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












39














The following result
$$
sum_{k=1}^inftyleft(psi^{(1)} (k)right)^2 = 3zeta(3)
$$ 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?






calculus integration sequences-and-series closed-form polygamma







share|cite|improve this question




















  • 2




    If one proceeds along the same lines as that result, it seems that all one needs is a general result for the integral $$int_0^1 ! !int_0^1 frac{u^a v^b ln uln v}{(1-uv)(1 - u)(1-v)}{rm{d}} u:{rm{d}} v.$$ (That, of course, presumes that this integral is tractable...)
    – Semiclassical
    Aug 3 '14 at 18:24








  • 3




    Maybe through the identity: $$sum_{k=1}^{+infty}psi'(k), x^k = frac{x}{6-6x}left(pi^2-6operatorname{Li}_2(x)right).$$
    – Jack D'Aurizio
    Aug 3 '14 at 22:42








  • 5




    Based on that Mathematica has suggested that, $$ S_n=sum_{k=1}^infty(psi^{(1)}(k+n))^2 = 3zeta(3)-kappa_{n+1} $$ 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






  • 6




    This gives $$ S_2=3zeta(3)-2zeta(2)^2+2zeta(2)-1\ S_3=3zeta(3)-3zeta(2)^2+frac{9}{2}zeta(2)-frac{41}{16}\ S_4=3zeta(3)-4zeta(2)^2+frac{65}{9}zeta(2)-frac{2861}{648}\ S_5=3zeta(3)-5zeta(2)^2+frac{725}{72}zeta(2)-frac{133577}{20736}\ $$ which seem to check out numerically.
    – Benedict W. J. Irwin
    Jun 27 '17 at 15:29






  • 1




    It is not too difficult to show that $$S_n = 3 zeta(3) - nzeta(2)^2 +2(nH_{n-1,2} - H_{n-1})zeta(2) - sum_{k=1}^{n-1}H_{k,2}^2,$$ where $H_n$ is the $n$-th Harmonic number and $H_{n,2} = sum_i^n 1/i^2$
    – Alexander Vlasev
    Jan 28 at 22:13


















39














The following result
$$
sum_{k=1}^inftyleft(psi^{(1)} (k)right)^2 = 3zeta(3)
$$ 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?






calculus integration sequences-and-series closed-form polygamma







share|cite|improve this question




















  • 2




    If one proceeds along the same lines as that result, it seems that all one needs is a general result for the integral $$int_0^1 ! !int_0^1 frac{u^a v^b ln uln v}{(1-uv)(1 - u)(1-v)}{rm{d}} u:{rm{d}} v.$$ (That, of course, presumes that this integral is tractable...)
    – Semiclassical
    Aug 3 '14 at 18:24








  • 3




    Maybe through the identity: $$sum_{k=1}^{+infty}psi'(k), x^k = frac{x}{6-6x}left(pi^2-6operatorname{Li}_2(x)right).$$
    – Jack D'Aurizio
    Aug 3 '14 at 22:42








  • 5




    Based on that Mathematica has suggested that, $$ S_n=sum_{k=1}^infty(psi^{(1)}(k+n))^2 = 3zeta(3)-kappa_{n+1} $$ 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






  • 6




    This gives $$ S_2=3zeta(3)-2zeta(2)^2+2zeta(2)-1\ S_3=3zeta(3)-3zeta(2)^2+frac{9}{2}zeta(2)-frac{41}{16}\ S_4=3zeta(3)-4zeta(2)^2+frac{65}{9}zeta(2)-frac{2861}{648}\ S_5=3zeta(3)-5zeta(2)^2+frac{725}{72}zeta(2)-frac{133577}{20736}\ $$ which seem to check out numerically.
    – Benedict W. J. Irwin
    Jun 27 '17 at 15:29






  • 1




    It is not too difficult to show that $$S_n = 3 zeta(3) - nzeta(2)^2 +2(nH_{n-1,2} - H_{n-1})zeta(2) - sum_{k=1}^{n-1}H_{k,2}^2,$$ where $H_n$ is the $n$-th Harmonic number and $H_{n,2} = sum_i^n 1/i^2$
    – Alexander Vlasev
    Jan 28 at 22:13
















39












39








39


26





The following result
$$
sum_{k=1}^inftyleft(psi^{(1)} (k)right)^2 = 3zeta(3)
$$ 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?






calculus integration sequences-and-series closed-form polygamma







share|cite|improve this question















The following result
$$
sum_{k=1}^inftyleft(psi^{(1)} (k)right)^2 = 3zeta(3)
$$ 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?






calculus integration sequences-and-series closed-form polygamma




calculus integration sequences-and-series closed-form polygamma






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 13 '17 at 12:20









Community♦

1




1










asked Aug 3 '14 at 17:51









Olivier Oloa

107k17175293




107k17175293








  • 2




    If one proceeds along the same lines as that result, it seems that all one needs is a general result for the integral $$int_0^1 ! !int_0^1 frac{u^a v^b ln uln v}{(1-uv)(1 - u)(1-v)}{rm{d}} u:{rm{d}} v.$$ (That, of course, presumes that this integral is tractable...)
    – Semiclassical
    Aug 3 '14 at 18:24








  • 3




    Maybe through the identity: $$sum_{k=1}^{+infty}psi'(k), x^k = frac{x}{6-6x}left(pi^2-6operatorname{Li}_2(x)right).$$
    – Jack D'Aurizio
    Aug 3 '14 at 22:42








  • 5




    Based on that Mathematica has suggested that, $$ S_n=sum_{k=1}^infty(psi^{(1)}(k+n))^2 = 3zeta(3)-kappa_{n+1} $$ 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






  • 6




    This gives $$ S_2=3zeta(3)-2zeta(2)^2+2zeta(2)-1\ S_3=3zeta(3)-3zeta(2)^2+frac{9}{2}zeta(2)-frac{41}{16}\ S_4=3zeta(3)-4zeta(2)^2+frac{65}{9}zeta(2)-frac{2861}{648}\ S_5=3zeta(3)-5zeta(2)^2+frac{725}{72}zeta(2)-frac{133577}{20736}\ $$ which seem to check out numerically.
    – Benedict W. J. Irwin
    Jun 27 '17 at 15:29






  • 1




    It is not too difficult to show that $$S_n = 3 zeta(3) - nzeta(2)^2 +2(nH_{n-1,2} - H_{n-1})zeta(2) - sum_{k=1}^{n-1}H_{k,2}^2,$$ where $H_n$ is the $n$-th Harmonic number and $H_{n,2} = sum_i^n 1/i^2$
    – Alexander Vlasev
    Jan 28 at 22:13
















  • 2




    If one proceeds along the same lines as that result, it seems that all one needs is a general result for the integral $$int_0^1 ! !int_0^1 frac{u^a v^b ln uln v}{(1-uv)(1 - u)(1-v)}{rm{d}} u:{rm{d}} v.$$ (That, of course, presumes that this integral is tractable...)
    – Semiclassical
    Aug 3 '14 at 18:24








  • 3




    Maybe through the identity: $$sum_{k=1}^{+infty}psi'(k), x^k = frac{x}{6-6x}left(pi^2-6operatorname{Li}_2(x)right).$$
    – Jack D'Aurizio
    Aug 3 '14 at 22:42








  • 5




    Based on that Mathematica has suggested that, $$ S_n=sum_{k=1}^infty(psi^{(1)}(k+n))^2 = 3zeta(3)-kappa_{n+1} $$ 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






  • 6




    This gives $$ S_2=3zeta(3)-2zeta(2)^2+2zeta(2)-1\ S_3=3zeta(3)-3zeta(2)^2+frac{9}{2}zeta(2)-frac{41}{16}\ S_4=3zeta(3)-4zeta(2)^2+frac{65}{9}zeta(2)-frac{2861}{648}\ S_5=3zeta(3)-5zeta(2)^2+frac{725}{72}zeta(2)-frac{133577}{20736}\ $$ which seem to check out numerically.
    – Benedict W. J. Irwin
    Jun 27 '17 at 15:29






  • 1




    It is not too difficult to show that $$S_n = 3 zeta(3) - nzeta(2)^2 +2(nH_{n-1,2} - H_{n-1})zeta(2) - sum_{k=1}^{n-1}H_{k,2}^2,$$ where $H_n$ is the $n$-th Harmonic number and $H_{n,2} = sum_i^n 1/i^2$
    – Alexander Vlasev
    Jan 28 at 22:13










2




2




If one proceeds along the same lines as that result, it seems that all one needs is a general result for the integral $$int_0^1 ! !int_0^1 frac{u^a v^b ln uln v}{(1-uv)(1 - u)(1-v)}{rm{d}} u:{rm{d}} v.$$ (That, of course, presumes that this integral is tractable...)
– Semiclassical
Aug 3 '14 at 18:24






If one proceeds along the same lines as that result, it seems that all one needs is a general result for the integral $$int_0^1 ! !int_0^1 frac{u^a v^b ln uln v}{(1-uv)(1 - u)(1-v)}{rm{d}} u:{rm{d}} v.$$ (That, of course, presumes that this integral is tractable...)
– Semiclassical
Aug 3 '14 at 18:24






3




3




Maybe through the identity: $$sum_{k=1}^{+infty}psi'(k), x^k = frac{x}{6-6x}left(pi^2-6operatorname{Li}_2(x)right).$$
– Jack D'Aurizio
Aug 3 '14 at 22:42






Maybe through the identity: $$sum_{k=1}^{+infty}psi'(k), x^k = frac{x}{6-6x}left(pi^2-6operatorname{Li}_2(x)right).$$
– Jack D'Aurizio
Aug 3 '14 at 22:42






5




5




Based on that Mathematica has suggested that, $$ S_n=sum_{k=1}^infty(psi^{(1)}(k+n))^2 = 3zeta(3)-kappa_{n+1} $$ 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




Based on that Mathematica has suggested that, $$ S_n=sum_{k=1}^infty(psi^{(1)}(k+n))^2 = 3zeta(3)-kappa_{n+1} $$ 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




6




6




This gives $$ S_2=3zeta(3)-2zeta(2)^2+2zeta(2)-1\ S_3=3zeta(3)-3zeta(2)^2+frac{9}{2}zeta(2)-frac{41}{16}\ S_4=3zeta(3)-4zeta(2)^2+frac{65}{9}zeta(2)-frac{2861}{648}\ S_5=3zeta(3)-5zeta(2)^2+frac{725}{72}zeta(2)-frac{133577}{20736}\ $$ which seem to check out numerically.
– Benedict W. J. Irwin
Jun 27 '17 at 15:29




This gives $$ S_2=3zeta(3)-2zeta(2)^2+2zeta(2)-1\ S_3=3zeta(3)-3zeta(2)^2+frac{9}{2}zeta(2)-frac{41}{16}\ S_4=3zeta(3)-4zeta(2)^2+frac{65}{9}zeta(2)-frac{2861}{648}\ S_5=3zeta(3)-5zeta(2)^2+frac{725}{72}zeta(2)-frac{133577}{20736}\ $$ which seem to check out numerically.
– Benedict W. J. Irwin
Jun 27 '17 at 15:29




1




1




It is not too difficult to show that $$S_n = 3 zeta(3) - nzeta(2)^2 +2(nH_{n-1,2} - H_{n-1})zeta(2) - sum_{k=1}^{n-1}H_{k,2}^2,$$ where $H_n$ is the $n$-th Harmonic number and $H_{n,2} = sum_i^n 1/i^2$
– Alexander Vlasev
Jan 28 at 22:13






It is not too difficult to show that $$S_n = 3 zeta(3) - nzeta(2)^2 +2(nH_{n-1,2} - H_{n-1})zeta(2) - sum_{k=1}^{n-1}H_{k,2}^2,$$ where $H_n$ is the $n$-th Harmonic number and $H_{n,2} = sum_i^n 1/i^2$
– Alexander Vlasev
Jan 28 at 22:13












1 Answer
1






active

oldest

votes


















3














We have
$$
int^{1}_{0}int^{1}_{0}frac{log ulog v}{(1-uv)(1-u)(1-v)}dudv=3zeta(3)
$$
Also
$$
int^{1}_{0}int^{1}_{0}frac{(uv)^klog ulog v}{(1-u)(1-v)}dudv=Z(2,k+1)psi'(k+1)textrm{, where }Re(k)>-1
$$
The function $Z(s,k)=sum^{infty}_{l=0}frac{1}{(l+k)^s}$ is Hurwitz zeta function.
Hence
$$
sum^{a-1}_{k=0}int^{1}_{0}int^{1}_{0}frac{(uv)^klog ulog v}{(1-u)(1-v)}dudv
=int^{1}_{0}int^{1}_{0}frac{(1-(uv)^a)log ulog v}{(1-uv)(1-u)(1-v)}dudv.
$$
Hence
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^2=int^{1}_{0}int^{1}_{0}frac{(uv)^alog u log v}{(1-u v)(1-u)(1-v)}dudv=3zeta(3)-sum^{a-1}_{k=0}Z(2,k+1)psi'(k+1)
$$
By trying to generalize the problem we have
$$
C_{nu}=underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{log u_1ldots log u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots(1-u_{nu})}=sum^{infty}_{l=0}left(int^{1}_{0}frac{u^llog u}{1-u}duright)^{nu}=
$$
$$
=sum^{infty}_{l=0}left(-Z(2,l+1)right)^{nu}
$$
Also
$$
underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{u_1^kldots u_{nu}^k log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1)ldots (1-u_{nu})}=(-1)^{nu}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
Hence
$$
sum^{a-1}_{k=0}underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{u_1^kldots u_{nu}^k log u_1ldotslog u_{nu}du_{1}ldots du_{nu}}{(1-u_1)ldots (1-u_{nu})}=(-1)^{nu}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}=
$$
$$
=underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{(1-u_1^aldots u_{nu}^a) log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots (1-u_{nu})}.
$$
Hence
$$
underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{(u_1^aldots u_{nu}^a) log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots (1-u_{nu})}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
And finaly
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^{nu}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$






share|cite|improve this answer























  • 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 at 8:36













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f886486%2fa-closed-form-of-sum-k-1-infty-psi1-ka-psi1-kb%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














We have
$$
int^{1}_{0}int^{1}_{0}frac{log ulog v}{(1-uv)(1-u)(1-v)}dudv=3zeta(3)
$$
Also
$$
int^{1}_{0}int^{1}_{0}frac{(uv)^klog ulog v}{(1-u)(1-v)}dudv=Z(2,k+1)psi'(k+1)textrm{, where }Re(k)>-1
$$
The function $Z(s,k)=sum^{infty}_{l=0}frac{1}{(l+k)^s}$ is Hurwitz zeta function.
Hence
$$
sum^{a-1}_{k=0}int^{1}_{0}int^{1}_{0}frac{(uv)^klog ulog v}{(1-u)(1-v)}dudv
=int^{1}_{0}int^{1}_{0}frac{(1-(uv)^a)log ulog v}{(1-uv)(1-u)(1-v)}dudv.
$$
Hence
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^2=int^{1}_{0}int^{1}_{0}frac{(uv)^alog u log v}{(1-u v)(1-u)(1-v)}dudv=3zeta(3)-sum^{a-1}_{k=0}Z(2,k+1)psi'(k+1)
$$
By trying to generalize the problem we have
$$
C_{nu}=underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{log u_1ldots log u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots(1-u_{nu})}=sum^{infty}_{l=0}left(int^{1}_{0}frac{u^llog u}{1-u}duright)^{nu}=
$$
$$
=sum^{infty}_{l=0}left(-Z(2,l+1)right)^{nu}
$$
Also
$$
underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{u_1^kldots u_{nu}^k log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1)ldots (1-u_{nu})}=(-1)^{nu}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
Hence
$$
sum^{a-1}_{k=0}underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{u_1^kldots u_{nu}^k log u_1ldotslog u_{nu}du_{1}ldots du_{nu}}{(1-u_1)ldots (1-u_{nu})}=(-1)^{nu}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}=
$$
$$
=underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{(1-u_1^aldots u_{nu}^a) log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots (1-u_{nu})}.
$$
Hence
$$
underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{(u_1^aldots u_{nu}^a) log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots (1-u_{nu})}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
And finaly
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^{nu}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$






share|cite|improve this answer























  • 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 at 8:36


















3














We have
$$
int^{1}_{0}int^{1}_{0}frac{log ulog v}{(1-uv)(1-u)(1-v)}dudv=3zeta(3)
$$
Also
$$
int^{1}_{0}int^{1}_{0}frac{(uv)^klog ulog v}{(1-u)(1-v)}dudv=Z(2,k+1)psi'(k+1)textrm{, where }Re(k)>-1
$$
The function $Z(s,k)=sum^{infty}_{l=0}frac{1}{(l+k)^s}$ is Hurwitz zeta function.
Hence
$$
sum^{a-1}_{k=0}int^{1}_{0}int^{1}_{0}frac{(uv)^klog ulog v}{(1-u)(1-v)}dudv
=int^{1}_{0}int^{1}_{0}frac{(1-(uv)^a)log ulog v}{(1-uv)(1-u)(1-v)}dudv.
$$
Hence
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^2=int^{1}_{0}int^{1}_{0}frac{(uv)^alog u log v}{(1-u v)(1-u)(1-v)}dudv=3zeta(3)-sum^{a-1}_{k=0}Z(2,k+1)psi'(k+1)
$$
By trying to generalize the problem we have
$$
C_{nu}=underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{log u_1ldots log u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots(1-u_{nu})}=sum^{infty}_{l=0}left(int^{1}_{0}frac{u^llog u}{1-u}duright)^{nu}=
$$
$$
=sum^{infty}_{l=0}left(-Z(2,l+1)right)^{nu}
$$
Also
$$
underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{u_1^kldots u_{nu}^k log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1)ldots (1-u_{nu})}=(-1)^{nu}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
Hence
$$
sum^{a-1}_{k=0}underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{u_1^kldots u_{nu}^k log u_1ldotslog u_{nu}du_{1}ldots du_{nu}}{(1-u_1)ldots (1-u_{nu})}=(-1)^{nu}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}=
$$
$$
=underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{(1-u_1^aldots u_{nu}^a) log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots (1-u_{nu})}.
$$
Hence
$$
underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{(u_1^aldots u_{nu}^a) log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots (1-u_{nu})}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
And finaly
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^{nu}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$






share|cite|improve this answer























  • 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 at 8:36
















3












3








3






We have
$$
int^{1}_{0}int^{1}_{0}frac{log ulog v}{(1-uv)(1-u)(1-v)}dudv=3zeta(3)
$$
Also
$$
int^{1}_{0}int^{1}_{0}frac{(uv)^klog ulog v}{(1-u)(1-v)}dudv=Z(2,k+1)psi'(k+1)textrm{, where }Re(k)>-1
$$
The function $Z(s,k)=sum^{infty}_{l=0}frac{1}{(l+k)^s}$ is Hurwitz zeta function.
Hence
$$
sum^{a-1}_{k=0}int^{1}_{0}int^{1}_{0}frac{(uv)^klog ulog v}{(1-u)(1-v)}dudv
=int^{1}_{0}int^{1}_{0}frac{(1-(uv)^a)log ulog v}{(1-uv)(1-u)(1-v)}dudv.
$$
Hence
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^2=int^{1}_{0}int^{1}_{0}frac{(uv)^alog u log v}{(1-u v)(1-u)(1-v)}dudv=3zeta(3)-sum^{a-1}_{k=0}Z(2,k+1)psi'(k+1)
$$
By trying to generalize the problem we have
$$
C_{nu}=underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{log u_1ldots log u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots(1-u_{nu})}=sum^{infty}_{l=0}left(int^{1}_{0}frac{u^llog u}{1-u}duright)^{nu}=
$$
$$
=sum^{infty}_{l=0}left(-Z(2,l+1)right)^{nu}
$$
Also
$$
underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{u_1^kldots u_{nu}^k log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1)ldots (1-u_{nu})}=(-1)^{nu}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
Hence
$$
sum^{a-1}_{k=0}underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{u_1^kldots u_{nu}^k log u_1ldotslog u_{nu}du_{1}ldots du_{nu}}{(1-u_1)ldots (1-u_{nu})}=(-1)^{nu}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}=
$$
$$
=underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{(1-u_1^aldots u_{nu}^a) log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots (1-u_{nu})}.
$$
Hence
$$
underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{(u_1^aldots u_{nu}^a) log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots (1-u_{nu})}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
And finaly
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^{nu}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$






share|cite|improve this answer














We have
$$
int^{1}_{0}int^{1}_{0}frac{log ulog v}{(1-uv)(1-u)(1-v)}dudv=3zeta(3)
$$
Also
$$
int^{1}_{0}int^{1}_{0}frac{(uv)^klog ulog v}{(1-u)(1-v)}dudv=Z(2,k+1)psi'(k+1)textrm{, where }Re(k)>-1
$$
The function $Z(s,k)=sum^{infty}_{l=0}frac{1}{(l+k)^s}$ is Hurwitz zeta function.
Hence
$$
sum^{a-1}_{k=0}int^{1}_{0}int^{1}_{0}frac{(uv)^klog ulog v}{(1-u)(1-v)}dudv
=int^{1}_{0}int^{1}_{0}frac{(1-(uv)^a)log ulog v}{(1-uv)(1-u)(1-v)}dudv.
$$
Hence
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^2=int^{1}_{0}int^{1}_{0}frac{(uv)^alog u log v}{(1-u v)(1-u)(1-v)}dudv=3zeta(3)-sum^{a-1}_{k=0}Z(2,k+1)psi'(k+1)
$$
By trying to generalize the problem we have
$$
C_{nu}=underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{log u_1ldots log u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots(1-u_{nu})}=sum^{infty}_{l=0}left(int^{1}_{0}frac{u^llog u}{1-u}duright)^{nu}=
$$
$$
=sum^{infty}_{l=0}left(-Z(2,l+1)right)^{nu}
$$
Also
$$
underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{u_1^kldots u_{nu}^k log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1)ldots (1-u_{nu})}=(-1)^{nu}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
Hence
$$
sum^{a-1}_{k=0}underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{u_1^kldots u_{nu}^k log u_1ldotslog u_{nu}du_{1}ldots du_{nu}}{(1-u_1)ldots (1-u_{nu})}=(-1)^{nu}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}=
$$
$$
=underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{(1-u_1^aldots u_{nu}^a) log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots (1-u_{nu})}.
$$
Hence
$$
underbrace{int^{1}_{0}ldotsint^{1}_{0}}_{nu}frac{(u_1^aldots u_{nu}^a) log u_1ldotslog u_{nu}du_1ldots du_{nu}}{(1-u_1ldots u_{nu})(1-u_1)ldots (1-u_{nu})}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
And finaly
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^{nu}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 25 at 7:19

























answered Nov 24 at 23:36









Nikos Bagis

2,157312




2,157312












  • 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 at 8:36




















  • 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 at 8:36


















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 at 8:36






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 at 8:36




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f886486%2fa-closed-form-of-sum-k-1-infty-psi1-ka-psi1-kb%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more