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.
 A: 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
$$
\begin{align}
\int_0^\infty \frac{|\sin(x)|}{x} dx &\ge \frac12 \int_0^\infty \left(\frac{\sin^2(x)}{x} + \frac{\cos^2(x)}{x+\pi/2} \right)dx \\
&= \frac12\int_{0}^\infty\left(\frac{1}{x+\pi/2} + \frac{\frac\pi2 \sin^2(x)}{x(x+\pi/2)} \right) dx \\
&\ge \frac12 \int_{0}^\infty\frac{1}{x+\pi/2} dx \\
&= \infty
\end{align}
$$
A: You can make integration by patrs to integral from 1 to t, and  to the integral that appears in the way you can make absolute convergence.
In this way you can prove convergence since the first term in parts has finit limit and the integral that appears in the development of parts is convergent as I wrote.
A: One nice visual way to see that it doesn't converge absolutely is by looking at triangles under each "hump" of $\frac{|sin(x)|}{x}$.  Make sure the base of each triangle goes from $\pi n$ to $\pi (n+1)$ on the x-axis, and the third point goes to $(\pi(n+\frac{1}{2}),\frac{sin(\pi(n+\frac{1}{2}))}{\pi(n+\frac{1}{2})})$.  Then the area under each triangle is
$\frac{sin(\pi/2)/2}{n+\frac{1}{2}}$
which decays like $\frac{1}{n}$ and so is not summable.
To know if it converges conditionally, I think of it like the alternating-series condition: breaking up each hump, we see the signs alternate and the absolute values of the areas decrease.  
