Derive zeta values of even integers from the Euler-Maclaurin formula.

Question

Euler showed:

\begin{equation} B_{2 k} = (-1)^{k+1} \frac{2 \, (2 \, k)!}{ (2 \, \pi)^{2 k}} \zeta(2 k) \end{equation}

for $k=1,2, \cdots$. We could from here find $\zeta(2k)$ in terms of the even Bernoulli coefficients $B_{2k}$.

How can we derive the equivalent representation by using the Euler Maclaurin formula ?

Thanks.

Update

Here is what I have done:

\begin{eqnarray*} \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} \quad , \quad \mathrm{Re}(s) > 1. \end{eqnarray*} Euler used the Euler-Maclaurin series to find values of the Riemann Zeta function. We want to use the Euler-Maclaurin formula with $f(x)=1/x^s=x^{-s}$. We know that

\begin{eqnarray*} f^{(0)}(x) &=& \frac{1}{x^s} \\ f^{(1)}(x) &=& -\frac{s}{x^{s+1}} \\ f^{(2)}(x) &=& \frac{s(s+1)}{x^{s+2}} \\ &\vdots& \\ f^{(i)}(x) &=& (-1)^{i+1} \frac{s(s+1) \dots (s+i-2)}{x^{s+i}} = (-1)^{i+1} \frac{\Gamma(s+i)}{ \Gamma(s)} \frac{1}{ x^{s+i}}. \end{eqnarray*}

We then write using $h=1$, $a=1$, $b=\infty$

\begin{eqnarray*} \sum_{i=1}^{\infty} \frac{1}{n^s} = \int_1^{\infty} \frac{dx}{x^s} + \frac{1}{2} + \left . \sum_{i=1}^{m} \frac{B_{2i}}{(2i)!} \frac{\Gamma(s+2 i-1)}{ \Gamma(s)} \frac{1}{ x^{s+2 i-1}} \right |_1^{\infty} + R_{2m} \end{eqnarray*} with

\begin{eqnarray*} R_{2m} = -\int_1^{\infty} \mathrm{B}_{2m} \left \{ x-1 \right \} \frac{\Gamma(s+2 m)}{(2m)! \, \Gamma(s)} \frac{dx}{x^{s+2m}}. \end{eqnarray*} That is

\begin{eqnarray} \zeta(s) = \frac{1}{s-1} + \frac{1}{2} - \sum_{i=1}^{m} \frac{B_{2i}}{(2i)!} \frac{\Gamma(s+2 i-1)}{ \Gamma(s)} + R_{2m} \label{zetazeta} \end{eqnarray}

This is an important equation since it establishes an analytic continuation for the $\zeta(s)$ function. We observe that the first fraction is an analytic function except for $s=1$, then the sum of quotients of $\Gamma$ functions is analytic except for isolated singularities in the negative integer arguments. Finally, since the Bernoulli polynomial $B_{2m}\{x-1\}$ is bounded the residual is a convergent integral for $s + 2m >1$, so we can extend the convergence as far as $s > 1-2m$, for any positive number $m$.

The question here is what "$m$" to choose. If I choose $m=1$, I did the computations and found something which made no sense.

do you know the common ways for proving those values of $\zeta$ at even (positive) integers ? — reuns, Apr 30 '16 at 00:00
this is a common way, applying the residue theorem to $\displaystyle\frac{z^{-2k}}{e^z-1}$, you can also use the Fourier series $\displaystyle f(x) = \sum_n \frac{\sin(2 \pi n x)}{n^k}$ and compute the $L^2([0,1])$ norm, or can prove the functional equation and then look at the poles of $\displaystyle\Gamma(s) \zeta(s) = \int_0^\infty \frac{x^s}{e^x-1} dx$ at negative integers (my favourite). that's all I Know. — reuns, Apr 30 '16 at 00:06
@user1952009 I know two different ways to prove that. I am curious about how to do this using the Euler-Maclaurin series. — Herman Jaramillo, Apr 30 '16 at 00:21
you mean you are curious to prove it using NEW ways ? why would there exist a way using in particular Euler-Maclaurin ? and your two ways are in my list ?.. — reuns, Apr 30 '16 at 00:24
@user1952009 : I updated my question. I want to get the formula following the path above — Herman Jaramillo, Apr 30 '16 at 00:29
look at this fr.wikipedia.org/wiki/Fonction_zêta_de_Riemann#Par_la_formule_d'Euler-MacLaurin (and at the rest of the article, if you have difficulties for translating I can help) — reuns, Apr 30 '16 at 00:33
@user1952009 : The two ways that I know is the contour integration over the function that you discribe above. The sum of all residues provides the infinite series. The other way is the way Euler did it with the infinite factorization of the sine function. Thanks. — Herman Jaramillo, Apr 30 '16 at 00:36
@user1952009 : I looked at your reference. It is good, but I would like to find the formula that Euler found. That is, a simple equation with only one term involving $\pi$, factorials, and $B_{2k}$. Thanks. — Herman Jaramillo, Apr 30 '16 at 00:39
you can look at this too en.wikipedia.org/wiki/Particular_values_of_Riemann_zeta_function#Even_positive_integers , looks promising to me : "$\zeta(2n)=\frac{1}{n+\frac{1}{2}} \sum_{k=1}^{n-1} \zeta(2k)\zeta(2n-2k), n>1 $ can be proved using $\frac{d}{dx} \cot(x) = -1-\cot^{2} (x) $ " since it is based on Euler's series for $\cot(x)$ — reuns, Apr 30 '16 at 00:42

score 3 · Accepted Answer · edited Apr 13 '17 at 12:20

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Should you accept to use the functional equation then let's start with the Euler Maclaurin formula for zeta in the specific case where the binomial sum is convergent (see for example this thread) :
$$\tag{1}\zeta(s)= \sum_{k=1}^N \frac 1{k^s} {+\frac 1{(s-1)\;N^{s-1}}} + \sum_{i=1}^{\infty} \frac{B_{i}\;s^{(i-1)}}{i!\;N^{s-1+i}},\quad N\in\mathbb{N}^+$$ with $s^{(i)}$ the "rising factorial" : $\;s^{(i)}=s(s+1)\cdots(s+k-1),\,s^{(0)}=1$ and $B_1=-\dfrac 12$.
(I used the general formula $(1)$ but your formula supposing $N=1$ should give the same result!)

For $\;s:=1-j\;$ with $\,j\,$ a positive integer the EMl formula is applied to the polynomial $\,f(k):=k^{1-j}\;$ so that $(1)$ is valid will a finite sum at the right from $\;(1-j)^{(j)}=0\;$ giving :

\begin{align} \zeta(1-j)&= \sum_{k=1}^N {k^{j-1}} -\frac {N^{j}}{j\;} + \sum_{i=1}^{j} \frac{B_{i}\;(1-j)^{(i-1)}N^{j-i}}{i!\;}\\ \zeta(1-j)&= \sum_{k=1}^N {k^{j-1}} -\frac {N^{j}}{j\;} + \sum_{i=1}^{j} (-1)^{i-1}\frac{(j-1)!}{i!\,(j-i)!}B_{i}\;N^{j-i}\\ \tag{2}\zeta(1-j)&= \sum_{k=1}^N {k^{j-1}} -\frac {N^{j}}{j\;} - \frac 1j\sum_{i=1}^{j} (-1)^{i}{j\choose{i}}B_{i}\;N^{j-i}\\ \end{align}

But the Faulhaber formula gives us : $$\tag{3}\sum_{k=1}^N {k^{j-1}}= {1\over{j}}\sum_{i=0}^{j-1} (-1)^i {j\choose{i}} B_i\; N^{j-i}$$

When we combine these two expressions all the terms of the sum disappear except for $i=0$ and $i=j\,$ and :

$$\zeta(1-j)= {1\over{j}}(-1)^0 {j\choose{0}} B_0\; N^{j} -\frac {N^{j}}{j\;} - \frac 1j(-1)^{j}{j\choose{j}}B_{j}\;N^{0}= -(-1)^{j}\frac {B_{j}}j$$ Since $\,B_{2n+1}=0\,$ for any integer $n>1$ this becomes the simple : $$\tag{4}\zeta(1-j)= -\frac {B_{j}}j$$

To conclude we will use the promised functional equation applied to $\,j:=1-2m$ : \begin{align} \zeta(1-2m) &= 2^{1-2m}\pi^{-2m}\ \sin\left(\frac{\pi (1-2m)}2\right)\ \Gamma(2m)\ \zeta(2m)\\ -\frac {B_{2m}}{2m}&= 2(2\pi)^{-2m}\ \cos\left(\pi m\right)\ \Gamma(2m)\;\zeta(2m) \\ B_{2m}&= 2(2\pi)^{-2m}\ (-1)^{m+1}\ \Gamma(2m+1)\;\zeta(2m) \\ \end{align} with the wished result : $$\tag{5}\boxed{\displaystyle B_{2m}= (-1)^{m+1}\,\frac{2\,(2m)!}{(2\pi)^{2m}}\;\zeta(2m) }$$ $$-$$ References to Euler's derivation of the Euler–Maclaurin formula are here.

edited Apr 13 '17 at 12:20

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1

answered Apr 30 '16 at 20:28

Raymond Manzoni

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Dear @Raymond Manzoni : Thank you very much for your answer. It looks at first sight excellent. I will need some time to review it. I will be back about this later. Thanks again. – Herman Jaramillo May 02 '16 at 22:34
Glad it helped anyway dear @Herman Jaramillo and excellent continuation, – Raymond Manzoni May 03 '16 at 00:04
I am still working in your answer. Let me give you a little feedback of what I have read so far. When you define the "raising factorial" you use write "k" on the right hand side and it should be "i" (just a typo). I am still trying to understand why the last two terms of your Euler-Maclaurin formula are not functions of $N+1$ instead of $N$ as you indicate. See that $\sum_{i=1}^{\infty} = \sum_1^N + \sum_{N+1}^{\infty}$ . Somehow in the center of equations (2) I got $(j-i)!/(j-1)!$ that is the multiplicative inverse of what you get. This might be my error. – Herman Jaramillo May 03 '16 at 12:23
I run out of characters so I need one more comment here: Your equation (4) is well known (but I believe it should be $j!$ instead of $j$ in the denominator. I derived your equation (4) with the $j!$ in the denominator using a different path. I am glad that you can connect the Euler-Maclaurin equation with your equation (4) (after correcting). This will make your path a valid path for my question. Thank you very much. – Herman Jaramillo May 03 '16 at 12:27
@HermanJaramillo: Well $(1)$ looks nicer than $$\zeta(s)= \sum_{k=1}^{N-1} \frac 1{k^s} {+\frac 1{(s-1);N^{s-1}}} + \frac 1{N^s}+ \frac{B_{1};s^{(0)}}{1!;N^{s}}+\sum_{i=2}^{\infty} \frac{B_{i};s^{(i-1)}}{i!;N^{s-1+i}}$$ (the two terms $\dfrac 1{N^s}$ may be combined to get $\dfrac 1{2,N^s}$ since $B_1=-\dfrac 12$ and since you consider only even values of $i$)
$$-$$ There was indeed a typo in my definition of the raising factorial but I don't see errors in $(4)$ (see here) – Raymond Manzoni May 03 '16 at 13:32
(except that I should have indicated "for $j>1$") nor in the rewriting of $;\displaystyle \frac{(1-j)^{(i-1)}}{i!}=(-1)^{i-1}\frac{(j-1)!}{i!,(j-i)!}=-\frac{(-1)^{i}{j \choose{i}}}j$. Excellent continuation, – Raymond Manzoni May 03 '16 at 14:03
I am sorry for going back to this. The equation that I have as Euler-Maclauren is: $ \sum_{i=0}^{n-1} f(a + i h) h = \int_a^b f(x) dx + \left . \sum_{k=1}^{m} \frac{B_k h^k}{k!} f^{(k-1)} (x) \right |_a^b + R_m $ . I do not get your $1/N^s$ above in your comment that starts "Well (1) looks nicer than".... I would appreciate your help here. – Herman Jaramillo May 03 '16 at 16:56
@HermanJaramillo: (you probably suppose that $;a+in=b$)
Set $;h=1,\ a=N,\ n=+\infty,\ f(k):=\dfrac1{k^s} ;$ then (supposing first that $,\Re(s)>1$) :
– Raymond Manzoni May 03 '16 at 18:22
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\begin{align} \sum_{i=0}^{n-1} f(a + i h) h = \int_a^b f(x) dx + \left . \sum_{k=1}^{m} \frac{B_k h^k}{k!} f^{(k-1)} (x) \right |a^b + R_m\ \sum{i=0}^{\infty} \frac 1{(N + i)^s} = \int_N^{\infty} \frac {dx}{x^s} + \left . \sum_{k=1}^{m} \frac{B_k}{k!} (-1)^{k-1}\frac{s^{(k-1)}}{k^{s+k-1}}\right |N^{\infty}+ R_m\ \sum{k=N}^{\infty} \frac 1{k^s} = \frac {1}{(s-1),N^{s-1}} - \sum_{k=1}^{m} \frac{B_k}{k!} (-1)^{k-1}\frac{s^{(k-1)}}{N^{s+k-1}}+ R_m\ \end{align} – Raymond Manzoni May 03 '16 at 18:22
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That is $$\zeta(s)- \sum_{k=1}^{N-1} \frac 1{k^s} = \frac {1}{(s-1),N^{s-1}} -\frac{B_{1}}{N^{s}} +\sum_{k=2}^{m} \frac{B_k}{k!} \frac{s^{(k-1)}}{N^{s+k-1}}+ R_m\$$ as wished. The derivation is in most ANT books like in Edwards' "Riemann's zeta function" p.$114$. Hoping this clarified things, – Raymond Manzoni May 03 '16 at 18:24
Your derivation here is correct. However if you compare your formula (and embrace the $B_1$ term into the sum) with your oriigina equation (1) they do not match. In your equation (1) the sum on the left goes all the way to $N$. Here it goes only up to $N-1$. This $1/N^s$ term is killing me. Thanks.e – Herman Jaramillo May 03 '16 at 18:35
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Well I explained that earlier ; since $B_1=-\dfrac 12$ : $$\text{(my initial)};\sum_{k=1}^{N}\frac 1{k^s}+\frac{B_{1}}{N^{s}}= \sum_{k=1}^{N-1} \frac 1{k^s} + \frac 1{N^s}+ \frac{B_{1}}{N^{s}}=\sum_{k=1}^{N-1} \frac 1{k^s} -\frac{B_{1}}{N^{s}};\text{(obtained here)}$$ Perhaps that writing things down or considering special cases like $N=1$ may help... – Raymond Manzoni May 03 '16 at 18:50
Yes. I see that. You explained that above. I was terribly confused with stupid things. Thanks again. – Herman Jaramillo May 03 '16 at 18:55
No problem @HermanJaramillo take it cool this may help too! All the best, – Raymond Manzoni May 03 '16 at 18:57

Derive zeta values of even integers from the Euler-Maclaurin formula.

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