Timeline for Are the zeros of the sum/difference of these integrals all on the critical line?
Current License: CC BY-SA 3.0
23 events
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| Aug 24, 2013 at 6:03 | comment | added | joro | @TomDickens Thanks. Sage is confused indeed, but Maple almost always gives corrects result for integrals of frac() and floor() using the OP method. Asked and gave examples of correct integrals here: mathoverflow.net/questions/140084/… | |
| Aug 23, 2013 at 19:27 | comment | added | Tom Dickens | @joro Don't know if you have cleared this up to your satisfaction yet. Sorry I haven't gotten back to you. What I mean is that I believe Sage/Maxima has made an incorrect assumption to do the integral, since the $\log()$ changes abruptly every time the phase goes through an integer multiple of $2\pi$. I hope to write a coherent reply this weekend! | |
| Aug 22, 2013 at 6:46 | vote | accept | Agno | ||
| Aug 22, 2013 at 6:03 | history | edited | joro | CC BY-SA 3.0 |
Zeros in the critical strip
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| Aug 21, 2013 at 12:08 | history | edited | joro | CC BY-SA 3.0 |
Found zeros after expressing I(s) with zeta.
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| Aug 21, 2013 at 11:28 | comment | added | joro | @TomDickens is breaking the integral from n to n+1 enough? When I compute $\int_{1/2}^1 \{1/x\} dx$ using the OP method my result disagrees with mathematica, why so? | |
| Aug 20, 2013 at 23:04 | comment | added | Tom Dickens | The problem is that, while the formula in the original post is correct, if you take the correct branch of the log, the integration by the CAS is not compatible with that. The integral would have to be broken up into a series of ranges $x=n $ to $x=n+1$, which just reproduces the sum for the zeta function. | |
| Aug 20, 2013 at 7:18 | comment | added | Agno | Joro. You are right. Sloppy work from my side. I should stop working this from my iPad and will fix this properly when I am back home on Wednesday. P.S. Does the Mathematica code work? | |
| Aug 20, 2013 at 6:37 | comment | added | joro | @LoïcTeyssier The OP fixed two critical typos and his mathematica code disagrees with his formula... | |
| Aug 19, 2013 at 19:36 | comment | added | Agno | Joro, Loic. Apologies for the errors. Unfortunately I am on business travel and away from my desktop. Therefore can't check my Maple code until Wednesday evening. However have fixed an error in the OP and at least it now works in Wolfram Alpha! | |
| Aug 19, 2013 at 17:45 | comment | added | Loïc Teyssier | @Joro: that's another issue altogether… | |
| Aug 19, 2013 at 13:24 | comment | added | joro | @LoïcTeyssier since log() is multivalued I agree with you. Don't get numerical support for the OP claims using the principal branch of the logarithm. | |
| Aug 19, 2013 at 13:20 | comment | added | Loïc Teyssier | @Joro: I disagree that $\log(\exp(t))=t$ for every purely imaginary number $t$, if $\log$ is to be understood as a function holomorphic on $\mathbb C$ with a real half-line removed. That's the whole point of writing the fractional part in that form: the cut in the domain of the logarithm provides the «integer part elimination». Your simplification only applies when you perform the analytic continuation of the logarithm, which I suspect is why the closed-form (given by the CAS using implicitely an analytic continuation, I presume) does not agree with the OP's formula. | |
| Aug 19, 2013 at 12:33 | comment | added | Agno | Dang There is a typo. The integrals should start at 1 and not at 0. Sorry for this :-( Fixed it in the post. | |
| Aug 19, 2013 at 11:34 | comment | added | joro | @Agno Can't confirm finite version of the integral approaches zeta - it appears to diverge. Might be wrong on this. | |
| Aug 19, 2013 at 11:26 | comment | added | joro | @Agno log() is multivalued and you don't specify which branch you take. The simplification is not multivalued. | |
| Aug 19, 2013 at 11:13 | comment | added | Agno | Could you confirm whether you got the finite version of the integral indeed approaching $\zeta(s)$? To your questions: I am not entirely sure the indefinite integral is evaluated correctly (I found it in Maple as well, but given the results of the increasing finite integral, I do not trust it yet). Are you sure your proposed simplification is allowed when the power contains an imaginary component? I believe it becomes multi-valued in that case and cannot be simplified this way. | |
| Aug 19, 2013 at 10:34 | comment | added | joro | @Agno since log(exp(x))==x, why don't you simplify the integral? Do you work with the principal branch of log() as all CAS do? (there is closed form too after the simplification). | |
| Aug 19, 2013 at 9:11 | comment | added | joro | @Agno do you agree that the indefinite integral is correct? | |
| Aug 19, 2013 at 9:10 | comment | added | Agno | Pretty sure there are no typos. Try using a finite integral up to say x=100 (I used Maple). Please note that it only works for $\re(s) \ge 0$ (should have stated that!), but it certainly should work in the critical strip. | |
| Aug 19, 2013 at 9:02 | comment | added | joro | @Agno are you sure you don't have typos and the formula for zeta is correct? For s=2 the limit of the integral at x=0 is infinity... Numerically and according to maple the integral doesn't converge, might be wrong. | |
| Aug 19, 2013 at 8:52 | comment | added | Agno | There is indeed seems to be a strange ambivalence in the integral occuring at infinity, however try using an ever increasing finite integral and you'll see that it perfectly approaches $\zeta(s)$. Don't know why the behavior "tilts" at infinity. | |
| Aug 19, 2013 at 7:58 | history | answered | joro | CC BY-SA 3.0 |