A closed form of $\int_0^1{\dfrac{1-x}{\log x}(x+x^2+x^{2^2}+x^{2^3}+\cdots)}\:dx$

Question

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!

Answer 1

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 $(1)$.