Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help for zeta functional equation

  1. Sep 23, 2009 #1
    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

  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

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

    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.


  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]?


  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.

  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)?

  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.

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

    thank you very much.

    Ilario M.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook