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

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).

• 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. – Daniel Fischer Dec 27 '13 at 19:38

## 4 Answers

Hint: For appropiate values of $x$ it holds that $x\ge \log (x)$ and $\dfrac 1{\log (x)}\ge \dfrac 1 x$.

• On the other side, a similar hint: $\frac{1}{\log x}>\frac{1}{x-1}$. – alex.jordan Dec 27 '13 at 21:49

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.

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*}
• 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? – Ryan Dec 28 '13 at 5:58
• @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. – Américo Tavares Dec 28 '13 at 14:09


• Nice!! %%%%%%%%%%%%%%% – Ryan Dec 28 '13 at 5:39