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Use a comparison test to determine whether or not the following improper integrals converge or diverge $$\int_{2}^\infty \frac{1}{\ln(x)}dx.$$ I'm stuck thinking of a function to compare it to.

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We have by the L'Hôpital's rule $$\lim_{x\to\infty}\frac{\ln x}{x}=0$$ so for $x$ sufficiently large we have $$\frac1{\ln x}\ge\frac1x$$ so the given integral is divergent.

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it obviously diverges because $\frac{1}{\log x } \geq \frac{1}{x}$ which diverges. Compare to harmonic sum/integral.

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Set $\ln x=y$ then $dx=e^ydy$ so the integral becomes $$\int_{\ln 2}^\infty \frac{e^y}{y}dy$$

Therefore the integral is divergent.

-mike

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