\infty}^2\sin^2(\pi y) \left(1-\frac{y}{\sqrt{y^2-4}}\right)\,dy+\frac{1}{2}\int_2^{\infty}\sin^2(\pi y) \left(1+\frac{y}{\sqrt{y^2-4}}\right)\,dy
$$
$$
=\frac{1}{2}\int_2^{\infty}\sin^2(\pi y) \left(-1+\frac{y}{\sqrt{y^2-4}}\right)\,dy+\frac{1}{2}\int_2^{\infty}\sin^2(\pi y) \left(1+\frac{y}{\sqrt{y^2-4}}\right)\,dy
$$
$$
=\int_2^{\infty}\sin^2(\pi y)\frac{y}{\sqrt{y^2-4}}\,dy 
$$This new integral is actually easier to work with, as it is clear $y/\sqrt{y^2-4}>1$ for $y>2$, and then your argument basically holds. One could also use
Cauchy-Bunyakovsky-Schwarz, or the First MVT for Integrals.
A: 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.
A: Alternative proof:
It suffices to prove that,
for any $M > 1$,
$\int_M^{M+4} \sin^2 (\pi(x + 1/x))\, \mathrm{d} x \ge \frac{1}{16}$.
Let $N \in (M + 1, M + 3)$ be a positive integer.
Let
$$x_1 = \frac{N}{2} + \frac{1}{8} + \frac{1}{8}\sqrt{16N^2 + 8N - 63}$$
and
$$x_2 = \frac{N}{2} + \frac{3}{8} + 
\frac{1}{8} \sqrt{16N^2 + 24N -55}.$$
We have $x_2 - x_1 \ge \frac{1}{4}$,
and $M < x_1 < x_2 < M + 4$ (easy to prove), and
$$x_1 + \frac{1}{x_1} = N + \frac{1}{4},
\quad x_2 + \frac{1}{x_2} = N + \frac{3}{4}.$$
Since $x\mapsto x + 1/x$ is strictly increasing on $(1, \infty)$,
we have $N + 1/4 \le x + 1/x \le N + 3/4$ for all $x \in [x_1, x_2]$ and thus
$\sin^2 (\pi(x + 1/x)) \ge \frac{1}{4}$
for all $x \in [x_1, x_2]$.
Thus,
$\int_M^{M+4} \sin^2 (\pi(x + 1/x))\, \mathrm{d} x \ge \int_{x_1}^{x_2} \sin^2 (\pi(x + 1/x))\, \mathrm{d} x \ge \frac{1}{4}(x_2 - x_1) \ge \frac{1}{16}$.
We are done.
