TITLE: A closed form of $\int_0^1{\dfrac{1-x}{\log x}(x+x^2+x^{2^2}+x^{2^3}+\cdots)}\:dx$
QUESTION [5 upvotes]: I need some hint to calculate this integral

$$\int_{0}^{1}{\dfrac{1-x}{\log x}\left(x+x^{2}+x^{2^2}+x^{2^3}+\cdots\right)}{\rm d} x$$

Regards!

REPLY [12 votes]: We have

$$
\int_{0}^{1}{\dfrac{1-x}{\log x}(x+x^{2}+x^{2^{2}}+x^{2^{3}}+\cdots)}\:dx=-\log 3. \tag1
$$

Proof. One may recall that, using Frullani's integral, we have
$$
\int_{0}^{1}\frac{x^{a-1}-x^{b-1}}{\log x}\:dx=\log\frac ab \quad (a,b>0). \tag2
$$
Considering a finite sum in the integrand, we get
$$
\begin{align}
&\int_{0}^{1}{\dfrac{1-x}{\log(x)}(x+x^{2}+x^{2^{2}}+\cdots+x^{2^N})}\:dx\qquad  (N=0,1,2,\cdots)
\\\\&=\sum_{n=0}^N\int_{0}^{1}\frac{x^{2^n}-x^{2^n+1}}{\log x}\:dx\\\\
&=\sum_{n=0}^N\log\frac{2^n+1}{2^n+2}\qquad \quad (\text{using}\, (2))\\\\
&=\sum_{n=0}^N\left(\log(2^n+1)-\log(2^{n-1}+1)-\log 2\right) \quad(\text{a telescoping sum})\\\\
&=\log(2^N+1)-\log(2^{-1}+1)-(N+1)\log 2\\\\
&=-\log 3+\log\left(1+1/2^N\right),
\end{align}
$$ then, letting $N \to+\infty$, leads to $(1)$.
Remark. One may see that we can generalize $(1)$.