A closed form of $sum_{k=1}^infty psi^{(1)} (k+a)psi^{(1)} (k+b)$?
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
edited Apr 13 '17 at 12:20
Community♦
1
asked Aug 3 '14 at 17:51
Olivier Oloa
107k17175293
|
show 9 more comments
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
edited Apr 13 '17 at 12:20
Community♦
1
asked Aug 3 '14 at 17:51
Olivier Oloa
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
|
show 9 more comments
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
edited Apr 13 '17 at 12:20
Community♦
1
asked Aug 3 '14 at 17:51
Olivier Oloa
107k17175293
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
edited Apr 13 '17 at 12:20
Community♦
1
asked Aug 3 '14 at 17:51
Olivier Oloa
107k17175293
edited Apr 13 '17 at 12:20
Community♦
1
asked Aug 3 '14 at 17:51
Olivier Oloa
107k17175293
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
asked Aug 3 '14 at 17:51
Olivier Oloa
107k17175293
asked Aug 3 '14 at 17:51
Olivier Oloa
107k17175293
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
|
show 9 more comments
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
|
show 9 more comments
1 Answer
1
active
oldest
votes
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
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^{nu}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
answered Nov 24 at 23:36
Nikos Bagis
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
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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votes
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
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^{nu}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
answered Nov 24 at 23:36
Nikos Bagis
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
add a comment |
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
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^{nu}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
answered Nov 24 at 23:36
Nikos Bagis
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
add a comment |
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
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^{nu}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
answered Nov 24 at 23:36
Nikos Bagis
2,157312
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
$$
sum^{infty}_{k=1}left(psi'(k+a)right)^{nu}=C_{nu}+(-1)^{nu-1}sum^{a-1}_{k=0}Z(2,k+1)(psi'(k+1))^{nu-1}
$$
answered Nov 24 at 23:36
Nikos Bagis
2,157312
edited Nov 25 at 7:19
answered Nov 24 at 23:36
Nikos Bagis
2,157312
answered Nov 24 at 23:36
Nikos Bagis
2,157312
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
add a comment |
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
add a comment |
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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