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How to show that $\int_1^\infty \frac{\ln(t)}{t}dt$ is divergent?

I know how to show that $\int_1^\infty \frac{1}{t}dt$ is divergent

So maybe $$ \int_1^\infty \frac{\ln(t)}{t}dt = \int_1^e \frac{\ln(t)}{t}dt + \int_e^\infty \frac{\ln(t)}{t}dt \geq \int_1^e \frac{\ln(t)}{t}dt + \int_e^\infty \frac{1}{t}dt = \infty $$

so it diverges by comparison test? is this right? THANKS!

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    $\begingroup$ Looks good to me! You might want to argue that the integral from $1$ to $e$ is finite, but that's easy. $\endgroup$ Sep 10, 2020 at 19:10

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Your solution is good, but perhaps this is simpler:

If $u = \ln t$ the $du = dt/t$ so $$ \int_1^\infty \frac{\ln t}{t} dt = \int_0^\infty udu = \left. \frac{u^2}{2} \right|_0^\infty $$

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Your way by direct comparison is fine and we can simplify it further as follows

$$\int_1^\infty \frac{\ln(t)}{t}dt \ge \int_e^\infty \frac{\ln(t)}{t}dt \geq \int_e^\infty \frac{1}{t}dt = \infty$$

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