check the convergence of the integral $\int_{0}^{\infty}\frac{1}{x\log x}\,dx$

Question

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!

Answer 1

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.

Answer 2

Use the substitution $u=\log x$. And compare with the convergence of $$\int\frac{1}{u} \,du$$

Answer 3

With the substitution $u=\log x, du=\frac{1}{x} dx$, we find the antiderivative $\frac{1}{(\log x)^2}$ to the indefinite integral. To find the definite integral, we split it into $\int_0^1 \frac{1}{x\log x} dx+\int_1^\infty \frac{1}{x\log x} dx$, replace the improper endpoints ($0, \infty$) with constants, use the antiderivative above, then take limits as those constants approach $0,\infty$ respectively.

Answer 4

We have $$\int\frac{dx}{x\log x}=\log(|\log x|)+C\to\infty\quad\text{when}\quad x\to\infty \quad\text{or}\quad x\to0$$ so for $a>0$ the both integrals $$\int_0^a\frac{dx}{x\log x}\quad;\quad \int_a^\infty\frac{dx}{x\log x}$$ are divergent.

Answer 5

The accepted answer is wrong and the book is right (no offence). Make the substition u=-log(x) into vadim's first integral $\int_0^1\frac{1}{x \log(x)}dx$ add it to vadim's second integral and you will see that it converges.