[Math] Study the convergence or divergence of the improper integral $\int_2^\infty \frac{1}{\sqrt x\cdot \ln x} dx.$

Question

I am stuck in solving the following exercise, please help me with the improper integral

$$\int_2^\infty \frac{1}{\sqrt x\cdot \ln x} dx.$$
I am asked to determine whether it is divergent or convergent.

Answer 1

Setting $t=\sqrt{x}$, we have $$ \int_2^\infty\frac{1}{\sqrt{x}\ln x}\,dx=\int_{\sqrt{2}}^\infty\frac{2t}{t\ln t^2}\,dt=\int_{\sqrt{2}}^\infty\frac{1}{\ln t}\,dt \ge\int_{\sqrt{2}}^\infty\frac{1}{t}\,dt=\infty. $$