Difficulty understanding way of evaluating integral

Question

I was trying to evaluate the definite integral

$$ \int_0^1 \frac{1}{1 - \log_2(x) } .$$

The solution was just one line and read

$$ \int_0^1 \frac{1}{1 - \log_2(x) } = \sum_{k=1}^\infty \frac{1}{k2^k} = \log(2). $$

Both of these steps are entirely non-obvious for me and I have no idea how to justify them. Any help would be greatly appreciated.

Answer 1

Let $z=1-\frac{\ln x}{\ln2}$, then $y=z\ln2$

\begin{align} \int\limits_{0}^{1} \frac{1}{1-\log_{2}x} dx &= 2\ln2 \int\limits_{1}^{\infty} \frac{1}{z} \mathrm{e}^{-z\ln2} dz \\ &= 2\ln2 \int\limits_{\ln2}^{\infty} \frac{\mathrm{e}^{-y}}{y} dy \\ &=-(2\ln2)\mathrm{Ei}(-\ln2) \approx 0.52495 \end{align}

\begin{equation} \mathrm{Ei}(z) = -\int\limits_{-z}^{\infty} \frac{\mathrm{e}^{-t}}{t} dt \end{equation} is the exponential integral function.