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