# Residue of f$f(z)=\Gamma \left(\frac{z+1}{a} \right)$

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.

Thanks

• Maybe duplicate math.stackexchange.com/questions/1757445/… – Nosrati Jan 14 '17 at 21:48
• @MyGlasses Very similar, but the computation of residue is not shown. :( – Lisa Jan 14 '17 at 21:49
• 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. – user243301 Jan 14 '17 at 21:53
• @user243301 I hope this is not something very difficult? – Lisa Jan 14 '17 at 22:00
• @user243301 Bleh, who needs that ;-) – Simply Beautiful Art Jan 14 '17 at 22:56

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

and more generally,

$$(x+a)\dots(x+2)(x+1)(x)\Gamma(x)=\Gamma(x+a+1)$$

Thus, to calculate the residue:

$$(x+a)\Gamma(x)=\frac{\Gamma(x+a+1)}{(x+a-1)\dots(x+2)(x+1)(x)}\to(-1)^a/a!$$

As $x\to-a$.

• Sorry for all the edits, phone plus MathJax is pretty painful. – Simply Beautiful Art Jan 14 '17 at 22:50
• You are right with respect your comment below the question, thanks I've read your answer, +1. – user243301 Jan 14 '17 at 23:23
• @SimpleArt Does your approach work when $a$ is any positive real number? – Lisa Jan 15 '17 at 4:28
• 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? – Lisa Jan 15 '17 at 4:44
• @SimpleArt So, is the answer $a \frac{(-1)^n}{n!}$ ??? – Lisa Jan 15 '17 at 17:38