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Math Help - Riemann Zeta Function

  1. #1
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    Riemann Zeta Function

    For  1<s< \infty define  \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} . Prove that  \zeta(s) = s \int_{1}^{\infty} \frac{[x]}{x^{s+1}} \ dx . So consider an interval  [1,N] . Let  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  g(s) = 0 . Then we take  N \to \infty and conclude that  g(s) = 0 on  [1, \infty) . The second series is convergent.
    Last edited by manjohn12; April 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  1<s< \infty define  \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} . Prove that  \zeta(s) = s \int_{1}^{\infty} \frac{[x]}{x^{s+1}} \ dx . So consider an interval  [1,N] . Let  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  g(s) = 0 . Then we take  N \to \infty and conclude that  g(s) = 0 on  [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 \lim_{N\to \infty} g(s)=0 may be enough (because both terms of the difference converge (*))

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

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

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

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

    Now :
    \begin{aligned}<br />
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} \\<br />
&=-\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)} \\<br />
&=-\frac{1}{N^s}+\sum_{k=1}^{N-1} \frac{k-1}{k^s}-\frac{k}{(k+1)^s}<br />
\end{aligned}

    But this is a telescoping sum !
    So we have :
    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 s>1)



    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    (*)
    \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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  3. #3
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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 \lim_{N\to \infty} g(s)=0 may be enough (because both terms of the difference converge (*))

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

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

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

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

    Now :
    \begin{aligned}<br />
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} \\<br />
&=-\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)} \\<br />
&=-\frac{1}{N^s}+\sum_{k=1}^{N-1} \frac{k-1}{k^s}-\frac{k}{(k+1)^s}<br />
\end{aligned}

    But this is a telescoping sum !
    So we have :
    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 s>1)



    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    (*)
    \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  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  g(s) not necessarily equal to 0?
    Because you want to prove that \lim_{n \to \infty} I(s)=\zeta(s)=\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k^s}

    That is to say :
    \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 :
    \lim_{n \to \infty} \left[I(s)-\sum_{k=1}^n \frac{1}{k^s}\right]=0

    \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...

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

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



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

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

    According to (1) and remembering that...

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

    ... we obtain...

    \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...

    \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}}\} =

    = \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  Re(s)>1 ...

    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)

    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

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