How to evaluate $\int_0^1 \frac{1-x}{\ln x}(x+x^2+x^{2^2}+x^{2^3}+x^{2^4}+\ldots) \, dx$?

Evaluate the definite integral: $$\int_0^1 \frac{1-x}{\ln x}(x+x^2+x^{2^2}+x^{2^3}+x^{2^4}+\ldots) \, dx$$

I think the series involving $x$ converges because $x\in[0,1]$, but I cannot form an expression for the series. If I let $$ u_n=x^{2^{n-1}} \\ \frac{\ln u_n}{\ln x}=2^{n-1} $$ but then this series does not converge. Even WolframAlpha cannot evaluate a definite integral together with an infinite series, so I am stuck on this.

Solution 1:

Hint. One may recall Frullani's integral, for $a,b>0$, $$ \int_{0}^{1}\frac{x^{a-1}-x^{b-1}}{\ln x}\:dx=\ln\frac ba \tag1 $$ giving, by telescoping terms, for $N=0,1,2,\cdots$, $$ \begin{align} &\int_{0}^{1}{\frac{1-x}{\ln x}(x+x^{2}+x^{2^{2}}+\cdots+x^{2^N})}\:dx \\\\&=\sum_{n=0}^N\int_{0}^{1}\frac{x^{2^n}-x^{2^n+1}}{\ln x}\:dx\\\\ &=\sum_{n=0}^N\ln\frac{2^n+1}{2^n+2}\\\\ &=\sum_{n=0}^N\left[\ln(2^n+1)-\ln(2^{n-1}+1)-\ln 2\right]\\\\ &=\ln(2^N+1)-\ln(2^{-1}+1)-(N+1)\ln 2\\\\ &=-\ln 3+\ln\left(1+1/2^N\right), \end{align} $$ then, by letting $N \to \infty$, one gets

$$ \int_{0}^{1}{\dfrac{1-x}{\ln x}(x+x^{2}+x^{2^{2}}+x^{2^{3}}+\cdots)}\:dx=-\ln 3. \tag2 $$

Edit. One may justify the interchange between $\displaystyle\int$ and $\displaystyle\sum$ by noticing that $$ \left|\frac{1-x}{\ln x}\right|\le1,\qquad 0<x<1,\tag3 $$ and that $$ x+x^{2}+x^{2^{2}}+x^{2^{3}}+\cdots \sim -\frac{1}{\ln 2}\:\ln (-\ln x),\quad \text{as} \quad x \to 1^-,\tag4 $$ proved here (example 12, p.31).

Solution 2:

Claim 1: For $k\geq 1$, we have that \begin{align} \int^1_0 \frac{1-x}{\log x}x^{2^k}\ dx = -\log \frac{2^k+2}{2^k+1}. \end{align}

Claim 2: We have \begin{align} \prod^\infty_{k=0}\left( 1+\frac{1}{2^k+1}\right) = 3 \end{align}

Using the claims, we have the series \begin{align} -\sum^\infty_{k=0} \log \left(\frac{2^k+2}{2^k+1}\right) =-\log\left(\prod^\infty_{k=0} \left(1+\frac{1}{2^k+1}\right) \right) =-\log 3. \end{align}

Solution 3:

First note that

$$\int_0^1 x^y\,dy=\frac{x-1}{\log(x)}\tag1$$

Next, using $(1)$ along with Fubini's Theorem reveals

$$\begin{align} \int_0^1 \frac{1-x}{\log(x)}\,x^{2^n}\,dx&=-\int_0^1 \left(\int_0^1 x^y\,dy\right)\,x^{2^n}\,dx\\\\ &=-\int_0^1 \left(\int_0^1 x^{y+2^n}\,dx\right)\,\,dy\\\\ &=-\int_0^1 \frac{1}{y+2^n}\,dy\\\\ &=-\log(2^n+2)+\log(2^n+1)\\\\ &=-\log(2^n+2)+\log(2^{n+1}+2)-\log(2)\tag2 \end{align}$$

which contains a telescoping term. Finally, summing $(2)$ over $n$ yields

$$\sum_{n=0}^\infty \int_0^1 \frac{1-x}{\log(x)}\,x^{2^n}\,dx=-\log(3)$$

Fubini's Theorem under the counting measure may be used again to justify interchanging the series with the integral.