Timeline for Inequality for a gamma function
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 17, 2014 at 14:18 | comment | added | Johannes Trost | The proof is more complicated, making use of a generalization of the Phragmén–Lindelöf principle (or theorem as it is also called). (links are en.wikipedia.org/wiki/… or encyclopediaofmath.org/index.php/…) Unfortunately, the above cited book of Rademacher with the proof of the inequality is hard to get. If I find some time, I will post a sketch of the proof here. | |
| Apr 14, 2014 at 15:50 | comment | added | Sergei | In fact $$ |\frac{\Gamma(s)}{\Gamma(2-s)}|=|\frac{\Gamma(s)\Gamma(s-1)}{\Gamma(2-s)\Gamma(s-1)}|=\frac{1}{\pi}|\sin\pi s||\Gamma(s)||\Gamma(s-1)|.$$ Due to the well-known inequality $$|\Gamma(x+iy)|\le|\Gamma(x)|$$ both gammas are $\le 1$. But why sinus is power-bounded? | |
| Apr 14, 2014 at 10:59 | comment | added | Johannes Trost | In Eq. (5.1) of arxiv.org/pdf/1301.1749.pdf the $c$ has to be replaced by $2(\sigma-1)$ (note that this matches the condition $0\leq c\leq 1$), then replace $s\rightarrow s-2(\sigma-1)$ and note that $s=\sigma + i t$ and (most importantly) that $|\Gamma(\sigma - i t)| = |\Gamma(\sigma + i t)|$. | |
| Apr 8, 2014 at 14:11 | history | answered | Sergei | CC BY-SA 3.0 |