Showing that $\int^{\infty}_{\pi} \frac{\sin(x)}{x} ~ dx$ has a improper Riemann integral

Solution 1:

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 $n_0>2/\varepsilon$.

Solution 2:

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.