Help me on checking the convergence of the integral $$\int_{0}^{\infty}\frac{1}{x\log x}\,dx$$
I have tried it in this way $$\int_{0}^{\infty}\frac{1}{x\log x}\,dx=\int_{0}^{\frac{1}{2}}\frac{1}{x\log x}\,dx+\int_{\frac{1}{2}}^{1}\frac{1}{x\log x}\,dx+\int_{1}^{2}\frac{1}{x\log x}\,dx+\int_{2}^{\infty}\frac{1}{x\log x}\,dx$$and then by taking comparison integral $\frac{1}{1-x}$, I came into conclusion that $\int_{\frac{1}{2}}^{1}\frac{1}{x\log x}\,dx$ is divergent, then I thought the given integral must be divergent, but when I saw it's answer it's given that it is convergent. Please help.
Thanks!
Best Answer
Integral convergent test says that $$\int_2^\infty \frac{dx}{x\log x}$$ converges if and only if this sum converges: $$\sum_{n = 2}^\infty \frac{1}{n\log n}$$ Use Cauchy condensation test we can see this sum does not converge. Your suggested answer is wrong.