- Thread starter
- #1

- Jan 17, 2013

- 1,667

Find Residue at $z =0 $ of

Try to find Residues for $ z=-n $

\(\displaystyle f(z) = \Gamma(z) \Gamma(z-1) x^{-z}\)

Try to find Residues for $ z=-n $

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- Thread starter ZaidAlyafey
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- Thread starter
- #1

- Jan 17, 2013

- 1,667

\(\displaystyle f(z) = \Gamma(z) \Gamma(z-1) x^{-z}\)

Try to find Residues for $ z=-n $

Last edited:

- Jan 31, 2012

- 253

$ \displaystyle \implies \Gamma(z) = \frac{1}{z} \Gamma(z+1) = \frac{1}{z} - \gamma + \mathcal{O} (z) $

$ \displaystyle \implies \Gamma(z-1) = \frac{\Gamma(z)}{z-1} = - \frac{1}{z} + (\gamma -1) + \mathcal{O}(z) $

$ \displaystyle \implies \Gamma(z) \Gamma(z-1) = - \frac{1}{z^{2}} + \frac{2 \gamma -1}{z} + \mathcal{O}(1) $

So $\displaystyle \text{Res} [ \Gamma(z) \Gamma(z-1) x^{-z},0] = \text{Res} \Big[ - \frac{1}{z^{2}} x^{-z},0 \Big] + \text{Res} \Big[ \frac{2 \gamma-1}{z} x^{-z}, 0 \Big] $

$ \displaystyle = - \lim_{z \to 0} \frac{d}{dz} x^{-z} + \lim_{z \to 0} \ (2 \gamma -1) x^{-z} = \ln(x) + 2 \gamma -1 $

which seems to be correct for different values of $x$ according to Maple

- Thread starter
- #3

- Jan 17, 2013

- 1,667

Nice solution , another way is shift the factorial then differentiate

\(\displaystyle \Gamma(z) \Gamma(z-1) x^{-z}= \frac{\Gamma(z+1)^2}{z^2(z-1)} x^{-z}\)

So we have a second order pole at $z=0$ so differentiating \(\displaystyle \frac{\Gamma(z+1)^2}{(z-1)} x^{-z}\) once will work and taking the limit to zero will work .

\(\displaystyle \Gamma(z) \Gamma(z-1) x^{-z}= \frac{\Gamma(z+1)^2}{z^2(z-1)} x^{-z}\)

So we have a second order pole at $z=0$ so differentiating \(\displaystyle \frac{\Gamma(z+1)^2}{(z-1)} x^{-z}\) once will work and taking the limit to zero will work .

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