Conditional convergence of $\int_1^{\infty} \frac{\sin(x)}{x}dx$?

Question

What's a simple way to display $$\int_1^{\infty} \frac{\sin(x)}{x}dx$$ conditionally convergent (i.e. convergent, but not absolutely)?

Answer 1

Dirichlet's test: $\sin x$ has a bounded primitive and $\frac{1}{x}$ is decreasing to zero, hence $\int_{1}^{+\infty}\frac{\sin x}{x}\,dx $ is conditionally convergent. It cannot be absolutely convergent since the mean value of $|\sin x|$ is $\frac{2}{\pi}$, so $\int_{1}^{M}\frac{|\sin x|}{x}\,dx $ behaves like $\frac{2}{\pi}\log M$ by integration by parts.