# A closed form of $\sum_{k=1}^\infty \psi^{(1)} (k+a)\psi^{(1)} (k+b)$?

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?

• 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 3 '14 at 18:24
• 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 3 '14 at 22:42
• 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
• 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
• 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

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 $$\sum^{\infty}_{k=1}\left(\psi'(k+a)\right)^{\nu}=C_{\nu}+(-1)^{\nu-1}\sum^{a-1}_{k=0}Z(2,k+1)(\psi'(k+1))^{\nu-1}$$
• 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 8:36