How to prove that an integral doesn't exist?

Question

$$\int_{0}^{\infty}\sin^2\left(\pi \left(x + \frac{1}{x} \right) \right) dx $$

Should I use any test for convergence?

Answer 1

\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.

Answer 2

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.