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Gamma of negative integers

  1. Jan 21, 2008 #1

    I know the Gamma function has a first order pole at the negative integers, and the residue at -k is (-1)^k/k!

    So now I want to compute Gamma(-k)/Gamma(-l) , k and l positive integers, and I have a feeling it should be [ (-1)^k/k! ] / [ (-1)^l/l! ] =(-1)^{k+l} l!/k!

    What would be an easy way to prove that...?

    I thought about considering Gamma(x)/Gamma(x-l+k) and then doing a Laurent expansion of the numerator and denominator around x=-k

    This would be


    However this doesnt seem to make much sense being divergent for x->-k ...
    Any tips?:smile:

  2. jcsd
  3. Jan 22, 2008 #2


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    Science Advisor

    Having said that neither [itex]\Gamma(-l)[/itex], not [itex]\Gamma(-k)[/itex] exist, you have already said that [itex]\Gamma(-l)/\Gamma(-k)[/itex] does not exist. Apparently you really mean [itex]\lim_{x\rightarrow -l}\Gamma(x)/\Gamma(x+l-k)[/itex]. If that limit existed, then Gamma would be analytic over all real numbers, and it is not.
  4. Jan 22, 2008 #3
    Thanks for your reply. Yes, I mean the limit, of course:wink:

    How does the existence of this limit imply analyticity of Gamma over the real numbers?

    Isn't it true that if g und f have first order poles at k, then [tex]\lim_{x
    \to k}\frac{f(x)}{g(x)}=

    I am not yet convinced ..

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