# Lebesgue integral $\int_{(0, \infty)} \frac{\sin x}{x} dm$ doesn't exist but improper Riemann integral exists

Show that the Lebesgue integral $\int_{(0, \infty)} \frac{\sin x}{x} dm$ doesn't exist but the improper Riemann integral $\int_{0}^\infty \frac{\sin x}{x} dx = \lim_{t\to\infty} \int_{0}^t \frac{\sin x}{x} dx$ does exist and is finite.

• What did you tried to do? – Aram Nov 6 '14 at 3:31

## 1 Answer

The Lebesgue integration does not exist. Why? consider $$\int \frac{(\sin x)^+}{x} \text{ v.s. }\int \frac{(\sin x)^-}{x}$$ What the value of those two integration? are they finite? infinite? if both infinite, then what is the definition of lebesgure integration say?

why they are infinite? Hint: $$\int \frac{(\sin x)^+}{x} \geq\sum_{k=1}^\infty \int_{2k\pi+6/\pi}^{2k\pi+\pi/3} \frac{(\sin x)^+}{x}=?$$

why riemann is finite? can you get an idea from above prove?

• What's the intuition for it to be improper Riemann integrable while not Lebesgue integrable? – user186073 Nov 6 '14 at 5:18