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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

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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.

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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).

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