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
    $\begingroup$ why not subing $2^y=x$ and use the geometric series? $\endgroup$ – tired Nov 20 '16 at 17:50
  • $\begingroup$ Take the geometric series. Integrate it. $\endgroup$ – Simply Beautiful Art Nov 20 '16 at 17:51
  • $\begingroup$ Is the base of ur logx 2 $\endgroup$ – Shiksharthi Sharma Nov 20 '16 at 18:27
  • $\begingroup$ @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. $\endgroup$ – aras Nov 20 '16 at 18:32
  • 1
    $\begingroup$ Integrate this:$$1+r+r^2+\dots=\frac1{1-r}$$ $\endgroup$ – 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
    $\begingroup$ :( t he answer is supposed to be $\ln(2)$? $\endgroup$ – Simply Beautiful Art Nov 20 '16 at 18:42
  • 1
    $\begingroup$ Not according to Wolfram Alpha: wolframalpha.com/input/… $\endgroup$ – poweierstrass Nov 20 '16 at 18:44
  • $\begingroup$ 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?) $\endgroup$ – aras Nov 20 '16 at 19:12
  • 1
    $\begingroup$ @aras try the Maclaurin series of arctan. $\endgroup$ – Simply Beautiful Art Nov 20 '16 at 21:18

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.