calculus - Problems with $int_{1}^{infty}frac{sin x}{x }dx$ convergence

I'd love your help with deciding whether the following integral converges or not and in what conditions: $\int_{1}^{\infty}\frac{\sin x}{x}$.

1. First, I wanted to use Dirichlet criterion: let $f,g: [a,w) \to R$ integrable function, $f$ is monotonic and $g$ is continuous and $f \in C^1[a,w]$. If in addition to these conditions, $G(x)=\int_{a}^{x}g(t)$ is bounded and $\lim_{x \to w}f(x)=0$ so $\int_{a}^{x}fg$ converges. I can choose $f=\frac{1}{x}$ and $g(x)=\sin x$, they applies all the conditions,(aren't they?) so why can't I use Dirichlet for this integral?

2. I used Wolfarm|Alpha and it says that $\int_{1}^{\infty}\frac{\sin x}{x}dx$ does converge to $\frac\pi2$ .is it only a conditional convergence? (and if so, does is count as non convergence?)

3. I was told that this integral does not absolute converges, meaning $\int_{1}^{\infty}\frac{|\sin x|}{x}dx$ does not converges, How can I prove it?

Thanks a lot.

Answer

Jonas Meyer has already pointed out a link where the properties of this integral are discussed. However, I would like to show you an alternative proof that this integral does not converge absolutely (I saw this in Analysis I/II by Zorich and it doesn't seem as well-known as it should be imho):

The idea is simple. We first see that the only issue with convergence is at $\infty$, since $\frac{\sin(x)}{x}$ is continuous at $0$.

Now $0\le |\sin(x)| \le 1$ implies

$$\frac{|\sin(x)|}{x} \ge \frac{\sin^2(x)}{x}$$

And we note that $\int_1^\infty \frac{\sin^2(x)}{x}dx$ should essentially have the same convergence-properties as $\int_1^\infty \frac{\cos^2(x)}{x}dx$ (with some hand-waving at this point). But if this is true, then $\int_0^\infty \frac{\sin^2(x)}{x}dx$ converges if and only if $$\int_1^\infty \left(\frac{\sin^2(x)}{x}+ \frac{\cos^2(x)}{x}\right) dx = \int_1^\infty \frac{dx}{x}$$
converges. The latter is clearly not true, so $\int_0^\infty \frac{\sin^2(x)}{x}dx$ doesn't converge, either.

More formally, we have

$$\begin{align} \int_0^\infty \frac{|\sin(x)|}{x} dx &\ge \int_{\pi/2}^\infty \frac{\sin^2(x)}{x} dx \\ &= \int_{0}^\infty\frac{\cos^2(x)}{x+\pi/2} dx \\ \end{align}$$

Hence

$$\begin{align} \int_0^\infty \frac{|\sin(x)|}{x} dx &\ge \frac12 \int_0^\infty \left(\frac{\sin^2(x)}{x} + \frac{\cos^2(x)}{x+\pi/2} \right)dx \\ &= \frac12\int_{0}^\infty\left(\frac{1}{x+\pi/2} + \frac{\frac\pi2 \sin^2(x)}{x(x+\pi/2)} \right) dx \\ &\ge \frac12 \int_{0}^\infty\frac{1}{x+\pi/2} dx \\ &= \infty \end{align}$$