How to find a residue of \begin{align} f(z)=\Gamma \left(\frac{z+1}{a} \right) \end{align} for $a>0$.

I know that the Gamma function has poles for non-positive integers so the polls happen at \begin{align} z_n= -ka-1, \ k=0,1,2,... \end{align}

but no sure how to compute the residue.


  • $\begingroup$ Maybe duplicate math.stackexchange.com/questions/1757445/… $\endgroup$
    – Nosrati
    Jan 14, 2017 at 21:48
  • $\begingroup$ @MyGlasses Very similar, but the computation of residue is not shown. :( $\endgroup$
    – Lisa
    Jan 14, 2017 at 21:49
  • 2
    $\begingroup$ I hope that some user provide you a detailed answer. I believe that you need the same technique showed in Section 3 of a lecture notes from Cornell University searching in Google residues of the gamma function. After you can do a comparison with the output of the online calculator of Wolfram Alpha, when you type the input residues Gamma((z+1)/a). Good luck. $\endgroup$
    – user243301
    Jan 14, 2017 at 21:53
  • $\begingroup$ @user243301 I hope this is not something very difficult? $\endgroup$
    – Lisa
    Jan 14, 2017 at 22:00
  • 1
    $\begingroup$ @user243301 Bleh, who needs that ;-) $\endgroup$ Jan 14, 2017 at 22:56

1 Answer 1


Hint: $$x\Gamma(x)=\Gamma(x+1)$$

and more generally,


Thus, to calculate the residue:


As $x\to-a$.

  • $\begingroup$ Sorry for all the edits, phone plus MathJax is pretty painful. $\endgroup$ Jan 14, 2017 at 22:50
  • $\begingroup$ You are right with respect your comment below the question, thanks I've read your answer, +1. $\endgroup$
    – user243301
    Jan 14, 2017 at 23:23
  • $\begingroup$ @SimpleArt Does your approach work when $a$ is any positive real number? $\endgroup$
    – Lisa
    Jan 15, 2017 at 4:28
  • $\begingroup$ So how do you use $x \Gamma(x)=\Gamma(x+1)$ on $\Gamma \left( \frac{z+1}{a} \right)$? How would the final answer look? $\endgroup$
    – Lisa
    Jan 15, 2017 at 4:44
  • 1
    $\begingroup$ @SimpleArt So, is the answer $ a \frac{(-1)^n}{n!}$ ??? $\endgroup$
    – Lisa
    Jan 15, 2017 at 17:38

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