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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.

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

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  • $\begingroup$ I see, thank you,my integral would become $\int_{e}^{A} \frac{1}{u^2} du$.I am correct? $\endgroup$
    – user137209
    Jun 16 '14 at 21:33
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    $\begingroup$ Not quite. We need change the bounds, to $1$ and $\log A$. $\endgroup$ Jun 16 '14 at 21:35
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    $\begingroup$ When you integrate, you will get $1-\frac{1}{\log A}$. Then find the limit of this as $A\to\infty$. $\endgroup$ Jun 16 '14 at 21:43
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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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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\large\tt\mbox{With}\quad 0\ <\ \delta\ <\ 1\quad\mbox{and}\quad \pars{~x\ \equiv\ \expo{-t}\quad\imp\quad t\ =\ -\ln\pars{x}~}:}$

\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}

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