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.

  • 1
    why not subing $2^y=x$ and use the geometric series? – tired Nov 20 '16 at 17:50
  • Take the geometric series. Integrate it. – Simply Beautiful Art Nov 20 '16 at 17:51
  • Is the base of ur logx 2 – Shiksharthi Sharma Nov 20 '16 at 18:27
  • @tired Could you possibly be more specific? When I do the substitution I get $\int_{-\infty}^0 \log(2) \frac{2^y}{1-y} dy$, but I don't see how to turn this into a geometric series to integrate. – aras Nov 20 '16 at 18:32
  • 1
    Integrate this:$$1+r+r^2+\dots=\frac1{1-r}$$ – Simply Beautiful Art Nov 20 '16 at 18:39

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.

  • 1
    :( t he answer is supposed to be $\ln(2)$? – Simply Beautiful Art Nov 20 '16 at 18:42
  • 1
    Not according to Wolfram Alpha: wolframalpha.com/input/… – poweierstrass Nov 20 '16 at 18:44
  • Well that is confusing...I got this problem from the 2014 MIT Integration Bee (math.mit.edu/~sswatson/pdfs/qualifying_round_2014.pdf), and none of their problems so far have involved $Ei(x)$. This solution makes sense to me though and obviously Wolfram Alpha rarely lies. Do you or @SimpleArt have any idea why the creator of the solution would have come up with the step to write the integral as a sum in the first place? (e.g. Could you point me to other integrals which can be evaluated by writing them as an infinite sum?) – aras Nov 20 '16 at 19:12
  • 1
    @aras try the Maclaurin series of arctan. – Simply Beautiful Art Nov 20 '16 at 21:18

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