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Math Help - Gamma function, residue

  1. #1
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    Gamma function, residue

    Show that for m \geq 0, the residue of \Gamma(z) at z = -m is \frac{(-1)^m}{m!}.

    \Gamma(z) is the gamma function. The gamma function is meromorphic. It is defined in the right half-plane by \Gamma(z)= \int_0^{\infty} e^{-t}t^{z-1}dt for \text{Re}(z)>0. There is also another representation of \Gamma(z)=\frac{\Gamma(z+m)}{(z+m-1) \cdots (z+1)z} where the right-hand side is defined and meromorphic for \text{Re}(z)>-m with simple poles at z=0, -1, \ldots, -m+1. However, I still do not see how to prove this. I need help with this. Thank you.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    In general, the residue of f(z) at z=z_0 is the residue of f(z+z_0) at z=0. Now we have \Gamma(z)=(z-1)\dots(z-m)\Gamma(z-m). Therefore what you are looking for is the residue of \frac{\Gamma(z)}{(z-1)\dots(z-m)} at z=0. The denominator is analytic at z=0 and its value is (-1)^mm!; so the solution will be \frac{(-1)^mR}{m!} where R is the residue of \Gamma(z) at z=0... can you finish now?
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