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Help for zeta functional equation

  1. Sep 23, 2009 #1
    hi,
    i'm studying the functional equation of riemann zeta function for Re(s)>1;
    my book(complex analysis by T. Gamelin) use contour integral in the proof, where the contour is taken on the usual 3 curves (real axis and a small circle [tex]C\epsilon[/tex] around the origin). i'm not able to figure why the integral on the circle vanish as epsilon->0; the text report:

    since [tex]e^{z -1}[/tex] has a simple zero at z=0, the integrand is bounded on the circle |z|=r by C [tex]\epsilon^{re(s)-2}[/tex]

    wich is the estimate that the author use in this assertion?
    i'm new to complex analysis and i want to say (if possible) what argument i've got to study
    thanks

    I.M.
     
  2. jcsd
  3. Sep 23, 2009 #2
    This appears to be a typo for [itex]{e^z}[/itex] - 1.

    What's the integrand?

     
  4. Sep 23, 2009 #3
    the integral that vanish is


    [tex]\frac{1}{2\pi i}\oint{\frac{(-z)^{s-1}}{e^{z}-1}dz}[/tex]


    taken on the circle |z|=epsilon as epsilon->0


    a little snapshot from Complex analysis by T. Gamelin:

    http://ilario.mazzei.googlepages.com/Immagine.GIF [Broken]
     
    Last edited by a moderator: May 4, 2017
  5. Sep 23, 2009 #4
    OK, let's concentrate on estimating the integral

    [tex]
    \frac{1}{2\pi i}\oint{\frac{(-z)^{s-1}}{e^{z}-1}dz}
    [/tex]

    We want to show that the integrand is bounded by [itex]C\epsilon^{Re(s) - 2}[/itex], where C is a constant that doesn't depend on [itex]\epsilon[/itex].

    The usual strategy for finding a bound on a fraction is to find an upper bound on the numerator and a lower bound on the denominator. I'll sketch the arguments, leaving it to you to fill in the details. Let me know if you want more details. I'll write [itex]s = \sigma + it[/itex]. First we consider the numerator:

    [tex]|(-z)^{s-1}| = |exp((s - 1)log (-z))|

    = |exp((s - 1) (log |z| + i arg (-z)))|

    = |exp((\sigma - 1) + it)(log |z| + i arg (-z))|[/tex]

    Now multiply out to show that this last expression equals

    [tex]|z}|^{\sigma-1} exp(-t arg(z))[/tex]

    and conclude that this expression is

    [tex]\leq C_1\epsilon^{\sigma-1}[/tex]

    for some constant [itex]C_1[/itex] independent of [itex]\epsilon[/itex].

    Next we consider the denominator [itex]e^{z} - 1[/itex]:

    Write [itex]f(z) = e^{z} - 1[/itex]. Since f(z) has a simple zero at z = 0, there exists a function g(z) such that

    [tex]f(z) = zg(z)[/tex]

    such that g is analytic and non-zero at z = 0. In fact, g is non-zero on some neighborhood of z = 0, which implies that g is nonzero on the circle |z| = [itex]\epsilon[/itex] for [itex]\epsilon[/itex] sufficiently small. (This is a standard result in complex analysis and should be in your book.)

    Now use standard theorems about continuous functions on compact sets to conclude that there exists a constant [itex]C_2[/itex] such that g(z) > [itex]C_2[/itex] on |z| = [itex]\epsilon[/itex]. Therefore

    [tex]|e^{z} - 1| > C_2|z|[/tex]

    Finally, combine these inequalities to conclude that

    [tex]\frac{|-z|^{s-1}}{|e^{z} - 1|} \leq C\epsilon^{s-2}[/tex]

    for some constant C.

    HTH

    Petek
     
  6. Sep 23, 2009 #5
    hi petek,
    thanks for your reply.
    i have some questions:

    why we can't use the ML estimate? (since L->0 as espilon->0 )


    i've multiplied but at the end i have:

    [tex]|z}|^{\sigma-1} exp(-t arg(-z)) exp(i (\sigma-1)arg(-z) + it log|z| )[/tex]


    what i'm missing?



    about the inequalities:


    [tex]\frac{|-z|^{s-1}}{|e^{z} - 1|} \leq C\epsilon^{s-2}[/tex]


    why this yields that the integral is less than [tex]C\epsilon^{s-1}[/tex]?



    thanks


    I.M.
     
  7. Sep 24, 2009 #6

    We will. See my reply to your last question.



    The above expression should be enclosed in "absolute value" marks. Then note that if x is a real number,

    [tex]|exp(ix)| = 1.[/tex]


    Use the right side of this inequality as the M in the ML estimate. As you noted, L goes to zero.

    Please let me know if my replies are unclear or if you have any other questions.

    Petek
     
  8. Sep 24, 2009 #7
    hi Petek,
    thank you very much for your help - very clear.

    I have one more question:

    if epsilon->0 the curve reduce to a point, and an every integral evaulated on a single point must be 0 since dz=0: this condition is not enough to proof that this curve integral is 0 (even if Re(s)<1)?


    I.M.
     
  9. Sep 25, 2009 #8
    I'm not sure that I understand your argument. However, in general, there's a difference between epsilon approaching zero and letting epsilon equal zero.

    Petek
     
  10. Sep 25, 2009 #9
    the fabulous world of discontinuity :))

    thank you very much.


    Ilario M.
     
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