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What are your thoughts on this series?

$$\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{\Gamma (k)^2 \Gamma (n) }{\Gamma (2 k+n)}((\psi ^{(0)}(n)-\psi ^{(0)}(2 k+n)) (\psi ^{(0)}(k)-\psi ^{(0)}(2 k+n))-\psi ^{(1)}(2 k+n)).$$

EDIT: Noting the interest for this series, I wanna add that the series gets reduced to the calculation of $$\int_0^1 \left(\frac{\text{Li}_3(x)}{x^2-2 x+2}+\frac{\text{Li}_3\left(x-x^2\right)}{x^2-2 x+2}-\frac{\text{Li}_3\left(\frac{x}{x-1}\right)}{x^2-2 x+2}-\frac{\text{Li}_3\left(\frac{(x-1) x}{x^2-x+1}\right)}{x^2-2 x+2}-\frac{\text{Li}_2\left(\frac{x}{x-1}\right) \log (1-x)}{x^2-2 x+2}-\frac{\text{Li}_2(x) \log (1-x)}{x^2-2 x+2}+\frac{\text{Li}_2\left(\frac{x}{x-1}\right) \log (x)}{x^2-2 x+2}+\frac{\text{Li}_2\left(\frac{(x-1) x}{x^2-x+1}\right) \log (x)}{x^2-2 x+2}-\frac{\text{Li}_2(x) \log (x)}{x^2-2 x+2}-\frac{\text{Li}_2\left(x-x^2\right) \log (x)}{x^2-2 x+2}-\frac{\text{Li}_2\left(x-x^2\right) \log \left(x^2-x+1\right)}{x^2-2 x+2}-\frac{\text{Li}_2\left(\frac{(x-1) x}{x^2-x+1}\right) \log \left(x^2-x+1\right)}{x^2-2 x+2}-\frac{\log ^3(1-x)}{3 \left(x^2-2 x+2\right)}-\frac{\log ^3\left(x^2-x+1\right)}{3 \left(x^2-2 x+2\right)}-\frac{\log ^2(x) \log (1-x)}{x^2-2 x+2}-\frac{\log ^2(x) \log \left(x^2-x+1\right)}{x^2-2 x+2}+\frac{\pi ^2 \log (1-x)}{6 \left(x^2-2 x+2\right)}+\frac{\pi ^2 \log \left(x^2-x+1\right)}{6 \left(x^2-2 x+2\right)}\right) \, dx$$ where we are pretty familiar with all the stuff in here. One can find the closed form by calculating the integral, a bit long, but it's a nice journey to go.

A 300 points bounty moment: After 2 years and 10 months since the problem has been posed no full solution has been provided. Is it possible to find a slick solution?

user 1591719
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  • The first part is a trinomial coefficient $($for which certain integral representations similar to that of the beta function are known to exist$)$. The latter are $($generalized$)$ harmonic numbers in disguise. – Lucian Jul 15 '15 at 19:14
  • @Lucian thanks for sharing some ideas. – user 1591719 Jul 15 '15 at 20:02
  • @Chris'ssistheartist Do you have the closed form for this? :-) – r9m Jul 16 '15 at 04:01
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    what is your reasonig behind the statement that this has "undoubtly" closed form? – tired Jul 16 '15 at 11:04
  • @tired The last integral looks amenable to having a closed form .. although wolfram is returning a numerical answer (no closed forms)! – r9m Jul 16 '15 at 11:45
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    Does anything interesting happen if the polygammas are reexpressed in terms of the Hurwitz zeta function? (I almost think we're looking at an infinite series of (q?)-hypergeometric functions here) – graveolensa Jul 16 '15 at 20:12
  • what a monster...i wouldn't call this a "nice journey " ;). Have you tried the various dilogarithmic identies? i think the number of independent integrals that you need to calculate could be reduced a little bit... – tired Jul 17 '15 at 09:39
  • @tired If the number of integrals could be reduce that would be great. I didn't have time to go into details, but from my experience all these integrals can be calculated with less or more effort. Some of them can be immediately calculated by Mathematica. – user 1591719 Jul 17 '15 at 10:01
  • @Chris'ssistheartist "Undoubtedly it has a closed form." Actually, I've been toying around with this problem for a bit and I'm starting to suspect this integral may be one of the subtle ones. For example, it seems part of the solution will require not only this one, but the trilogarithmic generalization of that integral as well. The search for a closed form could turn out to be quite the adventure! :-) – David H Aug 29 '15 at 20:16
  • @DavidH Yes, it's a great adventure :-) Unfortunately some of the integrals above are pretty difficult, they require much cleverness in order to be calculated, but, well, that makes the problems more appealing. – user 1591719 Aug 29 '15 at 20:22
  • The first thing to do is to downvote? – user 1591719 May 27 '18 at 11:45
  • What are your thoughts on this series ? - My thoughts on this series are that it looks significantly more doable than this integral. – Lucian May 30 '18 at 03:39

1 Answers1

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I'd try to use: $$\displaystyle \int\limits_{0<x+y<1} x^{u-1}y^{u-1}(1-x-y)^{v-1} \,dx\,dy= \frac{\Gamma(u)^2\Gamma(v)}{\Gamma(2u+v)}$$

for reals $u,v \in \mathbb{R}$.

Hence, $$\displaystyle \begin{align}\sum\limits_{n,k=1}^{\infty} \frac{\Gamma(k+u)^2\Gamma(n+v)}{\Gamma(2k+2u+n+v)} &= \int\limits_{0<x+y<1} \frac{(xy)^{u}(1-x-y)^v}{(1-xy)(x+y)}\,dx\,dy\end{align}$$

Since, $\displaystyle \frac{\partial^2}{\partial u\partial v} \left(\frac{\Gamma(u)^2\Gamma(v)}{\Gamma(2u+v)}\right) = \frac{2\Gamma(u)^2\Gamma(v)}{\Gamma(2u+v)}\left((\psi^{(0)}(u) - \psi^{(0)}(2u+v))(\psi^{(0)}(v) - \psi^{(0)}(2u+v)) - \psi^{(1)}(2u+v)\right)$

Hence, $$\begin{align}&2\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{\Gamma (k)^2 \Gamma (n) }{\Gamma (2 k+n)}((\psi ^{(0)}(n)-\psi ^{(0)}(2 k+n)) (\psi ^{(0)}(k)-\psi ^{(0)}(2 k+n))-\psi ^{(1)}(2 k+n))\\&= \lim\limits_{(u,v) \to (0,0)} \frac{\partial^2}{\partial u\partial v}\int\limits_{0<x+y<1} \frac{(xy)^{u}(1-x-y)^v}{(1-xy)(x+y)}\,dx\,dy\\&= \lim\limits_{(u,v) \to (0,0)} \int\limits_{0<x+y<1} \frac{(xy)^{u}(1-x-y)^v\log (xy)\log(1-x-y)}{(1-xy)(x+y)}\,dx\,dy\end{align}$$

We need to compute the integral and take the limit after that. This is only a partial attempt.

r9m
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  • May a change of variables $x+y=q$ and $xy=p$ be of some help? – tired Jul 16 '15 at 13:14
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    I've decided to give you a bounty to motivate you to finish this way. No need to make an update until you have a full solution with a final closed-form. No hurry. ;) – user 1591719 May 28 '18 at 11:38