Timeline for Reference request for a "well-known identity" in a paper of Shepp and Lloyd
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 28, 2010 at 18:02 | comment | added | John Jiang | Thanks Robin. I will write a short summary of the proof combining all the ingredients given so far. | |
| Mar 28, 2010 at 7:58 | comment | added | Robin Chapman | The integral $\int_0^\infty e^{-t}\log t\,dt$ equals $\Gamma'(1)$. This can be evauated as $-\gamma$ using the infinite product for the gamma function. | |
| Mar 27, 2010 at 20:51 | comment | added | John Jiang | Indeed, it makes it a lot easier. Using integration by part, one can show that $\int_1^{\infty} \exp(-y)/y dy - \int_0^1 \frac{1-\exp(-y)}{y}dy = \int_0^{\infty} \exp(-y) \log y dy which is listed as equal to $-\gamma$ in the following wiki page: en.wikipedia.org/wiki/… I am yet to figure out why that formula in the wiki page is true. | |
| Mar 27, 2010 at 19:55 | comment | added | Qiaochu Yuan | You can prove the identity up to a constant factor by differentiating with respect to x, so it only remains to prove it for x = 1. This should be a little easier. | |
| Mar 27, 2010 at 19:46 | comment | added | John Jiang | Thanks for the nice reference on wiki! I was looking at the wrong place. | |
| Mar 27, 2010 at 19:27 | history | answered | Ryan O'Donnell | CC BY-SA 2.5 |