Find convergence of improper integral.

Question

Hello I have to find the convergence of this improper integral: $$\int_{e}^{\infty} \frac{1}{x\log^2x} dx$$ So I started by doing the following: $\lim \limits_{x \to A} \int_{e}^{A} \frac{1}{x\log^2x} dx$, but I don't really know how to solve this integral:$\int_{e}^{A} \frac{1}{x\log^2x} dx$.

Any tips would be great, thank you.

Answer 1

Hint: Make the substitution $u=\log x$. Then $du=\frac{1}{x}\,dx$.

Answer 2

Integration by parts gives us: $$ \int_e^\infty \frac{dx}{x\ln ^2x} = \left[ \frac{1}{\ln x} \right]_e^\infty +2\int_e^\infty \frac{\ln x \,dx}{x \ln^3 x} = -1 +2\int_e^\infty \frac{dx}{x\ln^2 x} $$ Thus: $$ \int_e^\infty \frac{dx}{x\ln^2 x} = 1 $$

Answer 3

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\begin{align} &\color{#66f}{\large\lim_{\delta \to 0^{+}}\bracks{% \int_{\epsilon}^{1 - \delta}{\dd x \over x\ln^{2}\pars{x}} +\int_{1 + \delta}^{\infty}{\dd x \over x\ln^{2}\pars{x}}}} \\[3mm]&=\lim_{\delta \to 0^{+}}\bracks{% -\int_{-\ln\pars{\epsilon}}^{-\ln\pars{1 - \delta}}{\dd t \over t^{2}} -\int_{-\ln\pars{1 + \delta}}^{-\infty}{\dd t \over t^{2}}} \\[3mm]&=\lim_{\delta \to 0^{+}}\bracks{% -\,{1 \over \ln\pars{1 - \delta}} +{1 \over \ln\pars{\epsilon}} +{1 \over \ln\pars{1 + \delta}}} = \color{#66f}{\large +\infty} \end{align}