proving converge of an improper integral via riemann

Question

I need to show if the following integral converges: $$\int_{-\infty}^{\infty}\left|\sin{1 \over x}\right|\,\mathrm dx$$ my idea for the solution is to show that the serie of rectangles that are blocked within the sin function does not converge. but i'm having trouble with writing that sum...

Answer 1

1. Split the integral into two domains $(0;\infty)$ and $(-\infty;0)$ (the function does not exist by the point $0$, hence is allowably removed). The main integral exists (is finite) iff the two others exist.

2. The other integrals are the same. Thus the main one exists the integral over $(0;\infty)$ exists.

It holds, that

$\int_{x\in(0;\infty)}|sin(1/x)|~\text{d}x \geq \int_{x\in[1;\infty)}|sin(1/x)|~\text{d}x = \int_{x\in(0;1]}|sin(x)|\cdot|\frac{1}{x^{2}}|~\text{d}x \geq \int_{x\in(0;\delta)}|\frac{sin(x)}{x^{2}}|~\text{d}x$,

where $\delta$ is any sufficiently small real.

On $(0;\delta)$ it holds that $sin(x)/x^{2}>0$. Also $sin(x)=x+\frac{x^{3}\cdot cos(\theta(x))}{3!}$, where $\theta(x)\in(0;\delta)$. Hence $sin(x)≥x-\frac{x^{3}}{3!}$ and furthermore $\frac{sin(x)}{x^{2}}≥\frac{1}{x}-\frac{x}{3!}$.

Since $\int_{x\in(0;\delta)}\frac{1}{x}-\frac{x}{3!}~\text{d}x=\infty$, so too $\int_{x\in(0;\delta)}|\frac{sin(x)}{x^{2}}|~\text{d}x=\infty$ and thus also $\int_{x\in(0;\infty)}|sin(1/x)|~\text{d}x=\infty$

Hence $\int_{x\in\mathbb{R}}|sin(1/x)|~\text{d}x=\infty$.

Answer 2

The integral is improper only at $\infty$, since on $[-1,1]$ the integrand is bounded and continuous except at $x=0$.

For $x$ large, use that $\sin(1/x)\sim1/x$.