I'm doing all of the integrals that are in the MIT 2006 Integration Bee (video is posted on Youtube). I am stuck on one of the integrals.

At about the 55 minute mark, the following definite integral shows up.

$$\int_{\frac{1}{\pi}}^{\frac{1}{2}} \ln{\left \lfloor{\frac{1}{x}}\right \rfloor} \, dx $$

The answer that one of the commentators (an MIT math major) gave off the top of his head was: $$\frac{1}{6}{\ln\left(2\right)}+\left({\frac{1}{3} - \frac{1}{\pi}}\right){\ln\left(3\right)} \approx 0.13202947375$$

Is the commentator's answer correct?

The final answer turned out to be: $$\left({\frac{1}{2} - \frac{1}{\pi}}\right){\ln\left(3\right)} \approx 0.19960699176$$

How do I solve this integral step-by-step? Also, if the commentator's answer was correct, how does the commentator's answer simplify to the final answer? And why are the numeral approximations done by a calculator so wildly different from one another?

NOTE: The definite integral was written as: $$\int_{\frac{1}{\pi}}^{\frac{1}{2}} \log{\left \lfloor{\frac{1}{x}}\right \rfloor} \, dx $$ And the final answer was given as: $$\left({\frac{1}{2} - \frac{1}{\pi}}\right){\log\left(3\right)}$$ However, both of the commentators noted that log in all of these Integration Bee problems was actually natural log (ln), even though in other situations, it's base 10.

  • 2
    $\begingroup$ Your answer appears to be correct. The other answers are, therefore, not correct. $\endgroup$
    – Mark Viola
    Nov 28, 2019 at 4:34
  • $\begingroup$ Was there a floor function in the integrand somewhere? $\endgroup$
    – JimmyK4542
    Nov 28, 2019 at 4:50
  • $\begingroup$ @JimmyK4542 Yes! You are correct, the video is from 2006 and not the best quality so I missed the floor function. I will edit the question to reflect this. $\endgroup$
    – rplee
    Nov 28, 2019 at 5:16
  • $\begingroup$ With the correction, I think $\frac{1}{6}{\ln\left(2\right)}+\left({\frac{1}{3} - \frac{1}{\pi}}\right){\ln\left(3\right)} \approx 0.13202947375$ is the correct answer. On the interval $[\tfrac{1}{\pi},\tfrac{1}{3}]$, the integrand is $\ln 3$, and on $(\tfrac{1}{3},\tfrac{1}{2}]$, the integrand is $\ln 2$. Hence, the answer. $\endgroup$
    – JimmyK4542
    Nov 28, 2019 at 5:37

1 Answer 1


The commentator is correct. Note from $\frac{1}{\pi}$ to $\frac{1}{2}$, $\lfloor{\frac{1}{x}}\rfloor$ can take value of 3 and 2 and the cut off point is $x=\frac{1}{3}$. So the integration is really $\int_{\frac{1}{\pi}}^{\frac{1}{3}}ln(3)\ dx + \int_{\frac{1}{3}}^{\frac{1}{2}}ln(2)\ dx=\left(\frac{1}{3} - \frac{1}{\pi}\right)ln(3) + \frac{1}{6}ln(2).$

  • 1
    $\begingroup$ This explanation makes a lot of sense. As an aside, it's interesting that the commentator's answer off the top of his head was correct, but the final answer that was revealed must have been incorrect! $\endgroup$
    – rplee
    Nov 28, 2019 at 5:51
  • 1
    $\begingroup$ The so called "final answer" IS incorrect. $\endgroup$ Nov 28, 2019 at 5:56

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