Sum over all inverse zeta nontrivial zeros

Clash Royale CLAN TAG#URR8PPP

up vote
3
down vote

favorite

Starting from the Hadamard product for the Riemann Zeta Function (assuming the product is taken over matching pairs of zeros)
$$zeta(s)=frace^(log(2pi)-1-gamma/2)s2(s-1)Gamma(1+s/2)prod_rholeft(1-fracsrho right)e^s/rho$$
can one derive the exact value of $sum_rho frac1rho$
to be
$$sum_rho frac1rho = -log(2sqrtpi)+1+gamma/2$$
What implications does this have?

complex-analysis riemann-zeta

share|cite|improve this question

edited Jul 25 at 16:08

Arnaud Mortier

18.8k22159

asked Jul 25 at 16:05

Jose Ortega

485

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Basically none. Also note the sum doesn't converge absolutely.
  – Wojowu
  Jul 25 at 16:49
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

Starting from the Hadamard product for the Riemann Zeta Function (assuming the product is taken over matching pairs of zeros)
$$zeta(s)=frace^(log(2pi)-1-gamma/2)s2(s-1)Gamma(1+s/2)prod_rholeft(1-fracsrho right)e^s/rho$$
can one derive the exact value of $sum_rho frac1rho$
to be
$$sum_rho frac1rho = -log(2sqrtpi)+1+gamma/2$$
What implications does this have?

complex-analysis riemann-zeta

share|cite|improve this question

edited Jul 25 at 16:08

Arnaud Mortier

18.8k22159

asked Jul 25 at 16:05

Jose Ortega

485

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Basically none. Also note the sum doesn't converge absolutely.
  – Wojowu
  Jul 25 at 16:49
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Starting from the Hadamard product for the Riemann Zeta Function (assuming the product is taken over matching pairs of zeros)
$$zeta(s)=frace^(log(2pi)-1-gamma/2)s2(s-1)Gamma(1+s/2)prod_rholeft(1-fracsrho right)e^s/rho$$
can one derive the exact value of $sum_rho frac1rho$
to be
$$sum_rho frac1rho = -log(2sqrtpi)+1+gamma/2$$
What implications does this have?

complex-analysis riemann-zeta

share|cite|improve this question

edited Jul 25 at 16:08

Arnaud Mortier

18.8k22159

asked Jul 25 at 16:05

Jose Ortega

485

Starting from the Hadamard product for the Riemann Zeta Function (assuming the product is taken over matching pairs of zeros)
$$zeta(s)=frace^(log(2pi)-1-gamma/2)s2(s-1)Gamma(1+s/2)prod_rholeft(1-fracsrho right)e^s/rho$$
can one derive the exact value of $sum_rho frac1rho$
to be
$$sum_rho frac1rho = -log(2sqrtpi)+1+gamma/2$$
What implications does this have?

complex-analysis riemann-zeta

share|cite|improve this question

edited Jul 25 at 16:08

Arnaud Mortier

18.8k22159

asked Jul 25 at 16:05

Jose Ortega

485

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 16:08

Arnaud Mortier

18.8k22159

edited Jul 25 at 16:08

Arnaud Mortier

18.8k22159

edited Jul 25 at 16:08

Arnaud Mortier

18.8k22159

18.8k22159

asked Jul 25 at 16:05

Jose Ortega

485

asked Jul 25 at 16:05

Jose Ortega

485

asked Jul 25 at 16:05

Jose Ortega

485

485

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Basically none. Also note the sum doesn't converge absolutely.
  – Wojowu
  Jul 25 at 16:49
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Basically none. Also note the sum doesn't converge absolutely.
  – Wojowu
  Jul 25 at 16:49
  

  
  
  
  

1

1

Basically none. Also note the sum doesn't converge absolutely.
– Wojowu
Jul 25 at 16:49

Basically none. Also note the sum doesn't converge absolutely.
– Wojowu
Jul 25 at 16:49

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

Let's start with the logarithm of the Hadamard product and take the derivative relatively to $s$ (remembering that $;psi(z):=dfrac ddzlogGamma(z),$ with $psi$ the digamma function) :
beginalign
logzeta(s)&=(log(2pi)-1-gamma/2)s-logleft(2(s-1)Gamma(1+s/2)right)+sum_rhologleft(1-fracsrho right)+frac srho\
fraczeta'(s)zeta(s)&=log(2pi)-1-gamma/2-left(frac 1s-1+frac 12psileft(frac s2+1right)right)+sum_rhofrac 1s-rho+frac 1rho\
fraczeta'(s)zeta(s)+frac 1s-1&=log(2pi)-1-gamma/2-frac 12psileft(frac s2+1right)+sum_rhofrac 1s-rho+frac 1rho\
endalign
taking the limit as $sto 1$ and remembering that $;displaystyle zeta(s)-frac 1s-1=gamma+O(s-1);$ and $psileft(frac 32right)=psileft(frac 12right)+2=-gamma-log(4)+2;$ we get :
beginalign
gamma&=log(2pi)-1-gamma/2+left(gamma/2+log(2)-1right)+sum_rhofrac 11-rho+frac 1rho\
sum_rhofrac 11-rho+frac 1rho&=-log(4pi)+2+gamma
endalign
With the answer the double of your series (as it should since if $rho$ is a root then $1-rho$ will also be one).

I considered here only a formal derivation (the series is only conditionally convergent : the roots must be sorted with increasing absolute imaginary parts, the limit inside the series should be justified) ; for detailed proofs see for example Edwards' book p.67.

For the sum of integer powers of the roots see Mathworld starting at $(5)$.

Other references were proposed in this answer as this MO thread that may be more useful for your question concerning 'implications' (see the answer and comment to Micah Milinovich).

share|cite|improve this answer

edited Jul 26 at 11:28

answered Jul 26 at 11:03

Raymond Manzoni

37.1k562114

add a comment | 

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2862563%2fsum-over-all-inverse-zeta-nontrivial-zeros%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Let's start with the logarithm of the Hadamard product and take the derivative relatively to $s$ (remembering that $;psi(z):=dfrac ddzlogGamma(z),$ with $psi$ the digamma function) :
beginalign
logzeta(s)&=(log(2pi)-1-gamma/2)s-logleft(2(s-1)Gamma(1+s/2)right)+sum_rhologleft(1-fracsrho right)+frac srho\
fraczeta'(s)zeta(s)&=log(2pi)-1-gamma/2-left(frac 1s-1+frac 12psileft(frac s2+1right)right)+sum_rhofrac 1s-rho+frac 1rho\
fraczeta'(s)zeta(s)+frac 1s-1&=log(2pi)-1-gamma/2-frac 12psileft(frac s2+1right)+sum_rhofrac 1s-rho+frac 1rho\
endalign
taking the limit as $sto 1$ and remembering that $;displaystyle zeta(s)-frac 1s-1=gamma+O(s-1);$ and $psileft(frac 32right)=psileft(frac 12right)+2=-gamma-log(4)+2;$ we get :
beginalign
gamma&=log(2pi)-1-gamma/2+left(gamma/2+log(2)-1right)+sum_rhofrac 11-rho+frac 1rho\
sum_rhofrac 11-rho+frac 1rho&=-log(4pi)+2+gamma
endalign
With the answer the double of your series (as it should since if $rho$ is a root then $1-rho$ will also be one).

I considered here only a formal derivation (the series is only conditionally convergent : the roots must be sorted with increasing absolute imaginary parts, the limit inside the series should be justified) ; for detailed proofs see for example Edwards' book p.67.

For the sum of integer powers of the roots see Mathworld starting at $(5)$.

Other references were proposed in this answer as this MO thread that may be more useful for your question concerning 'implications' (see the answer and comment to Micah Milinovich).

share|cite|improve this answer

edited Jul 26 at 11:28

answered Jul 26 at 11:03

Raymond Manzoni

37.1k562114

add a comment | 

up vote
1
down vote



accepted

Let's start with the logarithm of the Hadamard product and take the derivative relatively to $s$ (remembering that $;psi(z):=dfrac ddzlogGamma(z),$ with $psi$ the digamma function) :
beginalign
logzeta(s)&=(log(2pi)-1-gamma/2)s-logleft(2(s-1)Gamma(1+s/2)right)+sum_rhologleft(1-fracsrho right)+frac srho\
fraczeta'(s)zeta(s)&=log(2pi)-1-gamma/2-left(frac 1s-1+frac 12psileft(frac s2+1right)right)+sum_rhofrac 1s-rho+frac 1rho\
fraczeta'(s)zeta(s)+frac 1s-1&=log(2pi)-1-gamma/2-frac 12psileft(frac s2+1right)+sum_rhofrac 1s-rho+frac 1rho\
endalign
taking the limit as $sto 1$ and remembering that $;displaystyle zeta(s)-frac 1s-1=gamma+O(s-1);$ and $psileft(frac 32right)=psileft(frac 12right)+2=-gamma-log(4)+2;$ we get :
beginalign
gamma&=log(2pi)-1-gamma/2+left(gamma/2+log(2)-1right)+sum_rhofrac 11-rho+frac 1rho\
sum_rhofrac 11-rho+frac 1rho&=-log(4pi)+2+gamma
endalign
With the answer the double of your series (as it should since if $rho$ is a root then $1-rho$ will also be one).

I considered here only a formal derivation (the series is only conditionally convergent : the roots must be sorted with increasing absolute imaginary parts, the limit inside the series should be justified) ; for detailed proofs see for example Edwards' book p.67.

For the sum of integer powers of the roots see Mathworld starting at $(5)$.

Other references were proposed in this answer as this MO thread that may be more useful for your question concerning 'implications' (see the answer and comment to Micah Milinovich).

share|cite|improve this answer

edited Jul 26 at 11:28

answered Jul 26 at 11:03

Raymond Manzoni

37.1k562114

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Let's start with the logarithm of the Hadamard product and take the derivative relatively to $s$ (remembering that $;psi(z):=dfrac ddzlogGamma(z),$ with $psi$ the digamma function) :
beginalign
logzeta(s)&=(log(2pi)-1-gamma/2)s-logleft(2(s-1)Gamma(1+s/2)right)+sum_rhologleft(1-fracsrho right)+frac srho\
fraczeta'(s)zeta(s)&=log(2pi)-1-gamma/2-left(frac 1s-1+frac 12psileft(frac s2+1right)right)+sum_rhofrac 1s-rho+frac 1rho\
fraczeta'(s)zeta(s)+frac 1s-1&=log(2pi)-1-gamma/2-frac 12psileft(frac s2+1right)+sum_rhofrac 1s-rho+frac 1rho\
endalign
taking the limit as $sto 1$ and remembering that $;displaystyle zeta(s)-frac 1s-1=gamma+O(s-1);$ and $psileft(frac 32right)=psileft(frac 12right)+2=-gamma-log(4)+2;$ we get :
beginalign
gamma&=log(2pi)-1-gamma/2+left(gamma/2+log(2)-1right)+sum_rhofrac 11-rho+frac 1rho\
sum_rhofrac 11-rho+frac 1rho&=-log(4pi)+2+gamma
endalign
With the answer the double of your series (as it should since if $rho$ is a root then $1-rho$ will also be one).

I considered here only a formal derivation (the series is only conditionally convergent : the roots must be sorted with increasing absolute imaginary parts, the limit inside the series should be justified) ; for detailed proofs see for example Edwards' book p.67.

For the sum of integer powers of the roots see Mathworld starting at $(5)$.

Other references were proposed in this answer as this MO thread that may be more useful for your question concerning 'implications' (see the answer and comment to Micah Milinovich).

share|cite|improve this answer

edited Jul 26 at 11:28

answered Jul 26 at 11:03

Raymond Manzoni

37.1k562114

Let's start with the logarithm of the Hadamard product and take the derivative relatively to $s$ (remembering that $;psi(z):=dfrac ddzlogGamma(z),$ with $psi$ the digamma function) :
beginalign
logzeta(s)&=(log(2pi)-1-gamma/2)s-logleft(2(s-1)Gamma(1+s/2)right)+sum_rhologleft(1-fracsrho right)+frac srho\
fraczeta'(s)zeta(s)&=log(2pi)-1-gamma/2-left(frac 1s-1+frac 12psileft(frac s2+1right)right)+sum_rhofrac 1s-rho+frac 1rho\
fraczeta'(s)zeta(s)+frac 1s-1&=log(2pi)-1-gamma/2-frac 12psileft(frac s2+1right)+sum_rhofrac 1s-rho+frac 1rho\
endalign
taking the limit as $sto 1$ and remembering that $;displaystyle zeta(s)-frac 1s-1=gamma+O(s-1);$ and $psileft(frac 32right)=psileft(frac 12right)+2=-gamma-log(4)+2;$ we get :
beginalign
gamma&=log(2pi)-1-gamma/2+left(gamma/2+log(2)-1right)+sum_rhofrac 11-rho+frac 1rho\
sum_rhofrac 11-rho+frac 1rho&=-log(4pi)+2+gamma
endalign
With the answer the double of your series (as it should since if $rho$ is a root then $1-rho$ will also be one).

I considered here only a formal derivation (the series is only conditionally convergent : the roots must be sorted with increasing absolute imaginary parts, the limit inside the series should be justified) ; for detailed proofs see for example Edwards' book p.67.

For the sum of integer powers of the roots see Mathworld starting at $(5)$.

Other references were proposed in this answer as this MO thread that may be more useful for your question concerning 'implications' (see the answer and comment to Micah Milinovich).

share|cite|improve this answer

edited Jul 26 at 11:28

answered Jul 26 at 11:03

Raymond Manzoni

37.1k562114

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 11:28

edited Jul 26 at 11:28

edited Jul 26 at 11:28

answered Jul 26 at 11:03

Raymond Manzoni

37.1k562114

answered Jul 26 at 11:03

Raymond Manzoni

37.1k562114

answered Jul 26 at 11:03

Raymond Manzoni

37.1k562114

37.1k562114

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2862563%2fsum-over-all-inverse-zeta-nontrivial-zeros%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

Name Email

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

Name Email

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

Name Email

Name Email

Name

Email

Name Email

Name

Email