A closed form of $sum_{k=1}^infty psi^{(1)} (k+a)psi^{(1)} (k+b)$?
up vote
39
down vote
favorite
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
up vote
39
down vote
favorite
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
up vote
39
down vote
favorite
up vote
39
down vote
favorite
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
up vote
3
down vote
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 |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
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 |
up vote
3
down vote
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 |
up vote
3
down vote
up vote
3
down vote
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 |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
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