last day (15 days later) » 

21:14
Yes.
@Marvis I guess we're in this new room now.
I've not done this before.
ya. I think I have used this chat in a new room before
Ok. So what are your thoughts on the question?
Anyway, I'm interested in your idea of writing up what you have.
Is it worth writing it up and sending it somewhere?
How much of it is new, do you think?
21:15
I donot know
I really like the integral representations.
I guess you have looked more into these things. There is a huge collection on Euler sums in the link I sent to you
I think, in order to be publishable, that the integral representations would need to lead to new evaluations of some Euler sums, though. And that's the part I'm not sure about.
and I did not see the integral representation in the few articles I managed to g othrough
By new evaluations do you mean sums which have not been evaluated before?
Yes.
Or, if not that, evaluations of a large class of sums that have already been evaluated.
21:18
Ok. Is there anyway to find out what class of sums have been evaluated before?
In my cursory search of the literature before I posted the question, I did not find any exhaustive lists.
I did find some lists. I can go dig those up again.
Mhenni also had a nice result for generic A(p,q) math.stackexchange.com/questions/275643/…
Yes, he did.
By the way, I think this is the third time you and I have discussed the possibility of writing a paper based on something from math.SE. :)
he he :) Apologizes for my laziness :)
Don't apologize! I think this one is the most promising, actually. :)
21:21
Nice :-). Good to know.
To start off I think I will go over the webpage I linked to
to see if there is something this integral representation can address
Were you ever able to find out how to evaluate those other integrals in your answer (the ones that probably involved polylogs)?
Well, those actually are the definitions of the polylogs
if we bring the poly log into picture then all those can be evaluated
I will updated the remaining three integrals in terms of Polylog
I was looking for a way to get around using the Polylog in the integral evaluation
I think this article is the one that I found that had the most explicit evaluations: davidhbailey.com/dhbpapers/eulsum-em.pdf. See page 19.
For instance the integral in the second one, $$\int \dfrac{\log^2(1+z)}{z} dz = log(-z) log^2(1+z)+2 log(1+z) Li_2(1+z)-2 Li_3(1+z)$$
There's a bunch of conjectures on page 26.
I don't know much about polylogs. I guess we would have to find out whether anyone else has used them to evaluate Euler sums.
Some of those conjectures may have been proven by now, as that article is almost 20 years old.
21:29
Oh those are interesting set of conjectures! A quick glance at the conjectures : there is no one conjecture on $\sigma_h$ which is what the integral representation gives us
everything is on $s_a$
This article also has some: algo.inria.fr/flajolet/Publications/FlSa98.pdf. I verified one of the alternating Euler sums in my question with it.
Is it possible to extend the integral idea to the $s_a$ instead of $\sigma_a$?
Let me also think on this
Here's a statement from that Flajolet and Salvy article: "Identities of low weight can sometimes be proved by special integral representations and functional properties of polylogarithms (de Doelder 1991)." So I guess we need to look up that de Doelder reference.
@Marvis I would think so.
Do you have access to Journal of Computational and Applied Mathematics?
I got hold of the De Doelder article
Let me check.
No, I don't have access to it. (I miss Princeton's journal collection sometimes! I'm sure Stanford's is really good.)
Does de Doelder have your integral representations?
Yes, thanks!
Yes. Stanford has some good access to journal articles, one thing that I might miss after leaving this place. And yes I think it should be possible to evaluate $s_a$ if we have access to $\sigma_b$'s
Could you clarify what you mean by "Does de Doelder have your integral representations?"?
I have not worked with these polylogs either
@Marvis I just meant that if de Doelder has the same derivations you do then that's our first solid reference for your derivations.
And maybe your work is just repeating his.
Or maybe your work could extend what he does.
Oh ok.
Or we could search for other articles that build more directly off of his so that we could find out exactly what is known about integral representations for Euler sums.
I'm not seeing the term "Euler sum" in his paper yet.
And the integrals I'm seeing involve logs. So they're more like the integral $\int_0^1 \log (1 + x) \log x \log (1- x) dx$ that Chris's sister asked about initially.
21:46
Exactly yes.
That's encouraging. :)
For your work being new, I mean.
Somehow these polylogs, I am unable to get my head around it. There are a lot of interesting properties, but somehow I do not seem to have a good feel/understanding about these stuff.
Same here. I imagine we just need to work with them more.
There are some nice formulas here for triple sums in this article
Yes, it does.
21:54
This has some integral representation for multiple suims
at the end of the second page
is that same/similar to what we have?
Not sure. There are the Crandall, Borwein, and Zagier references, too.
I guess this has it
It also has Mhenni's version
@Marvis Which formula?
Isn't formula 2 a rearrangement of ours?
equation 2 in introduction
I'll have to take a closer look. I guess it needs a variable transformation to remove the log?
22:08
It is not apparently clear to me as well. I too will have to take a closer look but looks like setting $y_j = 1/x_j$ and performing some cleaning up of the formula we have
equation 2 below introduction can be obtained
That's the transformation I was thinking of, too. I don't see yet how it removes the log. Maybe it isn't really the same as ours.
The transformation $y_j = e^{u_j}$ gives the integral representation in eq. (3). That removes the log but introduces exp.
The number of integral in this is only $k$, whereas ours has $s_1 + s_2 + \cdots s_k$ integrals. Mhenni's formula for instance collapses these integrals into one for $A(p,1)$
You're right. I guess ours is different.
Infact formula 5 and 6
capture the relation between the two
on the introduction page
there he discusses the $s_1 + s_2 + \cdots s_k$ integral as well :)
So I guess this has already been done :(
I think you're right. :(
Well, this would only be the umpteenth time I've worked on something only to discover someone else has already done it. :)
22:15
I think i should still look how exaclty his result is obtained from ours
It doesn't look like he uses those integral representations, though.
but do you agree that his is just a variable change of ours?
I'd have to work through the details to make sure, but I'm pretty sure that it is.
ok
:) That is fine. As Feynman says, the pleasure of finding things out ourselves :)
Yes. :)
22:17
Is there anything we can do with what we have now?
I don't think so.
You're probably right.
Ok cool.
Catch u later then. Now I can focus my attention on writing my thesis
Sounds good. Best wishes! :)
Will be graduating this June :)
thanks
see you then

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