Does $\int_0^{1/2} \frac{1}{x\ln x}dx$ converge? I tried this:
$$
\begin{align*}
\ln x &= t \\
\frac{1}{x} dx &= dt \\
\lim_{x \to 0^+} \ln x &= -\infty
\end{align*}
$$
So now we have
$$
\int_{-\infty}^{\ln(1/2)} \frac1t dt
$$
which does not converge, is this correct?
 A: Hint
$$\int\frac{1}{x\ln(x)}\mathrm d x=\ln(|\ln(x)|).$$
A: $\int _0^{\frac{1}{2}}\frac{1}{x\ln \left(x\right)}dx=-\infty \:$

$\mathrm{Apply\:Integral\:Substitution:}\:\int
f\left(g\left(x\right)\right)\cdot g^{'}\left(x\right)dx=\int
 f\left(u\right)du,\:\quad u=g\left(x\right)$
$u=\ln \left(x\right)\quad \:du=\frac{1}{x}dx$
$=\int \frac{1}{u}du$

$\mathrm{Substitute:}\:u=\ln \left(x\right)$
$\frac{du}{dx}=\frac{1}{x}$
$\Rightarrow \:du=\frac{1}{x}dx$
$\Rightarrow \:dx=xdu$
$=\int \frac{1}{xu}xdu$
$=\int \frac{1}{u}du$
$\mathrm{Use\:the\:common\:integral}:\quad \int \:\frac{1}{u}du=\ln \left(\left|u\right|\right)$

$\mathrm{Substitute\:back}\:u=\ln \left(x\right)$

$=\ln \left|\ln \left(x\right)\right|$

$Add\:a\:constant\:to\:the\:solution$

$=\ln \left|\ln \left(x\right)\right|+C$

$\mathrm{Compute\:the\:boundaries}:\quad \int _0^{\frac{1}{2}}\frac{1}{x\ln \left(x\right)}dx=\ln \left(\ln \left(2\right)\right)-\infty \:$
$\int _a^bf\left(x\right)dx=F\left(b\right)-F\left(a\right)=\lim _{x\to \:b-}\left(F\left(x\right)\right)-\lim _{x\to \:a+}\left(F\left(x\right)\right)$
$\lim _{x\to \:0+}\left(\ln \left|\ln \left(x\right)\right|\right)=\infty $
$\lim _{x\to \frac{1}{2}-}\left(\ln \left|\ln \left(x\right)\right|\right)=\ln \left(\ln \left(2\right)\right)$

$=\ln \left(\ln \left(2\right)\right)-\infty $
$=-\infty \:$


 $\text{Since infinity is not a real number, so given integral not convergent.}$


A: Your question was already answered, but I guess I have useful hints for you, enabling you to say whether a series/integral converges or not for many series.
1. Hint: Cauchy condensation test: 
   Is $(a_n)_n \subseteq [0,\infty)$ a monotone decreasing sequence, then $\sum_{n=1}^{\infty}a_n$ converges iff $\sum_{k=0}^\infty 2^ka_{2^k}$ converges.
2. Hint: Integral test for convergence: you already know that.
3. Hint: Abel's series For every $i \in \mathbb{N}$ the following series 
$$ \sum_{n=k}^{\infty}\frac{1}{n\ln(n)\ln(\ln(n))\cdots(\ln^{[i]}(n))^s}$$
(absolutely) converges iff $s > 1,$ for $k$ sufficiently large, where $\ln^{[1]}(n) \colon\!= \ln(n), \ln^{[2]}(n) \colon\!= $ $\ln(\ln(n)), \ln^{[3]}(n) \colon\!= $ $\ln(\ln(\ln(n)))...$ . That can be shown with the criteria I mentioned before.
