I read that the improper Riemann integral ##\int_0^1 \frac{1}{x}\sin\frac{1}{x}dx## converges and that ##\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx## does not.(adsbygoogle = window.adsbygoogle || []).push({});

I have tried comparison criteria for ##\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx##, but I cannot find a function ##f## with a divergent integral such that ##0\leq f(x)\leq|\frac{1}{x}\sin\frac{1}{x}|##.

I heartily thank you for any help!

EDIT: I have managed to use complex analysis to show that ##\int_0^{\infty} \frac{1}{x}\sin xdx=\frac{\pi}{2}##, but, with a change of variable, I see that ##\int_0^1 \frac{1}{x}\sin\frac{1}{x}dx=\int_1^{\infty}\frac{1}{x}\sin x dx##. ##\frac{1}{x}\sin x## is integrable on ##[0,1]## (it can be "made continuous" in 0 where it approaches 1), therefore ##\int_0^1 \frac{1}{x}\sin\frac{1}{x}dx## converges.

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# Two improper integrals

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