# Regarding convergence of improper integral to be used in Analytic number theory

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 .

• 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 $$\frac1{\log{(x)}}$$ – Peter Foreman Dec 26 '19 at 14:05

$$\int_x^{\infty} \frac{dt}{t (\log{t})^2} = \left [ -\frac1{\log{t}} \right ]_x^{\infty} = \frac1{\log{x}}$$
Note that the integral converges because $$\log{t} \to \infty$$ as $$t \to \infty$$.