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Gamma function convergence

  1. Dec 22, 2008 #1
    [tex]
    \Gamma(z) = \int\limits_0^{\infty} t^{z-1} e^{-t} dt
    [/tex]

    I can see that if [itex]\textrm{Re}(z)>0[/itex], then the integral converges, and that if [itex]\textrm{Re}(z)\leq 0[/itex] and [itex]\textrm{Im}(z)=0[/itex], then it diverges. However, I found the case [itex]\textrm{Re}(z)\leq 0[/itex] and [itex]\textrm{Im}(z)\neq 0[/itex] more difficult.

    [tex]
    t^{z-1} = t^{x-1}\big(\cos(y\log(t)) + i\sin(y\log(t))\big),\quad\quad z=x+iy
    [/tex]

    Clearly the positive and negative parts (of real and imaginary parts) are not integrable alone, but it could be, assuming that we don't yet know the right answer, that the rapid oscillation would make the integral

    [tex]
    \lim_{\delta\to 0^+} \int\limits_{\delta}^{\infty} t^{z-1} e^{-t} dt
    [/tex]

    convergent. So, indeed, is it convergent or not? Anyone knowing proofs?

    edit: Heuristically speaking, when [itex]t\to 0^+[/itex], then [itex]\log(t)\to-\infty[/itex] rather slowly, so my guess is that the oscillations are going to be too slow for convergence.

    edit edit: hmhmh... or then I could approximate [itex]e^{-t}=1+O(t)[/itex], and actually integrate [itex]t^{z-1}[/itex] with formula [itex]t^{z-1} = D_t \frac{1}{z}t^z[/itex], and see that it diverges...

    Thanks for the attention, and I hope you enjoyed the entertainment.
     
    Last edited: Dec 22, 2008
  2. jcsd
  3. Dec 22, 2008 #2
    You must use the analytic continuation in re z < 0.
     
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