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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

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    Maybe duplicate http://math.stackexchange.com/questions/1757445/sum-of-gamma-function-residues?rq=1 – 2017-01-14
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    @MyGlasses Very similar, but the computation of residue is not shown. :( – 2017-01-14
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    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. – 2017-01-14
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    @user243301 I hope this is not something very difficult? – 2017-01-14
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    @user243301 Bleh, who needs that ;-) – 2017-01-14
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    $\Gamma(z)\Gamma(1-z)\sin(\pi z) = \pi$. – 2017-01-15

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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 $x\to-a$.

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    Sorry for all the edits, phone plus MathJax is pretty painful. – 2017-01-14
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    You are right with respect your comment below the question, thanks I've read your answer, +1. – 2017-01-14
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    @SimpleArt Does your approach work when $a$ is any positive real number? – 2017-01-15
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    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? – 2017-01-15
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    @Lisa the Gamma function only has poles on the negative integers, and for yours, you have to evaluate the limit:$$(z+1)\Gamma((z+1)/a$$so let $s=(z+1)/a$. And it follows from there. – 2017-01-15
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    @SimpleArt So, is the answer $ a \frac{(-1)^n}{n!}$ ??? – 2017-01-15
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    @Lisa Yup :-) Good job! – 2017-01-15