Timeline for The cotangent sum $\sum_{k=0}^{n-1}(-1)^k\cot\Big(\frac{\pi}{4n}(2k+1)\Big)=n$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Aug 10, 2020 at 16:11 | comment | added | Iosif Pinelis | @bryanjaeho : Oops! The integral is of course from$0$ to $\pi/2$, not to $1$. This typo is now corrected | |
| Aug 10, 2020 at 16:08 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 6 characters in body
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| Aug 10, 2020 at 16:01 | vote | accept | bryanjaeho | ||
| Aug 10, 2020 at 16:00 | comment | added | bryanjaeho | I believe the integral should be over $\frac{1}{2}\Big(\frac{1}{x}-\frac{\pi}{2}\cdot \cot(\pi/2\cdot x)\Big)$ ranging from $0$ to $1$. Thank you for the approach! | |
| Aug 10, 2020 at 15:41 | comment | added | bryanjaeho | Just to make sure, the definite integral on the left gives $-\ln(\sin(1))$? | |
| Aug 10, 2020 at 13:59 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |