Finding Riemann Integrability for f(x) on [0,1]

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SUMMARY

The discussion focuses on determining the conditions under which the function f(x) = x^a * cos(1/x) for x > 0, with f(0) = 0, is Riemann integrable on the interval [0,1]. Participants emphasize the importance of continuity, boundedness, and differentiability in establishing Riemann integrability. A function is Riemann integrable if it is continuous almost everywhere and bounded on the interval. The challenge lies in demonstrating the non-integrability for certain values of 'a' and utilizing the definition involving partitions effectively.

PREREQUISITES
  • Understanding of Riemann integrability criteria
  • Familiarity with continuity and boundedness of functions
  • Knowledge of the properties of the cosine function
  • Basic concepts of partitions in the context of integration
NEXT STEPS
  • Study the Riemann integrability criteria in detail
  • Explore examples of functions that are continuous but not Riemann integrable
  • Learn about the implications of boundedness on integrability
  • Investigate the behavior of the function f(x) = x^a * cos(1/x) as 'a' varies
USEFUL FOR

Mathematics students, educators, and anyone studying real analysis, particularly those interested in Riemann integration and its applications.

Silviu
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Homework Statement


Find a such that f is Riemann integrable on [0,1], where:
##f = x^acos(1/x)##, x>0 and f(0) = 0

Homework Equations

The Attempt at a Solution


I found at previous points a such that f is continuous, bounded and derivable, but I am not sure how to use that (as all these implications work just one way). Also the definition with partitions seems hard to use here. Any hint?
 
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Silviu said:
I found at previous points a such that f is continuous, bounded and derivable, but I am not sure how to use that (as all these implications work just one way).
You should have a statement "a function is Riemann integrable if it is [...]" that helps with all a where it is integrable. For the rest you'll have to see how to show that it is not.
 
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