Convergence of Improper Integral: $\int_{e^2}^\infty {dx\over x\log\log x}$

Question

Test the convergence of the following integral$$\int_{e^2}^\infty {dx\over x\log\log x}$$

I understand that the problem is only at $\infty$ how to proceed ?

Answer 1

Hint: Your function $\frac{1}{x\log\log x}$ goes to $0$ more slowly than $\frac{1}{x\log x}$. You can comfortably find $\int_{e^2}^M \frac{dx}{x\log x}$ explicitly, and show that it (sedately) blows up as $M\to \infty$.

Answer 2

The integral converges iff the series

$$\sum_{n>e^2}\frac1{n\log\log n}\;\;\text{converges}$$

Applying the condensation test to this series, we get it converges iff

$$\sum_{n>e^2}\frac{2^n}{2^n\log\log2^n}=\sum_{n>e^2}\frac1{\log n+\log\log2}$$

and this last series diverges...

Answer 3

For $x>100$ $$\frac{dx}{x\ln\ln x} \ge \frac{dx}{x\ln x\ln\ln x}=\frac{d\ln x}{\ln x\ln \ln x}=\frac{d \ln \ln x}{\ln \ln x}=d \ln \ln \ln x,$$which implies the divergence, because the limits of integration are $[\ln \ln \ln 100,\infty)$.