TITLE: How to evaluate $\int_0^1 \frac{1-x}{\ln x}(x+x^2+x^{2^2}+x^{2^3}+x^{2^4}+\ldots) \, dx$?
QUESTION [11 upvotes]: 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.

REPLY [4 votes]: Another idea is to differentiate under the integral sign to kill the logarithm. By inserting a parameter , call it $\alpha$ , and then differentiate under the integral sign we have that ,  if we define
$$f\left ( \alpha \right ) = \int_{0}^{1} \frac{1-x^\alpha}{\log x} \sum_{n=0}^{\infty} x^{2^n} \, {\rm d}x \quad , \quad \alpha \geq 0$$
then
\begin{align*} 
\frac{\mathrm{d} }{\mathrm{d} \alpha} f(\alpha) &= \int_{0}^{1} \frac{\partial }{\partial \alpha} \frac{1-x^\alpha}{\log x} \sum_{n=0}^{\infty} x^{2^n} \, {\rm d}x \\ &=-\int_{0}^{1} x^\alpha \sum_{n=0}^{\infty} x^{2^n} \, {\rm d}x \\ &= -\sum_{n=0}^{\infty} \int_{0}^{1} x^{2^n} x^\alpha \, {\rm d}x\\ &= -\sum_{n=0}^{\infty} \int_{0}^{1} x^{2^n +\alpha} \, {\rm d}x \\ &=- \sum_{n=0}^{\infty} \frac{1}{\alpha +2^n +1} 
\end{align*}
While you cannot find a general form for the derivative ( at least not without special functions anyway ) , you can at least evaluate the original integral. How, you may ask? Just integrate from $0$ to $1$, thus:
$$f(1)=\int_0^1 f'(\alpha) \, {\rm d}\alpha$$
Therefore,
\begin{align*} 
f(1) &= \int_{0}^{1} f'(\alpha) \, {\rm d}\alpha \\ &= -\sum_{n=0}^{\infty} \int_{0}^{1} \frac{{\rm d}\alpha}{\alpha +2^n +1}\\ &= -\sum_{n=0}^{\infty} \log \left ( \frac{2^n +2}{2^n +1} \right )\\ &= - \lim_{N \rightarrow +\infty} \sum_{n=0}^{N} \log \left ( \frac{2^n +2}{2^n +1} \right ) \\ &= - \lim_{N \rightarrow +\infty} \log \left ( \frac{3 \cdot 2^N}{2^N +1} \right ) \\ &= - \log 3 
\end{align*}