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!