Timeline for Regularisation of $\sum_{n=0}^\infty \frac{1}{(a+n^2)^p}$
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 30, 2016 at 4:31 | comment | added | BLS | Don't worry. Anyway, if somebody can give me suitable references (about how to deal with this kind of zeta and how to regularise it) I would be much appreciated (in the general case, since I cannot take the $a>>1$ approx) | |
| Sep 30, 2016 at 2:23 | comment | added | reuns | @CarloBeenakker You are clearly dis-respectful. | |
| Sep 29, 2016 at 11:15 | comment | added | Carlo Beenakker | @BLS --- my mistake, apologies, I do not have a closed form expression for $p=3/2$ :( | |
| Sep 29, 2016 at 9:27 | comment | added | BLS | btw, @CarloBeenakker, how do you get the result you wrote for $p=3/2$, Mathematica can't find the explicit form... | |
| Sep 29, 2016 at 7:08 | comment | added | BLS | Ok, but what about general values for $a$? Isn't $F_a(s) - \frac{1}{s-1/2}$ finite for $s \rightarrow 1/2$ ? | |
| Sep 28, 2016 at 17:47 | comment | added | reuns | More precisely $\lim_{s \to 1/2^+} F_a(s) = + \infty$ and (if $|a| < 1$) $\lim_{s \to 1/2^+} F_a(s) - \frac{1}{s-1/2} = \frac{\gamma}{2} + \sum_{k=0}^\infty {-1/2 \choose k} a^k \zeta(2k+1)$ is finite | |
| Sep 28, 2016 at 6:12 | comment | added | BLS | Do I have to expand around $\epsilon$? | |
| Sep 28, 2016 at 6:11 | comment | added | BLS | Thank you guys! Now, how can I exactly regularise it by adding $\epsilon$ to $p$ (never done that, sorry for the dumb question)? I need to find a way to extract a finite result like in the Riemann's zeta case (where we can take express the zeta as div+finite parts and one can take the average to cancel the div part, for instance). Moreover, I want to keep $a$ as general as possible, without making any assumption like $a>>1,a<<1$. ( $a>>1$ has actually a physical meaning, but then I recover the standard zeta, when $p=1/2$, and I know how to regularise). I want to understand the general case. | |
| Sep 27, 2016 at 16:56 | answer | added | reuns | timeline score: 3 | |
| Sep 27, 2016 at 12:05 | comment | added | Carlo Beenakker | en.wikipedia.org/wiki/Integral_test_for_convergence | |
| Sep 27, 2016 at 11:43 | comment | added | Carlo Beenakker | @user1952009 --- it's divergent for $s\leq 1/2$. | |
| Sep 27, 2016 at 11:37 | comment | added | reuns | expanding $(x^2+a)^{-s} = x^{-2s}\sum_{k=0}^\infty {-s \choose k} a^k x^{-2k}$ I get that $\displaystyle\sum_{n=m}^\infty (n^2+a)^{-s} = \sum_{k=0}^\infty {-s \choose k} a^k (\zeta(2k+2s)-\sum_{n < m} n^{-2k-2s} ) $ converges for every $s$ provided $m > a$ | |
| Sep 27, 2016 at 10:19 | answer | added | Carlo Beenakker | timeline score: 3 | |
| Sep 27, 2016 at 9:41 | review | Low quality posts | |||
| Sep 27, 2016 at 10:53 | |||||
| Sep 27, 2016 at 8:38 | history | asked | BLS | CC BY-SA 3.0 |