$\int_1^\infty\frac 1{\ln x}dx$ 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).

 A: 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.
A: 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*}

A: $\newcommand{\+}{^{\dagger}}%
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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}
A: Hint: For appropiate values of $x$ it holds that $x\ge \log (x)$ and $\dfrac 1{\log (x)}\ge \dfrac 1 x$.
