# Find convergence of improper integral.

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.

Hint: Make the substitution $u=\log x$. Then $du=\frac{1}{x}\,dx$.
• I see, thank you,my integral would become $\int_{e}^{A} \frac{1}{u^2} du$.I am correct? Jun 16 '14 at 21:33
• Not quite. We need change the bounds, to $1$ and $\log A$. Jun 16 '14 at 21:35
• When you integrate, you will get $1-\frac{1}{\log A}$. Then find the limit of this as $A\to\infty$. Jun 16 '14 at 21:43
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: $$\int_e^\infty \frac{dx}{x\ln^2 x} = 1$$
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