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In many place one finds accounts of how to evaluate $$ \int_0^\infty \frac{\sin x} x\,dx = \underbrace{\lim_{a\to\infty}\int_0^a}_{\text{Why view it this way?}} \frac{\sin x} x\, dx. $$

And it gets asserted that the reason why the integral must first be evaluated over a bounded interval and then afterward the bound is allowed to go to $\infty$ is that $$ \int_0^\infty \left|\frac{\sin x} x\right| \, dx\ \underbrace{{}\ =\infty\ {}}_{\text{That's why.}} \tag 1 $$ and therefore the ways of thinking about integrals introduced by Henri Lebesgue are not applicable on the unbounded interval. If it is asked how we know that $(1)$ is true, my first thought it that one ought to compare it somehow with a harmonic series because it ultimately declines to $0$ at the same rate.

It seems as if this question is worth having among our stock of questions and answers, but a couple of days ago I found the answer posted as a mere comment. Hence this present posting. I'll notify the person who posted the comment. Doubtless some variety of ways to prove this result exists, so others should post their own versions.


marked as duplicate by Did, user7530, Claude Leibovici, user147263, Venus Jun 7 '15 at 5:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


We have $$\int_{n\pi}^{(n+1)\pi}\left|\frac{\sin x}{x}\right|\,dx\gt \frac{1}{(n+1)\pi}\int_{n\pi}^{(n+1)\pi} |\sin x|\,dx=\frac{2}{(n+1)\pi}.$$

The series $\sum_0^\infty \frac{2}{(n+1)\pi}$ diverges to $\infty$, so by comparison $\lim_{B\to\infty} \int_0^B \left|\frac{\sin x}{x}\right|\,dx=\infty$.


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