The Student Room Group
Reply 1
Heuristically: it's continuous on (0, 1), and the endpoints are basically irrelevant, so it's integrable. To prove it properly, I guess you have to bound it above and below by suitable Riemann sums, by taking suitable dissections.
Reply 2
Don't actually try to construct a dissection.

Use the Riemann criterion for integrability, and split your original integral into two parts: one from 0 to epsilon and another from epsilon to 1.

Note that for any interval [epsilon, 1] sin(1/x) is continuous, and you can bound the other bit.

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