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Analytic continuation of the zeta function

  1. Jan 3, 2012 #1
    I was reading through the first chapter of Edwards' book on the zeta function, and I'm a little confused about Riemann's original continuation of zeta to all of the complex plane... The zeta function is supposed to be defined for all s in the set of complex numbers by

    [tex]\zeta \left( s \right) = \frac{{\Gamma \left( {1 - s} \right)}}{{2\pi i}}\oint\limits_\gamma {\frac{{{{\left( { - u} \right)}^s}}}{{{e^u} - 1}}} \cdot \frac{{du}}{u}[/tex]

    where

    [tex]\begin{array}{l}
    \gamma = {L^ + } \cup {C_\delta } \cup {L^ - } \cup {C_R}\\
    {L^ + } = \left\{ {u = x + i\delta |R \ge x \ge \delta } \right\}\\
    {C_\delta } = \left\{ {u = \delta {e^{i\vartheta }}|0 \le \vartheta < 2\pi } \right\}\\
    {L^ - } = \left\{ {u = x - i\delta |\delta \le x \le R} \right\}\\
    {C_R} = \left\{ {u = R{e^{i\varphi }}|2\pi > \varphi \ge 0} \right\}
    \end{array}[/tex]

    Its easy enough to show that the integrands along both circles are constant and therefore the second and fourth integrals approach zero. Then, as

    [tex]\delta \to 0 \wedge R \to \infty [/tex]

    we have

    [tex]\zeta \left( s \right) = \frac{{\Gamma \left( {1 - s} \right)}}{{2\pi i}}\int\limits_{ + \infty }^{ + \infty } {\frac{{{{\left( { - x} \right)}^s}}}{{{e^x} - 1}}} \cdot \frac{{dx}}{x}[/tex]

    and since the two lines are on either side of the branch cut,

    [tex]\zeta \left( s \right) = \Gamma \left( {1 - s} \right)\frac{{\left( {{e^{i\pi s}} - {e^{ - i\pi s}}} \right)}}{{2i\pi }}\int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{{e^x} - 1}}dx} [/tex]

    [tex]\zeta \left( s \right) = \frac{{\sin \left( {\pi s} \right)}}{\pi }\Gamma \left( {1 - s} \right)\int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{{e^x} - 1}}dx} [/tex]

    Next,

    [tex]\begin{array}{l}
    \zeta \left( s \right) = \frac{{\sin \left( {\pi s} \right)}}{\pi }\Gamma \left( {1 - s} \right)\int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{{e^x} - 1}}dx} \\
    \zeta \left( s \right) = \frac{{\sin \left( {\pi s} \right)}}{\pi }\Gamma \left( {1 - s} \right)\int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{{e^x}}} \cdot \frac{1}{{1 - {e^{ - x}}}}dx} \\
    \end{array}[/tex]

    in which the geometric series converges to the integrand on the interval and also converges to zero absolutely, therefore it is sufficient to use it to carry out the integration and also to interchange the integral and summation signs.

    [tex]\begin{array}{l}
    \zeta \left( s \right) = \frac{{\sin \left( {\pi s} \right)}}{\pi }\Gamma \left( {1 - s} \right)\int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{{e^x}}}\sum\limits_{k = 0}^\infty {{e^{ - kx}}} dx} ,\left| {{e^{ - x}}} \right| < 1 \Rightarrow x > 0\\
    \zeta \left( s \right) = \frac{{\sin \left( {\pi s} \right)}}{\pi }\Gamma \left( {1 - s} \right)\int\limits_0^\infty {{x^{s - 1}}\sum\limits_{k = 0}^\infty {{e^{ - \left( {k + 1} \right)x}}} dx} \\
    \zeta \left( s \right) = \frac{{\sin \left( {\pi s} \right)}}{\pi }\Gamma \left( {1 - s} \right)\sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {k + 1} \right)}^{s - 1}}}}} \int\limits_0^\infty {{{\left[ {\left( {k + 1} \right)x} \right]}^{s - 1}}{e^{ - \left[ {(k + 1)x} \right]}}dx} \end{array}[/tex]

    Then,

    [tex]\begin{array}{l}
    \zeta \left( s \right) = \sum\limits_{k = 0}^\infty {\frac{{\sin \left( {\pi s} \right)}}{\pi }\Gamma \left( {1 - s} \right)\frac{1}{{{{\left( {k + 1} \right)}^{s - 1}}}}} \cdot \frac{{\Gamma (s)}}{{k + 1}}\\
    \zeta \left( s \right) = \sum\limits_{k = 0}^\infty {\frac{{\sin \left( {\pi s} \right)}}{\pi }\Gamma \left( {1 - s} \right)\Gamma (s)\frac{1}{{{{\left( {k + 1} \right)}^s}}}} \\
    \zeta \left( s \right) = \sum\limits_{k = 0}^\infty {\frac{{\sin \left( {\pi s} \right)}}{\pi }\pi \csc \left( {\pi s} \right)\frac{1}{{{{\left( {k + 1} \right)}^s}}}} \\
    \zeta \left( s \right) = \sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {k + 1} \right)}^s}}}} = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^s}}}}
    \end{array}[/tex]

    Riemann's continuation of zeta is supposed to converge to the Dirichlet series for only Re>1. What I don't understand is that I placed no restraint on s, but from the math it would seem that it converges to the Dirichlet series for all s on which the function was defined which would imply that it is not an analytic continuation to the entire complex plane.

    Obviously, I either missed something or I am not understanding something. I was wondering if anyone could help me to understand?

    Thanks in advance
     
    Last edited: Jan 3, 2012
  2. jcsd
  3. Jan 4, 2012 #2
    can anyone help me out?
     
  4. Jan 4, 2012 #3
    nevermind i figured it out. the integral about the circle of radius delta doesn't approach zero for all of s
     
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