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The following result $$ \sum_{k=1}^\infty\left(\psi^{(1)} (k)\right)^2 = 3\zeta(3) $$ where $\psi^{(1)}$ is the polygamma function makes me think there is a nice sum for the series

$$ \sum_{k=1}^\infty \left(\psi^{(1)} (k+a)\right)^2 $$

or

$$ \sum_{k=1}^\infty \psi^{(1)} (k+a)\psi^{(1)} (k+b) $$

where $a$ and $b$ are any real numbers such that $a >-1, b>-1.$

Could you help me to find it?

Community
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Olivier Oloa
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  • Didn' t you answer it in your given link ?. – Felix Marin Aug 03 '14 at 18:23
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    If one proceeds along the same lines as that result, it seems that all one needs is a general result for the integral $$\int_0^1 ! !\int_0^1 \frac{u^a v^b \ln u\ln v}{(1-uv)(1 - u)(1-v)}{\rm{d}} u:{\rm{d}} v.$$ (That, of course, presumes that this integral is tractable...) – Semiclassical Aug 03 '14 at 18:24
  • @FelixMarin: I think that result only covered the $a=b=0$ case? – Semiclassical Aug 03 '14 at 18:26
  • @Semiclassical It looks like a hard one even with ${\rm Li_{, s}\left(z\right)}$ machinery. Thanks. – Felix Marin Aug 03 '14 at 18:29
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    Maybe through the identity: $$\sum_{k=1}^{+\infty}\psi'(k), x^k = \frac{x}{6-6x}\left(\pi^2-6\operatorname{Li}_2(x)\right).$$ – Jack D'Aurizio Aug 03 '14 at 22:42
  • A trick which might come in handy is the Feynman parametrization, since it allows one to convert the denominator into a linear combination of $u$ and $v$ (at the cost of introducing new integrations which are hopefully easier than those before.) – Semiclassical Aug 04 '14 at 19:09
  • You know of at last two special cases –  Dec 24 '15 at 02:35
  • Also appear to have $\sum_{k=1}^\infty\left(\psi^{(1)} (k+1)\right)^2 = 3\zeta(3) - \zeta(2)^2$ – Benedict W. J. Irwin May 29 '17 at 16:49
  • @BenedictWilliamJohnIrwin Sure, by putting $n=k+1$, one gets $\sum_{k=1}^\infty\left(\psi^{(1)} (k+1)\right)^2=\sum_{n=2}^\infty\left(\psi^{(1)} (n)\right)^2=\sum_{n=1}^\infty\left(\psi^{(1)} (n)\right)^2-\left(\psi^{(1)} (1)\right)^2$ $=3\zeta(3) - \zeta(2)^2$ which is equivalent to the hyperlink given at the beginning of my above question. Thank you. – Olivier Oloa May 29 '17 at 18:46
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    Based on that Mathematica has suggested that, $$ S_n=\sum_{k=1}^\infty(\psi^{(1)}(k+n))^2 = 3\zeta(3)-\kappa_{n+1} $$ with $\kappa_n$ given by the recurrence $$ \kappa_1=0\ \kappa_2=\zeta(2)^2\ \kappa_3=\zeta(2)^2+(\zeta(2)-1)^2\ \kappa_n=\frac{13 + 2ng_5 + g_3^4 g_2^2 \kappa_{n-3} - g_3^2 g_2^2 (22-16n+3n^2) \kappa_{n-2} + g_3^2 g_2^2 (17-14n+3n^2) \kappa_{n-1}}{g_3^2 g_2^4} $$ where $g_k=(n-k)$. – Benedict W. J. Irwin Jun 27 '17 at 15:29
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    This gives $$ S_2=3\zeta(3)-2\zeta(2)^2+2\zeta(2)-1\ S_3=3\zeta(3)-3\zeta(2)^2+\frac{9}{2}\zeta(2)-\frac{41}{16}\ S_4=3\zeta(3)-4\zeta(2)^2+\frac{65}{9}\zeta(2)-\frac{2861}{648}\ S_5=3\zeta(3)-5\zeta(2)^2+\frac{725}{72}\zeta(2)-\frac{133577}{20736}\ $$ which seem to check out numerically. – Benedict W. J. Irwin Jun 27 '17 at 15:29
  • @BenedictWilliamJohnIrwin Wow. That is a pretty amazing result. I would love to see some development of this discovery! – Brevan Ellefsen Sep 26 '17 at 20:51
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    It is not too difficult to show that

    $$S_n = 3 \zeta(3) - n\zeta(2)^2 +2(nH_{n-1,2} - H_{n-1})\zeta(2) - \sum_{k=1}^{n-1}H_{k,2}^2,$$

    where $H_n$ is the $n$-th Harmonic number and $H_{n,2} = \sum_i^n 1/i^2$

     – Alexander Vlasev Jan 28 '18 at 22:13
  • Perhaps there could be a similar formula for general $a$ that uses generalized Harmonic numbers? – Alexander Vlasev Jan 28 '18 at 22:19

1 Answers1

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We have $$ \int^{1}_{0}\int^{1}_{0}\frac{\log u\log v}{(1-uv)(1-u)(1-v)}dudv=3\zeta(3) $$ Also $$ \int^{1}_{0}\int^{1}_{0}\frac{(uv)^k\log u\log v}{(1-u)(1-v)}dudv=Z(2,k+1)\psi'(k+1)\textrm{, where }Re(k)>-1 $$ The function $Z(s,k)=\sum^{\infty}_{l=0}\frac{1}{(l+k)^s}$ is Hurwitz zeta function. Hence $$ \sum^{a-1}_{k=0}\int^{1}_{0}\int^{1}_{0}\frac{(uv)^k\log u\log v}{(1-u)(1-v)}dudv =\int^{1}_{0}\int^{1}_{0}\frac{(1-(uv)^a)\log u\log v}{(1-uv)(1-u)(1-v)}dudv. $$ Hence $$ \sum^{\infty}_{k=1}\left(\psi'(k+a)\right)^2=\int^{1}_{0}\int^{1}_{0}\frac{(uv)^a\log u \log v}{(1-u v)(1-u)(1-v)}dudv=3\zeta(3)-\sum^{a-1}_{k=0}Z(2,k+1)\psi'(k+1) $$ By trying to generalize the problem we have $$ C_{\nu}=\underbrace{\int^{1}_{0}\ldots\int^{1}_{0}}_{\nu}\frac{\log u_1\ldots \log u_{\nu}du_1\ldots du_{\nu}}{(1-u_1\ldots u_{\nu})(1-u_1)\ldots(1-u_{\nu})}=\sum^{\infty}_{l=0}\left(\int^{1}_{0}\frac{u^l\log u}{1-u}du\right)^{\nu}= $$ $$ =\sum^{\infty}_{l=0}\left(-Z(2,l+1)\right)^{\nu} $$ Also $$ \underbrace{\int^{1}_{0}\ldots\int^{1}_{0}}_{\nu}\frac{u_1^k\ldots u_{\nu}^k \log u_1\ldots\log u_{\nu}du_1\ldots du_{\nu}}{(1-u_1)\ldots (1-u_{\nu})}=(-1)^{\nu}Z(2,k+1)(\psi'(k+1))^{\nu-1} $$ Hence $$ \sum^{a-1}_{k=0}\underbrace{\int^{1}_{0}\ldots\int^{1}_{0}}_{\nu}\frac{u_1^k\ldots u_{\nu}^k \log u_1\ldots\log u_{\nu}du_{1}\ldots du_{\nu}}{(1-u_1)\ldots (1-u_{\nu})}=(-1)^{\nu}\sum^{a-1}_{k=0}Z(2,k+1)(\psi'(k+1))^{\nu-1}= $$ $$ =\underbrace{\int^{1}_{0}\ldots\int^{1}_{0}}_{\nu}\frac{(1-u_1^a\ldots u_{\nu}^a) \log u_1\ldots\log u_{\nu}du_1\ldots du_{\nu}}{(1-u_1\ldots u_{\nu})(1-u_1)\ldots (1-u_{\nu})}. $$ Hence $$ \underbrace{\int^{1}_{0}\ldots\int^{1}_{0}}_{\nu}\frac{(u_1^a\ldots u_{\nu}^a) \log u_1\ldots\log u_{\nu}du_1\ldots du_{\nu}}{(1-u_1\ldots u_{\nu})(1-u_1)\ldots (1-u_{\nu})}= $$ $$ =C_{\nu}+(-1)^{\nu-1}\sum^{a-1}_{k=0}Z(2,k+1)(\psi'(k+1))^{\nu-1} $$ And finaly $$ \sum^{\infty}_{k=1}\left(\psi'(k+a)\right)^{\nu}=C_{\nu}+(-1)^{\nu-1}\sum^{a-1}_{k=0}Z(2,k+1)(\psi'(k+1))^{\nu-1} $$

Nikos Bagis
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  • Thank you very much for your contribution! (+1) I think an interesting case would be a formula for $\sum^{\infty}_{k=1}\left(\psi'(k+a)\right)^{2} $ where $a$ could be any positive real number. – Olivier Oloa Nov 25 '18 at 08:36