# Evaluate the following integral: $\int_{\frac{1}{\pi}}^{\frac{1}{2}}\ln{\lfloor\frac{1}{x} \rfloor}\, dx$

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: $$\frac{1}{6}{\ln\left(2\right)}+\left({\frac{1}{3} - \frac{1}{\pi}}\right){\ln\left(3\right)} \approx 0.13202947375$$

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.

• Your answer appears to be correct. The other answers are, therefore, not correct. Nov 28, 2019 at 4:34
• Was there a floor function in the integrand somewhere? Nov 28, 2019 at 4:50
• @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. Nov 28, 2019 at 5:16
• 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. Nov 28, 2019 at 5:37

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 $$\int_{\frac{1}{\pi}}^{\frac{1}{3}}ln(3)\ dx + \int_{\frac{1}{3}}^{\frac{1}{2}}ln(2)\ dx=\left(\frac{1}{3} - \frac{1}{\pi}\right)ln(3) + \frac{1}{6}ln(2).$$