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Bartle has a statement:

Although the absolute value of a proper Riemann integrable function is Riemann integrable, this may no longer be the case for a function which has an improper Riemann integral (for example, consider $f(x)=x^{-1} \sin(x)$ on the interval $1\le x\lt +\infty$).

Could I get a little help with this statement? I am assuming I need to show that the improper integral is integrable, but the absolute value is not.

Thanks.

Zev Chonoles
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David
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  • Riemann-integrable functions $f$ are bounded and defined on a closed interval $[a,b]$. Fix $f$. Then $|f|$ is bounded and defined on $[a,b]$. The integral has an upper bound $(b-a) \sup_{x \in [a,b]} |f(x)|$. Improper Riemann-integral of the first kind is defined for bounded functions $f$ on an interval $[a,\infty)$ as the limit $\lim_{M \rightarrow \infty} \int_a^M f(x) dx$ whenever it exists. Now fix $f(x)=x^{-1} \sin(x)$. Integration by parts shows that the integral converges. However, the absolute integral diverges because it has a harmonic lower limit estimate, that diverges. – Juha-Matti Vihtanen Jun 10 '13 at 12:36
  • Sometimes it is said, that you cannot ever get anything without losing something. This time you get the existence of the Dirichlet integral. Good for you. It is useful if you want to show the Fourier inversion theorem using the Riemann-integral (of the first kind). – Juha-Matti Vihtanen Jun 10 '13 at 13:48

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