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Riemann zeta function and Bernoulli polynomial

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Riemann zeta function and Bernoulli polynomial

Thu Oct 24, 2013 1:12 am

Posts: 45
Location: China
I don't know whether it's proper to post here, but I real encounter with a problem.
Show that
$\displaystyle \zeta(2n+1)=\frac{(-1)^{n+1}(2\pi)^{2n+1}}{2(2n+1)!}\int_0^{1}B_{2n+1}(x)\cot({\pi}x)dx$
where $\displaystyle B_{2n+1}(x)$ is the Bernoulli polynomial.

This problem can be found, for example, in
http://people.math.sfu.ca/~cbm/aands/page_807.htm
Last edited by Roger209 on Thu Oct 24, 2013 12:30 pm, edited 1 time in total.

Re: Riemann zeta function and Bernoulli polynomial

Thu Oct 24, 2013 10:44 am
galactus
Global Moderator

Posts: 902
I have ran into this one before except the upper limit of integration was 1/2.

I have to go to work now, but if no one responds I believe I can write up a proof later today or tomorrow.

Re: Riemann zeta function and Bernoulli polynomial

Thu Oct 24, 2013 12:54 pm

Posts: 45
Location: China
galactus wrote:
I have ran into this one before except the upper limit of integration was 1/2.

I have to go to work now, but if no one responds I believe I can write up a proof later today or tomorrow.

I want to clear the problem you noted above. In fact, whether the upper limit of the integration is 1 or 1/2 is allowed once you notice the fact that
$\displaystyle \int_0^{1}B_{2n+1}(x)\cot({\pi}x)dx=2\int_0^{1/2}B_{2n+1}(x)\cot({\pi}x)dx$
The equation also can be rewriten as
$\displaystyle \int_0^{1/2}B_{2n+1}(x)\cot({\pi}x)dx=\int_{1/2}^{1}B_{2n+1}(x)\cot({\pi}x)dx$
If we substitute $\displaystyle x$ with $\displaystyle 1-u$,then the right hand of the identity is($\displaystyle I$)
$\displaystyle I=\int_{1/2}^{0}B_{2n+1}(1-u)\cot({\pi}(1-u))d(-u)=-\int_0^{1/2}B_{2n+1}(1-u)\cot({\pi}u)du$
Utilizing the identity
$\displaystyle B_{n}(1-x)=(-1)^{n}B_{n}(x)$
we obtain
$\displaystyle I=-\int_0^{1/2}(-1)^{2n+1}B_{2n+1}(u)\cot({\pi}u)du=\int_0^{1/2}B_{2n+1}(x)\cot({\pi}x)dx$

I hope what I have explained can do some help to you, and I also suggest you to read the remarkable article below(maybe you have already read it):
http://www.sciencedirect.com/science/article/pii/S0377042702003588

Re: Riemann zeta function and Bernoulli polynomial

Sat Oct 26, 2013 10:40 am
galactus
Global Moderator

Posts: 902
There is a useful identity I think we can begin with.

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^{n}}\sum_{k=1}^{n}\int_{a}^{b}\binom{n}{k}p(x)\sin(2\pi kx)dx=\int_{a}^{b}p(x)\cot(\pi x)dx$

Now replace $\displaystyle p(x)=B_{2n+1}$. Where p(x) is a polynomial, so we can replace p(x) with the Bernoulli polynomial.

Note now that $\displaystyle B_{2n+1}(0)=B_{2n+1}(1/2)=0$, so the range 0 to 1/2 works OK. Also, $\displaystyle B_{2n+1}(1)=0$. I only checked the first few Bernoulli polynomials. So, maybe I am in error. But, it looks like a upper limit of 1 or 1/2 gives the same result due to the property of the Bernoulli polys.

Integrate the above formula by parts:

$\displaystyle \int_{0}^{1/2}B_{2N+1}(x)\sin(2\pi kx)dx= -B_{2n+1}(x)\frac{\cos(2\pi kx)}{2\pi k}|_{0}^{1/2}+\frac{2N+1}{2\pi k}\int_{0}^{1/2}B_{2N}(x)\cos(2\pi kx)dx$

The left part of the right side is 0.

Now, integrate by parts again 2N-1 times and it whittles down to:

$\displaystyle I_{k}=\frac{(-1)^{N-1}(2N+1)!}{2(2\pi k)^{2N+1}}$

I think if we sub this into the formula at the top with $\displaystyle p(x)=B_{2n+1}$ it begins to take on the needed form. Play around a little and see if it begins to take shape.

***So, it would appear the reason we have the same formula is due to the fact that the Bernoulli polynomials are 0 rather we use x=1 or x=1/2. For n>0.

Here are the first few odd Bern polys:

$\displaystyle B_{3}(x)=x^{3}-\frac{3}{2}x^{2}+\frac{1}{2}x$

$\displaystyle B_{5}(x)=x^{5}-\frac{5}{2}x^{4}+\frac{5}{3}x^{3}-\frac{1}{6}x$

If we sub in x=1/2 or x=1 we still get 0.

Re: Riemann zeta function and Bernoulli polynomial

Wed Aug 12, 2015 4:39 am

Posts: 70
Here is an approach. Use the power series identity of $\displaystyle \cot x$ which has Bernoulli coefficients
$\displaystyle \sum_k a_k B_{2k} z^{2k-1}$ (just check the exact coefficients) and then use some properties of Bernoulli numbers.