Improper integral $\sin(x)/x $ converges absolutely, conditionally or diverges? Improper integral of $\sin(x)/x $ converges absolutely, conditionally or diverges?
We have
$$\int_1^{\infty}\frac{\sin x}{x}\text{d}x$$
Integrating by parts
$$u=\frac{1}{x}$$
$$\text{d}u=-\frac{1}{x^2}\text{d}x$$
$$\text{d}v=\sin x\;\text{d}x$$
$$v=-\cos x$$
$$
\begin{aligned}
\int_1^{\infty} \frac{\sin x}{x} \text{d}x
& = \frac{-\cos x}{x} \Big|_1^{\infty} 
    - \int_1^{\infty} \frac{\cos x}{x^2} \text{d}x \\
& = \cos 1 - \int_1^{\infty} \frac{\cos x}{x^2} \text{d}x
\end{aligned}
$$
$\int_1^{\infty} \frac{\cos x}{x^2} \text{d}x$ converges absolutely
(using the Comparison Test For Improper Integrals):
$$ 
\int_1^{\infty} \frac{|\cos x|}{x^2} \text{d}x < 
\int_1^{\infty} \frac{1}{x^2} \text{d}x
$$
So $\int_1^{\infty} \frac{\sin x}{x} \text{d}x$
converges.
Now I need to find out if
$\int_1^{\infty} |\frac{\sin x}{x}| \text{d}x$
converges or diverges.
 A: Let $N \in \Bbb N, N > 1$, we have:
\begin{align}
\int_0^{2\pi N} \left|\frac{\sin x}{x}\right|\,dx &= \sum_{n=0}^{N-1} \int_{2\pi n}^{2\pi(n+1)} \left|\frac{\sin x}{x}\right|\,dx \\
&\ge \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{2\pi n}^{2\pi(n+1)} \left|\sin x\right|\,dx \\
&= \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{0}^{2\pi} \left|\sin x\right|\,dx \\
&= \sum_{n=0}^{N-1} \frac{2}{\pi (n+1)}
\end{align}
The last sum diverges as $N \to \infty$, and so does the original integral.
Your integral is on $[1, \infty]$, but it also diverges because $\left|\frac{\sin{x}}{x}\right|$ is continuous on $[0, 1]$. My proof is on $[0, \infty]$ because it makes managing the summation slightly easier.
A: I would like to provide an intuitive graphic explanation.

Since $\left|\frac{\sin(x)}{x}\right|$ is continuous on $[0, \infty)$, thus it suffices to show the absolute divergence of the tail, namely $\int_{\pi}^{\infty} \left|\frac{\sin(x)}{x}\right| dx$.
Consider bounding this integral below by the infinite sum of area of triangles, namely the n-th triangle has width $\pi$ and height $\frac{1}{(n +1/2) \pi}$, thus the total area (of triangles) =  $$\sum_{n = 1}^{\infty} \frac{1}{n + 1/2} = \infty$$ By a comparison test to the harmonic series $\sum_{k = 2}^{\infty} \frac{1}{k}$, thus the original integral diverges.
A: If one's allowed to use the Absolute Divergent Theorem for Improper Integrals, then one could use what follows:
$\:|\sin x|>\frac{\:1}{\sqrt2}\:,\:\:\:\forall x\in\left]\pi(j+\frac{1}{4}),\pi(j+\frac{3}{4})\right[=I_j\:,\:\:\:j\in\mathbf N.$
$\:\:\forall x\in I_j,\: \forall q\in\:]0,1]\:$ we have $$\sum_{j\in\mathbf N}\int_{I_j}{{\text{d}x}\over x^q}\:\le\:\int_1^\infty{{\text{d}x}\over x^q}=\infty\:,$$by comparison.
So with $\{I_j\}$ being an increasing sequence and $|I_j|=\pi/2\:$ with $\:\lim_{j\in\mathbf N}\:\pi(j+3/4)=\infty,$  $$\int_{1}^\infty{|\sin x|\over x^q}\:\text{d}x=\infty$$
A: Ayman's proof shows the original improper integral is not absolutely convergent. But, working without absolute values, we can show that the series is conditionally convergent. Work with the integral on $ [2 \pi, \infty)$ , and break up the integral into regions where the integrand is $+$ve and $-$ve
