Show that \[ \lim_{t\to\infty}\int_1^t \frac{\sin x}{x}\,dx \] exists. Does \[ f(x)=\frac{\sin x}{x} \] belong to \(L^1((1,\infty))\)?
Solution
1. The improper integral converges.
For \(1\le A<B\), integration by parts gives \[ \int_A^B\frac{\sin x}{x}\,dx =\left[-\frac{\cos x}{x}\right]_A^B -\int_A^B\frac{\cos x}{x^2}\,dx. \] Hence \[ \left|\int_A^B\frac{\sin x}{x}\,dx\right| \le \frac1A+\frac1B+\int_A^\infty\frac{dx}{x^2} \le \frac3A. \] Thus the tails satisfy the Cauchy criterion, so \[ \boxed{\lim_{t\to\infty}\int_1^t\frac{\sin x}{x}\,dx\text{ exists}.} \]
2. The convergence is not absolute.
For each integer \(k\ge1\), let \[ I_k=\left[k\pi+\frac\pi6,\,k\pi+\frac{5\pi}6\right]. \] On \(I_k\) one has \[ |\sin x|\ge\frac12. \] Therefore \[ \int_{I_k}\frac{|\sin x|}{x}\,dx \ge \frac12\int_{I_k}\frac{dx}{x} \ge \frac12\cdot\frac{|I_k|}{(k+1)\pi} =\frac1{3(k+1)}. \] The intervals \(I_k\) are disjoint, so \[ \int_1^\infty\frac{|\sin x|}{x}\,dx \ge \sum_{k=1}^\infty\frac1{3(k+1)} =\infty. \] Hence \[ \boxed{\frac{\sin x}{x}\notin L^1((1,\infty)).} \]