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I am self studying Tom M Apostol Introduction to Analytic number theory.

In theorem 4.12 Apostol uses that improper integral $\int_x^{\infty} \frac {1} { t (logt) ^2 } \ , dt $ converges, x>2 .

I tried using comparison test by comparing with $t^{3/2} $ and $ t^2 $ but I don't get non-zero finite limit of there ratios as t tends to $\infty $ .

Can someone please tell how to prove this integral to be convergent .

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    $\begingroup$ Hint: The integrand is exactly the derivative of$$-\frac1{\log{(t)}}$$which can be seen by substituting $u=\log{(t)}$. So the given integral converges to $$\frac1{\log{(x)}}$$ $\endgroup$ – Peter Foreman Dec 26 '19 at 14:05
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The integrand has a simple antiderivative:

$$\int_x^{\infty} \frac{dt}{t (\log{t})^2} = \left [ -\frac1{\log{t}} \right ]_x^{\infty} = \frac1{\log{x}}$$

Note that the integral converges because $\log{t} \to \infty$ as $t \to \infty$.

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