How do I prove that $\int_{0}^{1} \frac{1}{\log(x)}dx$ diverges? This is an exercise of a book I'm using to study... The book gives a hint: compare to $f(x)=\frac{1}{1-x}$. However I was not able to realize how could I compare these two functions. I tried changing the variable $x$ to $(1-u)$ in the integral, but it didn't solve my problem. I also tried using the Taylor expansion of $\frac{1}{\log(x)}$, but it was not helpful either...
 A: Let $t=\log x$ then $dx=e^t dt$ and then
$$\int_0^1\frac{dx}{\log x}=\int_{-\infty}^0\frac{e^tdt}{t}$$
and   the last integral is divergent since
$$\frac{e^t}{t}\sim_0\frac{1}{t}$$
A: $$
y=\log x, \qquad dy =\frac{dx}{x}, \qquad dx = x\,dy = e^y\,dy
$$
$$
\int_0^1 \frac{dx}{\log x} = \int_{-\infty}^0 \frac{e^y \, dy}{y}
$$
This obviously converges at the $-\infty$ end since $0< e^y/y \le e^y$.
The problem is at the $0$ end. Suppose $y$ is near $0$?  How near?  Let's say $y\ge -1$, so that $e^y\ge e^{-1}>0$.  Then
$$
\int_{-1}^0 \frac{e^y \, dy}{y} \ge e^{-1}\int_{-1}^0\frac{dy}y = \infty.
$$
A: Working from
$$
e^{-x}\ge1-x
$$
it follows that on $(0,1)$
$$
\frac1{\log(x)}\le1+\frac1{x-1}
$$
Therefore,
$$
\begin{align}
\lim_{\epsilon\to0^+}\int_0^{1-\large\epsilon}\frac1{\log(x)}\,\mathrm{d}x
&\le\lim_{\epsilon\to0^+}\int_0^{1-\large\epsilon}1+\frac1{x-1}\,\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}1-\epsilon+\log(\epsilon)\\[4pt]
&=-\infty
\end{align}
$$
A: Better use $$  1-\frac 1 x\leq \log x$$
To clarify, note that the above gives $$\frac{1}{{\log x}} \leqslant \frac{x}{{x - 1}} \Rightarrow \int\limits_0^t {\frac{{dx}}{{\log x}}}  \leqslant \int\limits_0^t {dx}  + \int\limits_0^t {\frac{{dx}}{{x - 1}}}  \to  - \infty \;,\; t\to 1^{-}$$
ADD The inequality above is really fundamental, following from the definition of the logarithm. Note that $$1-\frac 1x=\int_1^x\frac{dt}{t^2}\leq \int_1^x\frac{dt}t=\log x\leq \int_1^x 1 dt=x-1$$
give the widely used tight intequalities $$1-\frac 1 x\leq \log x\leq x-1$$ which are used to prove, for example, that $$\lim_{x\to 0}\frac{\log(1+x)}x=1$$
A: Suppose to reach a contradiction that $\int_0^1\dfrac{1}{\log x}\,dx$ converges.  Making the substitution $x=u^2$, we get 
$$\int_0^1 \dfrac{1}{\log x}\,dx=\int_0^1\dfrac{u}{\log u}\,du.$$  In other words, 
$$\int_0^1 \dfrac{1}{\log x}\,dx=\int_0^1\dfrac{x}{\log x}\,dx,$$ or
$$\int_0^1 \dfrac{x-1}{\log x}\,dx=0.$$  The integral of a strictly positive function on $(0,1)$ cannot be $0$, so this is impossible.
