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Thread: Riemann Zeta Function

  1. #1
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    Riemann Zeta Function

    For $\displaystyle 1<s< \infty $ define $\displaystyle \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} $. Prove that $\displaystyle \zeta(s) = s \int_{1}^{\infty} \frac{[x]}{x^{s+1}} \ dx $. So consider an interval $\displaystyle [1,N] $. Let $\displaystyle g(s) = s \int_{1}^{N} \frac{[x]}{x^{s+1}} \ dx - \sum_{n=1}^{N} \frac{1}{n^s} $. Now we want to show that $\displaystyle g(s) = 0 $. Then we take $\displaystyle N \to \infty $ and conclude that $\displaystyle g(s) = 0 $ on $\displaystyle [1, \infty) $. The second series is convergent.
    Last edited by manjohn12; Apr 24th 2009 at 01:42 PM.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by manjohn12 View Post
    For $\displaystyle 1<s< \infty $ define $\displaystyle \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} $. Prove that $\displaystyle \zeta(s) = s \int_{1}^{\infty} \frac{[x]}{x^{s+1}} \ dx $. So consider an interval $\displaystyle [1,N] $. Let $\displaystyle g(s) = s \int_{1}^{N} \frac{[x]}{x^{s+1}} \ dx - \sum_{n=1}^{N} \frac{1}{n^s} $. Now we want to show that $\displaystyle g(s) = 0 $. Then we take $\displaystyle N \to \infty $ and conclude that $\displaystyle g(s) = 0 $ on $\displaystyle [1, \infty) $. The second series is convergent.
    Ha ! I think I found it \o/

    But no, g(s) is not necessarily equal to 0. I'm not too sure (I hate those limit/convergence theorems), but proving that $\displaystyle \lim_{N\to \infty} g(s)=0$ may be enough (because both terms of the difference converge (*))

    Consider the integral $\displaystyle I(s)=s \int_1^N \frac{\lfloor x \rfloor}{x^{s+1}} ~dx$
    Then divide the integral on intervals which boundaries are integers :

    $\displaystyle I(s)=s \sum_{k=1}^{N-1} \int_k^{k+1} \frac{\lfloor x \rfloor}{x^{s+1}} ~dx$

    But $\displaystyle x \in [k,k+1) \Rightarrow \lfloor x\rfloor =k$

    So :
    $\displaystyle \begin{aligned}
    I(s)
    &=\sum_{k=1}^{N-1} sk \int_k^{k+1} \frac{1}{x^{s+1}} ~dx \\
    &=\sum_{k=1}^{N-1} sk \cdot \left. \frac{1}{x^s \cdot (-s)}\right|_k^{k+1} \\
    &=\sum_{k=1}^{N-1} k \cdot \left(\frac{1}{k^s}-\frac{1}{(k+1)^s}\right)
    \end{aligned}$

    Now :
    $\displaystyle \begin{aligned}
    g(s) &=\sum_{k=1}^{N-1} k \cdot \left(\frac{1}{k^s}-\frac{1}{(k+1)^s}\right)-\sum_{k=1}^N \frac{1}{k^s} \\
    &=-\frac{1}{N^s}+\sum_{k=1}^{N-1} k \cdot \left(\frac{1}{k^s}-\frac{1}{(k+1)^s}\right)-\frac{1}{k^s} \quad \text{(because these are finite sums)} \\
    &=-\frac{1}{N^s}+\sum_{k=1}^{N-1} \frac{k-1}{k^s}-\frac{k}{(k+1)^s}
    \end{aligned}$

    But this is a telescoping sum !
    So we have :
    $\displaystyle g(s)=-\frac{1}{N^s}+\frac{1-1}{1^s}-\frac{N-1}{N^s}=-\frac{N}{N^s}$

    And this obviously goes to 0 as N goes to infinity (because $\displaystyle s>1$)



    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    (*)
    $\displaystyle \int_1^\infty \frac{\lfloor x \rfloor}{x^{s+1}} ~dx \leq \int_1^\infty \frac{x+1}{x^{s+1}} ~dx=\int_1^\infty \frac{1}{x^s}+\frac{1}{x^{s+1}} ~dx$
    and this converges
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    Quote Originally Posted by Moo View Post
    Hello,

    Ha ! I think I found it \o/

    But no, g(s) is not necessarily equal to 0. I'm not too sure (I hate those limit/convergence theorems), but proving that $\displaystyle \lim_{N\to \infty} g(s)=0$ may be enough (because both terms of the difference converge (*))

    Consider the integral $\displaystyle I(s)=s \int_1^N \frac{\lfloor x \rfloor}{x^{s+1}} ~dx$
    Then divide the integral on intervals which boundaries are integers :

    $\displaystyle I(s)=s \sum_{k=1}^{N-1} \int_k^{k+1} \frac{\lfloor x \rfloor}{x^{s+1}} ~dx$

    But $\displaystyle x \in [k,k+1) \Rightarrow \lfloor x\rfloor =k$

    So :
    $\displaystyle \begin{aligned}
    I(s)
    &=\sum_{k=1}^{N-1} sk \int_k^{k+1} \frac{1}{x^{s+1}} ~dx \\
    &=\sum_{k=1}^{N-1} sk \cdot \left. \frac{1}{x^s \cdot (-s)}\right|_k^{k+1} \\
    &=\sum_{k=1}^{N-1} k \cdot \left(\frac{1}{k^s}-\frac{1}{(k+1)^s}\right)
    \end{aligned}$

    Now :
    $\displaystyle \begin{aligned}
    g(s) &=\sum_{k=1}^{N-1} k \cdot \left(\frac{1}{k^s}-\frac{1}{(k+1)^s}\right)-\sum_{k=1}^N \frac{1}{k^s} \\
    &=-\frac{1}{N^s}+\sum_{k=1}^{N-1} k \cdot \left(\frac{1}{k^s}-\frac{1}{(k+1)^s}\right)-\frac{1}{k^s} \quad \text{(because these are finite sums)} \\
    &=-\frac{1}{N^s}+\sum_{k=1}^{N-1} \frac{k-1}{k^s}-\frac{k}{(k+1)^s}
    \end{aligned}$

    But this is a telescoping sum !
    So we have :
    $\displaystyle g(s)=-\frac{1}{N^s}+\frac{1-1}{1^s}-\frac{N-1}{N^s}=-\frac{N}{N^s}$

    And this obviously goes to 0 as N goes to infinity (because $\displaystyle s>1$)



    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    (*)
    $\displaystyle \int_1^\infty \frac{\lfloor x \rfloor}{x^{s+1}} ~dx \leq \int_1^\infty \frac{x+1}{x^{s+1}} ~dx=\int_1^\infty \frac{1}{x^s}+\frac{1}{x^{s+1}} ~dx$
    and this converges
    Why is $\displaystyle g(s) $ not necessarily equal to 0?
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  4. #4
    Moo
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    Quote Originally Posted by manjohn12 View Post
    Why is $\displaystyle g(s) $ not necessarily equal to 0?
    Because you want to prove that $\displaystyle \lim_{n \to \infty} I(s)=\zeta(s)=\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k^s}$

    That is to say :
    $\displaystyle \lim_{n \to \infty} I(s)-\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k^s}=0$

    Since the two converge, that's :
    $\displaystyle \lim_{n \to \infty} \left[I(s)-\sum_{k=1}^n \frac{1}{k^s}\right]=0$

    $\displaystyle \lim_{n \to \infty} g(s)=0$

    That's the limit as n goes to infinity, not the value for any n.
    You can check it if you take n=1. It's not equal to 0.
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  5. #5
    MHF Contributor chisigma's Avatar
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    According to the defintion reported in...

    Floor Function -- from Wolfram MathWorld

    ... the 'fractional part of x' is defined as...

    $\displaystyle \{x\}= x - \lfloor x \rfloor$ (1)

    ... and it is the function you see in figure...



    In the interval $\displaystyle n\le x< n+1$ is...

    $\displaystyle \{x\}= x-n$ (2)

    According to (1) and remembering that...

    $\displaystyle \int \frac{dx}{x^{s}}= -\frac{1}{(s-1)\cdot x^{s-1}} + c$ (3)

    ... we obtain...

    $\displaystyle \int_{n}^{n+1} \frac{\lfloor x \rfloor}{x^{s+1}}\cdot dx= \frac{n}{s}\cdot \{\frac{1}{n^{s}}-\frac{1}{(n+1)^{s}}\}$ (4)

    ... so that...

    $\displaystyle \int_{1}^{n} \frac{\lfloor x \rfloor}{x^{s+1}}\cdot dx= \frac{1}{s}\cdot (\frac{1}{1^{s}}- \frac{1}{2^{s}}) + \frac{2}{s}\cdot (\frac{1}{2^{s}}- \frac{1}{3^{s}}) +\dots + \frac{n-1}{s}\cdot \{\frac{1}{(n-1)^{s}}- \frac{1}{n^{s}}\} = $

    $\displaystyle = \frac{1}{s}\cdot \{\frac{1}{1^{s}} + \frac{1}{2^{s}} + \dots + \frac{1}{(n-1)^{s}} - \frac{n-1}{n^{s}}\}$ (5)

    From (5) it derives that for $\displaystyle Re(s)>1$ ...

    $\displaystyle s\cdot \int_{1}^{\infty} \frac{\lfloor x \rfloor}{x^{s+1}}\cdot dx= \sum_{n=1}^{\infty}\frac{1}{n^{s}} = \zeta(s) $ (6)

    $\displaystyle g(s)= s\cdot \int_{1}^{n} \frac{\lfloor x \rfloor}{x^{s+1}}\cdot dx - \sum_{k=1}^{n}\frac{1}{k^{s}} = - \frac{n}{n^{s}}$ (7)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; Apr 27th 2009 at 05:25 PM. Reason: error in (7)
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