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A trick to proof RH using asymptotic Analysis

  1. Jun 5, 2006 #1
    We have using Perron,s formula an integral for M(x) Mertens function:

    [tex] \int_{c-i\infty}^{c+i\infty}ds\frac{x^{s}}{\zeta(s) s}= 2i\pi M(x) [/tex]

    Of course we can,t calculate this integral directly, let,s put x=exp(t) and s=c+iu then our integral becomes:

    [tex] \int_{-\infty}^{\infty}du \frac{e^{iut+ict}}{\zeta(c+iu)(c+iu)} [/tex]

    now we could writte the complex integrand involving \zeta(c+iu)(c+iu) as a sum of a real and complex part U(c,u)+iV(c,u), so we should make an approach for big t--->oo of the integral:

    [tex] \int_{-\infty}^{\infty}duU(c,u)e^{iut+ict} [/tex]

    so we can calculate M(e^{t}) of course RH is true iff for big t the factor:

    [tex] M(e^{t})e^{-t(1/2+e)} [/tex] 0<e<<<<<<1 (epsilon)

    tends to 0, so we could give a "proof" of Riemann Hypothesis.
  2. jcsd
  3. Jun 6, 2006 #2
    Or simply solve Perron,s integral for [tex] 1/\zeta(s) [/tex] so

    [tex] M(e^{t})= -\sum_{n=1}^{\infty}e^{-nt}/d\zeta(-2n)(2n)+1/d\zeta(0)+\frac{1}{2i\pi}\oint_{D}dse^{st}/\zeta(s)s [/tex]

    for this last complex integral we could use the contour D given by the rectangle {0-iT,0+iT,1-iT,1+iT} with T---->oo and use "steepest descend " method to calculate the asymptotic evaluation for the integral.

    where df means df/dt or df/dx
    Last edited: Jun 6, 2006
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