How to prove that an integral doesn't exist? $$\int_{0}^{\infty}\sin^2\left(\pi \left(x + \frac{1}{x} \right) \right) dx $$
Should I use any test for convergence?
 A: \begin{align}
\int_0^{\infty}\sin^2 \left(\pi \left(x + \frac{1}{x} \right) \right) dx & \geq \int_1^{\infty}\sin^2 \left(\pi \left(x + \frac{1}{x} \right) \right) dx\\
& = \dfrac12\int_2^{\infty}\sin^2 \left(\pi y \right)\left(\frac{y}{\sqrt{y^2-4}}+1\right) dy & \left(\text{By setting }y=x+\dfrac1x \right)\\
& \geq \dfrac12 \int_2^{\infty} \sin^2(\pi y)dy\\
& = \dfrac12 \sum_{n=2}^{\infty} \int_n^{n+1} \sin^2(\pi y) dy\\
& = \dfrac12 \sum_{n=2}^{\infty}\dfrac12\\
& = \infty
\end{align}
Hence, the integral diverges.
A: Since $\left|\sin^2(a)-\sin^2(b)\right|=|\sin(a)+\sin(b)|\,|\sin(a)-\sin(b)|\le2|a-b|$, we have
$$
\begin{align}
&\left|\int_a^b\sin^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x
-\int_{a+1/2}^{b+1/2}\cos^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x\right|\\
&=\left|\int_a^b\sin^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x
-\int_a^b\sin^2\left(\pi x+\frac\pi{x+1/2}\right)\,\mathrm{d}x\right|\\
&\le\int_a^b2\left(\frac\pi{x}-\frac\pi{x+1/2}\right)\,\mathrm{d}x\\
&\le\frac\pi{a}\tag{1}
\end{align}
$$
Since $\sin^2(a)+\cos^2(a)=1$, the sum of the integrands is $1$ on $[a+1/2,b]$, therefore,
$$
\begin{align}
&\int_a^b\sin^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x
+\int_{a+1/2}^{b+1/2}\cos^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x\\
&\ge b-a-1/2\tag{2}
\end{align}
$$
The triangle inequality applied to $(1)$ and $(2)$ yields
$$
\int_a^b\sin^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x
\ge\frac12\left(b-a-1/2-\frac\pi{a}\right)\tag{3}
$$
Inequality $(3)$ shows that $\int_0^\infty\sin^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x$ diverges.
