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Simplifying $prodlimits_{kneq j=0}^{n-1}frac1{lambda_{n,k}-lambda_{n,j}}$ for $lambda_{n,k}=expfrac{ipi(2k+1)}{n}$

What is the closed form for $sum_{n=1}^infty frac1n - frac1{n+1/p}$?Computing $limlimits_{ntoinfty} Big(sumlimits_{i = 1}^n sumlimits_{j = 1}^n frac1{i^2+j^2}-frac{pi}{2} log(n)Big)$.How to find $int_0^1!! left(frac1{log^2left(1-xright)}-frac1{x^2}+frac1x-frac1{12}right) frac{mathrm{d}x}x$Challenging recurrence relation that depends on composition of integersClosed form for the series $sum_{k=1}^infty (-1)^k ln left( tanh frac{pi k x}{2} right)$Evaluating $intfrac{x^b}{1+x^a}~dx$ for $a,binBbb N$Show that $int_0^{infty}frac{operatorname{Li}_s(-x)}{x^{alpha+1}}mathrm dx=-frac1{alpha^s}frac{pi}{sin(pi alpha)}$closed form for $sum_{ngeq1}pi nlogfrac{4n+1}{4n-1}prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}$?General closed form for $L(phi)=int_0^phi log(sin x)mathrm dx$ when $phiin(0,pi)$?Prove closed form for $sum_{ninBbb N}frac1{5n(5n-1)}$

3

2

$begingroup$

I have been able to show that for $ninBbb N_{geq2}$ $$phi(n)=int_0^1frac{dx}{x^n+1}=sum_{k=0}^{n-1}Gamma_{n,k}logfrac{lambda_{n,k}-1}{lambda_{n,k}}$$
Where $$lambda_{n,k}=expfrac{ipi(2k+1)}{n}$$
And $$Gamma_{n,k}=prod_{kneq j=0}^{n-1}frac1{lambda_{n,k}-lambda_{n,j}}$$
And I was wondering: how do we simplify $Gamma_{n,k}$ to ease the manual calculation of $phi(n)$ values. The integral is always real, so I am sure there is a major way we can simplify $Gamma_{n,k}$, but I have been so far unable to find it. I do suspect however that the product $$P_n=prod_{k=0}^{n-1}Gamma_{n,k}$$
May play a significant role in finding the simplification I seek.

---

For those interested, a proof.

Note that $x^n+1$ bay be factored as
$$x^n+1=prod_{k=0}^{n-1}(x-lambda_{n,k})$$
Hence $$phi(n)=int_0^1prod_{k=0}^{n-1}frac1{x-lambda_{n,k}}dx$$
Then define $Gamma_{n,k}$ by saying that
$$prod_{k=0}^{n-1}frac1{x-lambda_{n,k}}=sum_{k=0}^{n-1}frac{Gamma_{n,k}}{x-lambda_{n,k}}$$
Multiplying both sides by $prod_{j=0}^{n-1}(x-lambda_{n,j})$:
$$1=sum_{k=0}^{n-1}frac{Gamma_{n,k}}{x-lambda_{n,k}}prod_{j=0}^{n-1}(x-lambda_{n,j})$$
$$1=sum_{k=0}^{n-1}Gamma_{n,k}prod_{kneq j=0}^{n-1}(x-lambda_{n,j})$$
So for any integer $0leq mleq n-1$ we may plug in $x=lambda{n,m}$ and simplify to get
$$Gamma_{n,m}=prod_{mneq j=0}^{n-1}frac1{lambda_{n,m}-lambda_{n,j}}$$
And our result follows directly.

Perhaps another motivation for easing manual calculation of this product would be that $$sum_{k=0}^{infty}frac{(-1)^k}{nk+1}=phi(n)$$
Which brings about a plethora of interesting closed forms.

---

Edit: A little progress

We define $$c_{n,j}=operatorname{Re}lambda_{n,j}=cosfrac{pi(2j+1)}{n}$$
And $$s_{n,j}=operatorname{Im}lambda_{n,j}=sinfrac{pi(2j+1)}{n}$$
So $$logfrac{lambda_{n,k}-1}{lambda_{n,k}}=logleft(1-lambda_{n,k}^{-1}right)=logleft(1-c_{n,k}+is_{n,k}right)$$
And we also see that
$$begin{align}
prod_{kneq j=0}^{n-1}frac1{lambda_{n,k}-lambda_{n,j}}&=prod_{kneq j=0}^{n-1}frac1{e^{ipi(2k+1)/n}-e^{ipi(2j+1)/n}}\
&=prod_{kneq j=0}^{n-1}frac{e^{-ipi(2k+1)/n}}{1-e^{ipi(2j-2k)/n}}\
&=e^{i(2k+1)(2-n)/n}prod_{kneq j=0}^{n-1}frac12left(1+icotfrac{pi(j-k)}nright)\
Gamma_{n,k}&=frac{lambda_{n,k}^{2-n}}{2^{n-2}}prod_{kneq j=0}^{n-1}left(1+icotfrac{pi(j-k)}nright)
end{align}$$
But the remaining product I do not know how to deal with.

sequences-and-series definite-integrals complex-numbers closed-form products

share|cite|improve this question

edited Mar 12 at 17:10

clathratus

asked Mar 12 at 4:48

clathratusclathratus

5,0351338

$endgroup$

add a comment | 

3

2

$begingroup$

I have been able to show that for $ninBbb N_{geq2}$ $$phi(n)=int_0^1frac{dx}{x^n+1}=sum_{k=0}^{n-1}Gamma_{n,k}logfrac{lambda_{n,k}-1}{lambda_{n,k}}$$
Where $$lambda_{n,k}=expfrac{ipi(2k+1)}{n}$$
And $$Gamma_{n,k}=prod_{kneq j=0}^{n-1}frac1{lambda_{n,k}-lambda_{n,j}}$$
And I was wondering: how do we simplify $Gamma_{n,k}$ to ease the manual calculation of $phi(n)$ values. The integral is always real, so I am sure there is a major way we can simplify $Gamma_{n,k}$, but I have been so far unable to find it. I do suspect however that the product $$P_n=prod_{k=0}^{n-1}Gamma_{n,k}$$
May play a significant role in finding the simplification I seek.

---

For those interested, a proof.

Note that $x^n+1$ bay be factored as
$$x^n+1=prod_{k=0}^{n-1}(x-lambda_{n,k})$$
Hence $$phi(n)=int_0^1prod_{k=0}^{n-1}frac1{x-lambda_{n,k}}dx$$
Then define $Gamma_{n,k}$ by saying that
$$prod_{k=0}^{n-1}frac1{x-lambda_{n,k}}=sum_{k=0}^{n-1}frac{Gamma_{n,k}}{x-lambda_{n,k}}$$
Multiplying both sides by $prod_{j=0}^{n-1}(x-lambda_{n,j})$:
$$1=sum_{k=0}^{n-1}frac{Gamma_{n,k}}{x-lambda_{n,k}}prod_{j=0}^{n-1}(x-lambda_{n,j})$$
$$1=sum_{k=0}^{n-1}Gamma_{n,k}prod_{kneq j=0}^{n-1}(x-lambda_{n,j})$$
So for any integer $0leq mleq n-1$ we may plug in $x=lambda{n,m}$ and simplify to get
$$Gamma_{n,m}=prod_{mneq j=0}^{n-1}frac1{lambda_{n,m}-lambda_{n,j}}$$
And our result follows directly.

Perhaps another motivation for easing manual calculation of this product would be that $$sum_{k=0}^{infty}frac{(-1)^k}{nk+1}=phi(n)$$
Which brings about a plethora of interesting closed forms.

---

Edit: A little progress

We define $$c_{n,j}=operatorname{Re}lambda_{n,j}=cosfrac{pi(2j+1)}{n}$$
And $$s_{n,j}=operatorname{Im}lambda_{n,j}=sinfrac{pi(2j+1)}{n}$$
So $$logfrac{lambda_{n,k}-1}{lambda_{n,k}}=logleft(1-lambda_{n,k}^{-1}right)=logleft(1-c_{n,k}+is_{n,k}right)$$
And we also see that
$$begin{align}
prod_{kneq j=0}^{n-1}frac1{lambda_{n,k}-lambda_{n,j}}&=prod_{kneq j=0}^{n-1}frac1{e^{ipi(2k+1)/n}-e^{ipi(2j+1)/n}}\
&=prod_{kneq j=0}^{n-1}frac{e^{-ipi(2k+1)/n}}{1-e^{ipi(2j-2k)/n}}\
&=e^{i(2k+1)(2-n)/n}prod_{kneq j=0}^{n-1}frac12left(1+icotfrac{pi(j-k)}nright)\
Gamma_{n,k}&=frac{lambda_{n,k}^{2-n}}{2^{n-2}}prod_{kneq j=0}^{n-1}left(1+icotfrac{pi(j-k)}nright)
end{align}$$
But the remaining product I do not know how to deal with.

sequences-and-series definite-integrals complex-numbers closed-form products

share|cite|improve this question

edited Mar 12 at 17:10

clathratus

asked Mar 12 at 4:48

clathratusclathratus

5,0351338

$endgroup$

add a comment | 

$begingroup$

I have been able to show that for $ninBbb N_{geq2}$ $$phi(n)=int_0^1frac{dx}{x^n+1}=sum_{k=0}^{n-1}Gamma_{n,k}logfrac{lambda_{n,k}-1}{lambda_{n,k}}$$
Where $$lambda_{n,k}=expfrac{ipi(2k+1)}{n}$$
And $$Gamma_{n,k}=prod_{kneq j=0}^{n-1}frac1{lambda_{n,k}-lambda_{n,j}}$$
And I was wondering: how do we simplify $Gamma_{n,k}$ to ease the manual calculation of $phi(n)$ values. The integral is always real, so I am sure there is a major way we can simplify $Gamma_{n,k}$, but I have been so far unable to find it. I do suspect however that the product $$P_n=prod_{k=0}^{n-1}Gamma_{n,k}$$
May play a significant role in finding the simplification I seek.

---

For those interested, a proof.

Note that $x^n+1$ bay be factored as
$$x^n+1=prod_{k=0}^{n-1}(x-lambda_{n,k})$$
Hence $$phi(n)=int_0^1prod_{k=0}^{n-1}frac1{x-lambda_{n,k}}dx$$
Then define $Gamma_{n,k}$ by saying that
$$prod_{k=0}^{n-1}frac1{x-lambda_{n,k}}=sum_{k=0}^{n-1}frac{Gamma_{n,k}}{x-lambda_{n,k}}$$
Multiplying both sides by $prod_{j=0}^{n-1}(x-lambda_{n,j})$:
$$1=sum_{k=0}^{n-1}frac{Gamma_{n,k}}{x-lambda_{n,k}}prod_{j=0}^{n-1}(x-lambda_{n,j})$$
$$1=sum_{k=0}^{n-1}Gamma_{n,k}prod_{kneq j=0}^{n-1}(x-lambda_{n,j})$$
So for any integer $0leq mleq n-1$ we may plug in $x=lambda{n,m}$ and simplify to get
$$Gamma_{n,m}=prod_{mneq j=0}^{n-1}frac1{lambda_{n,m}-lambda_{n,j}}$$
And our result follows directly.

Perhaps another motivation for easing manual calculation of this product would be that $$sum_{k=0}^{infty}frac{(-1)^k}{nk+1}=phi(n)$$
Which brings about a plethora of interesting closed forms.

---

Edit: A little progress

We define $$c_{n,j}=operatorname{Re}lambda_{n,j}=cosfrac{pi(2j+1)}{n}$$
And $$s_{n,j}=operatorname{Im}lambda_{n,j}=sinfrac{pi(2j+1)}{n}$$
So $$logfrac{lambda_{n,k}-1}{lambda_{n,k}}=logleft(1-lambda_{n,k}^{-1}right)=logleft(1-c_{n,k}+is_{n,k}right)$$
And we also see that
$$begin{align}
prod_{kneq j=0}^{n-1}frac1{lambda_{n,k}-lambda_{n,j}}&=prod_{kneq j=0}^{n-1}frac1{e^{ipi(2k+1)/n}-e^{ipi(2j+1)/n}}\
&=prod_{kneq j=0}^{n-1}frac{e^{-ipi(2k+1)/n}}{1-e^{ipi(2j-2k)/n}}\
&=e^{i(2k+1)(2-n)/n}prod_{kneq j=0}^{n-1}frac12left(1+icotfrac{pi(j-k)}nright)\
Gamma_{n,k}&=frac{lambda_{n,k}^{2-n}}{2^{n-2}}prod_{kneq j=0}^{n-1}left(1+icotfrac{pi(j-k)}nright)
end{align}$$
But the remaining product I do not know how to deal with.

sequences-and-series definite-integrals complex-numbers closed-form products

share|cite|improve this question

edited Mar 12 at 17:10

clathratus

asked Mar 12 at 4:48

clathratusclathratus

5,0351338

$endgroup$

share|cite|improve this question

edited Mar 12 at 17:10

clathratus

asked Mar 12 at 4:48

clathratusclathratus

5,0351338

share|cite|improve this question

edited Mar 12 at 17:10

clathratus

asked Mar 12 at 4:48

clathratusclathratus

5,0351338

asked Mar 12 at 4:48

clathratusclathratus

5,0351338

asked Mar 12 at 4:48

clathratusclathratus

5,0351338

1 Answer
1

active

oldest

votes

2

$begingroup$

Defining the polynomial
begin{align}
P(x)&=x^n+1\
&=prod_{j=0}^{n-1}left( x- lambda_{n,j}right)
end{align}
we can express its derivative at $x=lambda_{n,k}$ as:
begin{align}
P'(lambda_{n,k})&=prod_{kneq j=0}^{n-1}left( lambda_{n,k}-lambda_{n,j} right)\
&=frac{1}{Gamma_{n,k}}
end{align}
But we have also $P'(x)=nx^{n-1}=ntfrac{x^n}{x}$. Thus, as $left(lambda_{n,k} right)^n=-1$,
begin{equation}
P'(lambda_{n,k})=nfrac{-1}{lambda_{n,k}}
end{equation}
Finally,
begin{equation}
Gamma_{n,k}=-frac{lambda_{n,k}}{n}
end{equation}
This trick comes rather naturally if the integral is evaluated by the residue method, for the function $f(z)=(1+z^n)^{-1}lnleft(tfrac z{1-z}right)$ along the keyhole contour.

share|cite|improve this answer

answered Mar 13 at 16:25

Paul EntaPaul Enta

5,35111434

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You, sir, are a genius. That was so elegant. Many thanks.
  $endgroup$
  – clathratus
  Mar 13 at 16:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Too kind. Happy you like it!
  $endgroup$
  – Paul Enta
  Mar 13 at 16:28
  

  
  
  
  

add a comment | 

Your Answer

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2

$begingroup$

Defining the polynomial
begin{align}
P(x)&=x^n+1\
&=prod_{j=0}^{n-1}left( x- lambda_{n,j}right)
end{align}
we can express its derivative at $x=lambda_{n,k}$ as:
begin{align}
P'(lambda_{n,k})&=prod_{kneq j=0}^{n-1}left( lambda_{n,k}-lambda_{n,j} right)\
&=frac{1}{Gamma_{n,k}}
end{align}
But we have also $P'(x)=nx^{n-1}=ntfrac{x^n}{x}$. Thus, as $left(lambda_{n,k} right)^n=-1$,
begin{equation}
P'(lambda_{n,k})=nfrac{-1}{lambda_{n,k}}
end{equation}
Finally,
begin{equation}
Gamma_{n,k}=-frac{lambda_{n,k}}{n}
end{equation}
This trick comes rather naturally if the integral is evaluated by the residue method, for the function $f(z)=(1+z^n)^{-1}lnleft(tfrac z{1-z}right)$ along the keyhole contour.

share|cite|improve this answer

answered Mar 13 at 16:25

Paul EntaPaul Enta

5,35111434

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You, sir, are a genius. That was so elegant. Many thanks.
  $endgroup$
  – clathratus
  Mar 13 at 16:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Too kind. Happy you like it!
  $endgroup$
  – Paul Enta
  Mar 13 at 16:28
  

  
  
  
  

add a comment | 

2

$begingroup$

Defining the polynomial
begin{align}
P(x)&=x^n+1\
&=prod_{j=0}^{n-1}left( x- lambda_{n,j}right)
end{align}
we can express its derivative at $x=lambda_{n,k}$ as:
begin{align}
P'(lambda_{n,k})&=prod_{kneq j=0}^{n-1}left( lambda_{n,k}-lambda_{n,j} right)\
&=frac{1}{Gamma_{n,k}}
end{align}
But we have also $P'(x)=nx^{n-1}=ntfrac{x^n}{x}$. Thus, as $left(lambda_{n,k} right)^n=-1$,
begin{equation}
P'(lambda_{n,k})=nfrac{-1}{lambda_{n,k}}
end{equation}
Finally,
begin{equation}
Gamma_{n,k}=-frac{lambda_{n,k}}{n}
end{equation}
This trick comes rather naturally if the integral is evaluated by the residue method, for the function $f(z)=(1+z^n)^{-1}lnleft(tfrac z{1-z}right)$ along the keyhole contour.

share|cite|improve this answer

answered Mar 13 at 16:25

Paul EntaPaul Enta

5,35111434

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You, sir, are a genius. That was so elegant. Many thanks.
  $endgroup$
  – clathratus
  Mar 13 at 16:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Too kind. Happy you like it!
  $endgroup$
  – Paul Enta
  Mar 13 at 16:28
  

  
  
  
  

add a comment | 

$begingroup$

Defining the polynomial
begin{align}
P(x)&=x^n+1\
&=prod_{j=0}^{n-1}left( x- lambda_{n,j}right)
end{align}
we can express its derivative at $x=lambda_{n,k}$ as:
begin{align}
P'(lambda_{n,k})&=prod_{kneq j=0}^{n-1}left( lambda_{n,k}-lambda_{n,j} right)\
&=frac{1}{Gamma_{n,k}}
end{align}
But we have also $P'(x)=nx^{n-1}=ntfrac{x^n}{x}$. Thus, as $left(lambda_{n,k} right)^n=-1$,
begin{equation}
P'(lambda_{n,k})=nfrac{-1}{lambda_{n,k}}
end{equation}
Finally,
begin{equation}
Gamma_{n,k}=-frac{lambda_{n,k}}{n}
end{equation}
This trick comes rather naturally if the integral is evaluated by the residue method, for the function $f(z)=(1+z^n)^{-1}lnleft(tfrac z{1-z}right)$ along the keyhole contour.

share|cite|improve this answer

answered Mar 13 at 16:25

Paul EntaPaul Enta

5,35111434

$endgroup$

share|cite|improve this answer

answered Mar 13 at 16:25

Paul EntaPaul Enta

5,35111434

answered Mar 13 at 16:25

Paul EntaPaul Enta

5,35111434

answered Mar 13 at 16:25

Paul EntaPaul Enta

5,35111434