<s>[INST] 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.
 [/INST] 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.
</s>