Prove that $I = \int_0^{1/2} (\ln|\ln(r)|)^2rdr$ is convergent I need to prove that the following integral is convergent:
$I = \int_0^{1/2} (\ln|\ln(r)|)^2rdr$.
I can write the  integral as  $I = \int_0^{1/2} (\ln|\ln(r)|)^2rdr = \lim_{\varepsilon \to 0}  \int_\varepsilon^{1/2} (\ln|\ln(r)|)^2r$
If I use a Taylor Expansion around $r = 1$:  $\ln(r)=\ln(1+r-1)=(r-1)+o(r-1)$, the integrand becomes $(\ln|\ln(r)|)^2r = (\ln|(r-1)+o(r-1)|)^2r  $
I am not so sure if this is correct so far, since I am expanding around $r=1$, but in the integrand I should care of $r = 0$, being that the "problematic" integration limit , but certainly I can't expand around $0$, so if my calculation is correct, why does it make sense? I am stuck here too, I don't know how to get rid of the $|\cdot|$ and how to proceed from here. Can someone shed some light?
 A: Nothe that as $r \to 0$
$$(\ln|\ln(r)|)^2r \to 0$$
indeed
$$\ln|\ln(r)|<-\ln r$$
and
$$\sqrt r \ln r \to 0$$
therefore the given integral converges.
A: $\newcommand{\d}{\,\mathrm{d}}$Taylor expansions can be useful for assessing convergence. However I don't think that's the easiest path of approach here: rather, some elementary bounds can be made.
$0\le|\ln(r)|=\ln(1/r)\le1/r$ for all $1>r>0$. If, in addition, $|\ln(r)|>1$ then that means you can bound $0<\ln|\ln(r)|<\ln(1/r)$. The integrand is continuous on $(0,1/2)$, so to assess convergence it remains to assess convergence on $(\varepsilon,\delta)$ for all small $0<\varepsilon<\delta$ as $\varepsilon\to0^+$. In particular we can affix a $\delta$ small enough that $|\ln(r)|>1$ for all $r\in(0,\delta)$.
Then: $$0<\int_{\varepsilon}^{\delta}(\ln|\ln r|)^2r\d r<\int_{\varepsilon}^{\delta}(\ln(1/r))^2r\d r=\int_{\varepsilon}^{\delta}\ln^2(r)r\d r$$But: $$\lim_{r\to0^+}\ln^2(r)r\overset{r=e^{-x}}{=}\lim_{x\to\infty}x^2e^{-x}=0$$So it follows that: $$0<\lim_{\varepsilon\to0^+}\int_{\varepsilon}^{\delta}\ln^2(r)r\d r<\infty$$Which also means that your original integral is convergent.
If you are familiar with the gamma function, you can show the integral converges by (this is overkill) computing: $$\int_0^1(\ln|\ln r|)^2r\d r=\frac{\d^2}{\d s^2}[\Gamma(s)2^{-s}]\Big|_{s=1}=\frac{1}{2}\left[(\gamma+\ln2)^2+\frac{\pi^2}{6}\right]$$If my mental maths holds.
A: To get rid of the absolute value, note that
\begin{equation*}
\left \vert \ln r\right \vert =\ln \frac{1}{r}\text{ whenever }r\in (0,\frac{1%
}{2})\text{.}
\end{equation*}
So, the problem turns into
\begin{equation*}
\int_{0}^{1/2}r\left( \ln \left \vert \ln r\, \right \vert \right)
^{2}dr=\int_{0}^{1/2}r\left( \ln \left( \ln \frac{1}{r}\right) \, \right)
^{2}dr\text{.}
\end{equation*}
Define $x=1/r$. Then, by chance of variables, it becomes
\begin{equation*}
\int_{2}^{\infty }\frac{1}{x}\left( \frac{\ln \left( \ln x\right) }{x}
\, \right) ^{2}dx\text{,}
\end{equation*}
which is more convenient to study. Notice that since the integrand is always
positive, so is the integration itself:
\begin{equation*}
0\leq \frac{1}{x}\left( \frac{\ln \left( \ln x\right) }{x}\, \right)
^{2}\Longrightarrow 0\leq \int_{2}^{\infty }\frac{1}{x}\left( \frac{\ln
\left( \ln x\right) }{x}\, \right) ^{2}dx\text{.}
\end{equation*}
Since the question only asks the convergence and we know it has a lower
limit, it will suffice for our purposes to find an upper limit for the
integral. Here, a simple trick would do the job. Notice that
\begin{equation*}
\frac{\ln \left( \ln x\right) }{x}\leq \frac{1}{\sqrt{x}}\text{,}
\end{equation*}
which, given $x\geq 2$, holds if and only if
\begin{equation*}
0\leq \sqrt{x}-\ln \ln x\text{.}
\end{equation*}
It is a simple matter to check that
\begin{equation*}
\frac{\partial (\sqrt{x}-\ln \ln x)}{\partial x}=\ \frac{\sqrt{x}
\ln x-2}{2x\ln x}\left \{ 
\begin{array}{cc}
>0 & \text{if }x>x^{\ast } \\ 
<0 & \text{if }x<x^{\ast }
\end{array}
\right. \text{,}
\end{equation*}
where $x^{\ast }\approx 3.\,109$ uniquely solves
\begin{equation*}
\sqrt{x^{\ast }}\ln x^{\ast }-2=0\text{.}
\end{equation*}
Thus, we conclude that $\sqrt{x}-\ln \ln x$ attains its minimum at $x^{\ast }
$. That is, we have
\begin{equation*}
\sqrt{x^{\ast }}-\ln \ln x^{\ast }\leq \sqrt{x}-\ln \ln x\text{ for all }
x\geq 2\text{.}
\end{equation*}
Note that $\sqrt{x^{\ast }}\ln x^{\ast }-2=0$ implies that
\begin{equation*}
\ln x^{\ast }=\frac{2}{\sqrt{x^{\ast }}}\text{,}
\end{equation*}
which when plugged into $\sqrt{x^{\ast }}-\ln \ln x^{\ast }$ gives
\begin{equation*}
\sqrt{x^{\ast }}-\ln \ln x^{\ast }=\sqrt{x^{\ast }}-\ln \frac{2}{\sqrt{
x^{\ast }}}>0\text{ \ (since it is decreasing in }x^{\ast }\text{ and }\sqrt{
2}-\ln \frac{2}{\sqrt{2}}>0\text{).}
\end{equation*}
Thus, we have shown that
\begin{equation*}
0<\sqrt{x}-\ln \ln x\text{ for all }x\geq 2\text{,}
\end{equation*}
which ensures that
\begin{equation*}
\frac{\ln \left( \ln x\right) }{x}<\frac{1}{\sqrt{x}}\text{.}
\end{equation*}
Now, using this last inequality, we obtain
\begin{equation*}
0\leq \int_{2}^{\infty }\frac{1}{x}\left( \frac{\ln \left( \ln x\right) }{x}
\, \right) ^{2}dx<\int_{2}^{\infty }\frac{1}{x}\left( \frac{1}{\sqrt{x}}%
\, \right) ^{2}dx=\int_{2}^{\infty }\frac{1}{x^{2}}dx=\frac{1}{2}\text{,}
\end{equation*}
which insures convergence and ends the proof. Hope this helps.
