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Residue challenge

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Find Residue at $z =0 $ of

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

Try to find Residues for $ z=-n $
 
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Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
$ \displaystyle \Gamma(z+1) = \Gamma(1) + \Gamma'(1)x + O(z) = 1 - \gamma z + \mathcal{O} (z^{2})$

$ \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
 

ZaidAlyafey

Well-known member
MHB Math Helper
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 .
 
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