Showing that $\int^{\infty}_{\pi} \frac{\sin(x)}{x} ~ dx$ has a improper Riemann integral Observe that,
$$\int_{\pi}^{\infty}\frac{\sin(x)}{x}~ dx  = \sum^{\infty}_{k=1}\int^{(k+1)\pi}_{k\pi}\frac{\sin(x)}{x} dx.$$
Let, $\alpha_k = \int^{(k+1)\pi}_{k\pi}\frac{\sin(x)}{x} dx$ for $k \in \mathbb{N}$. Observe that, $\{|\alpha_k|\}_{k \ge 1}$ is monotonically decreasing, and
$$\lim_{k \rightarrow \infty} |\alpha_k| \le \lim_{k \rightarrow \infty} \int^{(k+1)\pi}_{k\pi}\frac{1}{x} ~ dx  = \lim_{k \rightarrow \infty} \ln(\frac{\pi k + \pi}{\pi k} ) = 0.$$
By the alternating series test we can conclude that,
$$\int^{\infty}_{\pi}\frac{\sin(x)}{x}dx \in \mathbb{R}.$$
Any mistakes in this proof? Thanks!
 A: In the OP it is rigorously shown that the limit
$$
\lim_{n\to\infty}\int_0^{n\pi}\frac{\sin x\,dx}{x}
$$
exists in $\mathbb R$. What is necessary to prove thought is that
the limit
$$
\lim_{M\to\infty}\int_0^{M}\frac{\sin x\,dx}{x}
$$
exists in $\mathbb R$.
To complete the proof, say
$\lim_{n\to\infty}\int_0^{n\pi}\frac{\sin x\,dx}{x}=\ell$ and
let $\varepsilon>0$. Then it is already established that there exists an $n_0\in\mathbb N$, such that
$$
n\ge n_0 \quad\Longrightarrow\quad \left|\int_0^{n\pi}\frac{\sin x\,dx}{x}-\ell\,\right|<\frac{\varepsilon}{2}
$$
If now $M>n_0\pi$, then, for $n_M=[M/\pi]\ge n_0$, we have
$$
\left|\int_0^{M}\frac{\sin x\,dx}{x}-\ell\,\right|\le
\left|\int_{n_M\pi}^{M}\frac{\sin x\,dx}{x}\right|+\left|\int_0^{n_M\pi}\frac{\sin x\,dx}{x}-\ell\,\right| \\
\le \frac{M-n_M\pi}{n_M\pi}+\frac{\varepsilon}{2}\le
\frac{\pi}{n_M\pi}+\frac{\varepsilon}{2} \le
\frac{1}{n_0}+\frac{\varepsilon}{2}<\varepsilon,
$$
provided that $n_0>2/\varepsilon$.
A: An elementary approach: integrate by parts, so that
$\displaystyle \int_{\pi}^{\infty}\frac{\sin x}{x}dx=-\frac{1}{\pi}-\int_{\pi}^{\infty}\frac{\cos x}{x^2}dx$
and observe that the second integral in this expression converges absolutely.
