<s>[INST] 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?
 [/INST] 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.
</s>