Aug 17, 2013 Thread starter #1 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 $ Last edited: Aug 17, 2013
Find Residue at $z =0 $ of \(\displaystyle f(z) = \Gamma(z) \Gamma(z-1) x^{-z}\) Try to find Residues for $ z=-n $
Aug 17, 2013 #2 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
$ \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
Aug 17, 2013 Thread starter #3 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 . Last edited: Aug 17, 2013
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 .