Sine integral and Lebesgue integration

Question

It can be proved in a wide variety of ways that \begin{equation} \lim_{n\to\infty}\int_{0}^n\frac{\sin(x)}{x}\mathrm{d}x=\frac{\pi}{2}\text{, } \end{equation} but that not necessarily mean that $\int_{0}^\infty\frac{\sin(x)}{x}\mathrm{d}x=\frac{\pi}{2}$ in the context of Lebesgue integration. Neither the monotone convergence theorem nor the dominated convergence theorem seems to help proving this. Any idea?

Answer 1

Formalizing Dominik's idea. Note that $\lvert\sin(x)\rvert-\frac{1}{2}$ is continuous and $\pi$-periodic, being $\lvert\sin(\frac{\pi}{2}+k\pi)\rvert-\frac{1}{2}=\frac{1}{2}>0$ for every integer $k$. As long as the function is continuous, we can find some $\delta>0$ such that if $\lvert x-(\frac{\pi}{2}+k\pi)\rvert<\delta$, then $\lvert\sin(x)\rvert>\frac{1}{2}$ for every integer $k$. \begin{equation} \begin{aligned} \int_{0}^\infty\frac{\lvert\sin(x)\rvert}{x}\mathrm{d} x&\geq\sum_{k=0}^\infty\int_{\frac{\pi}{2}+k\pi-\delta}^{\frac{\pi}{2}+k\pi+\delta}\frac{\lvert\sin(x)\rvert}{x}\mathrm{d}x\\&\geq\sum_{k=0}^\infty\frac{1}{2}\int_{\frac{\pi}{2}+k\pi-\delta}^{\frac{\pi}{2}+k\pi+\delta}\frac{1}{\frac{\pi}{2}+k\pi+\delta}\mathrm{d}x\\&=\sum_{k=0}^\infty\frac{1}{2}2\delta\frac{1}{\frac{\pi}{2}+k\pi+\delta}=\sum_{k=0}^\infty\frac{2\delta}{\pi+2k\pi+2\delta}\\&\geq\frac{\delta}{\pi}\sum_{k=0}^\infty\frac{1}{k+1}=\frac{\delta}{\pi}\sum_{k=1}^\infty\frac{1}{k}=\infty\text{, } \end{aligned} \end{equation} as the harmonic series diverges.

This shows that a function which is Riemann-integrable in the improper sense, may not be Lebesgue-Integrable.