Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is easy, but I couldn't find some example of a function that is not integrable but its Riemann improper integral exists and is finite

share|improve this question
add comment

2 Answers

up vote 6 down vote accepted

The standard example is $f(x)={\sin(x)\over x}$. The integral of the positive part diverges by comparison with the harmonic series, while the improper Riemann integral exists by use of the alternating series theorem.

share|improve this answer
add comment

Let $f(x) := \frac{\sin x}{x}$, then

$$\int_{\mathbb{R}} f(x) \, d\lambda(x)$$

(where $\lambda$ denotes the Lebesgue-Measure) does not exist, whereas the improper Riemann integral

$$\int_{\mathbb{R}} f(x) \, dx$$

exists (and is finite).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.