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How do I show that $\int_1^\infty\frac 1{\ln x}dx$ diverges? I'm thinking to break up the integral into two parts, say, over $[1,2]$ and $[2,\infty)$, but how do I integrate the integrand?

I'm stumped partly because the author claims part (f) is proved similarly as part (a)/(b). But how can this be!? Part (f) is so much more non-trivial than (a)/(b).

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    $\begingroup$ On the one hand, you have a non-integrable singularity at $1$, on the other, the integrand is larger than $\frac1x$, and the divergence of the latter shows the divergence. $\endgroup$ – Daniel Fischer Dec 27 '13 at 19:38
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Hint: For appropiate values of $x$ it holds that $x\ge \log (x)$ and $\dfrac 1{\log (x)}\ge \dfrac 1 x$.

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    $\begingroup$ On the other side, a similar hint: $\frac{1}{\log x}>\frac{1}{x-1}$. $\endgroup$ – alex.jordan Dec 27 '13 at 21:49
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Notice that $$\frac{1}{\ln x}\geq\frac{1}{x\ln x}\,\;\forall x\geq1$$ and $$\int_1^a \frac{1}{x\ln x}dx=\ln(\ln a)\to+\infty$$ so your integral is divergent.

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Apply the Limit Comparison Test for improper integrals to the functions $f(x)=\frac{1}{\log x}$ and $g(x)=\frac{1}{x}$. Since

  • by L'Hôpital \begin{equation*} \lim_{x\rightarrow \infty }\frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty } \frac{x}{\log x}=\lim_{x\rightarrow \infty }\frac{1}{1/x}=\infty , \end{equation*}
  • at the singularity of $f(x)$, i.e. at the lower limit of integration
    \begin{equation*} \lim_{x\rightarrow 1^+ }\frac{f(x)}{g(x)}=\lim_{x\rightarrow 1^+ } \frac{x}{\log x}=\infty , \end{equation*}
  • and by (a) \begin{equation*} \int_{1}^{\infty }g(x)\, dx=\int_{1}^{\infty }\frac{1}{x}dx \end{equation*} diverges, then so does \begin{equation*} \int_{1}^{\infty }f(x)\, dx=\int_{1}^{\infty }\frac{1}{\log x}dx. \end{equation*}
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  • $\begingroup$ Thank you so much for the link. I had been looking for a formal statement of the comparison test for improper integrals. Speaking of which, are you merely offering your answer as an alternative to the more straightforward direct comparison test? $\endgroup$ – Ryan Dec 28 '13 at 5:58
  • $\begingroup$ @Ryan This was an alternative to the direct comparison test (DCT), but sometimes it is easier to find the limit than to deal directly with the inequality of the DCT. $\endgroup$ – Américo Tavares Dec 28 '13 at 14:09
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $x = \expo{z}$: \begin{align} \int_{1}^{\infty}{\dd x\over \ln\pars{x}} &=\int_{0}^{\infty}{\expo{z}\,\dd z \over z}\quad\mbox{diverges !!!} \end{align}

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  • $\begingroup$ Nice!! %%%%%%%%%%%%%%% $\endgroup$ – Ryan Dec 28 '13 at 5:39

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