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This is a follow-up to a previous question that I deleted because I realised that it was badly posed. Here's a (hopefully) better phrased version.

I need to compute the residue $\text{Res}[F(z),z_0]$ of a function $F(z)$ at a point $z=z_0$, where $F(z)$ is given by

$$ F(z)=\frac{p(z)\Gamma(z-z_0)}{z-z_0} $$

While the Gamma function and therefore $F(z)$ has infinitely many poles $z=z_0-k$, $k=0,1\cdots\infty$, the one at $z=z_0$ makes it a double pole, owing to the $z-z_0$ in the denominator. Most textbooks or notes only deal with simple examples wherein the double pole is due only to a term like $(z-z_0)^2$ in the denominator, in which case the residue evaluation is straightforward using the standard double-pole formula.

In the case above however, I'm tempted to use something along the lines of the standard formula like

$$ \text{Res}[F(z),z_0]=\lim_{z\rightarrow z_0}\Bigl[\frac{\text{d}}{\text{d}z}\frac{(z-z_0)}{\Gamma(z-z_0)}F(z)\Bigr]=\frac{\text{d}p}{\text{d}z}\Big|_{z=z_0} $$

Is this correct? Thanks

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  • $\begingroup$ This is quite confusing. Why do you put $r(z)$ in the denominator and then define it as a reciprocal? $F(z)=p(z)\Gamma(z-z_0)/(z-z_0)$ would be simpler. Also, the poles you list are the poles of the gamma function, and thus zeros of its reciprocal $r$, not poles. Which is also what you need if you want a double pole in $F$. Please clarify this. $\endgroup$
    – joriki
    Commented Mar 20, 2020 at 1:30
  • $\begingroup$ @joriki, very fair observation. I have rephrased my question accordingly. The question remains the same however. $\endgroup$ Commented Mar 20, 2020 at 1:42

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I don’t quite understand what tempts you to write this. It’s not along the lines of the standard formula at all; you’re dividing by the gamma function for no apparent reason.

The standard formula is perhaps best viewed as a bookkeeping device to get the desired coefficient of $z^{-1}$ in the Laurent series of the function. In the present case, since all you have is the gamma function and a polynomial, it’s perhaps easier to work directly with the Laurent series of the gamma function (see How to obtain the Laurent expansion of gamma function around $z=0$?):

$$ \Gamma(z)=\frac1z-\gamma+O(z)\;. $$

Thus

\begin{eqnarray} F(z+z_0) &=& \frac{p(z+z_0)\Gamma(z)}z \\ &=& p(z+z_0)\left(\frac1{z^2}-\frac\gamma z+O(1)\right)\;, \end{eqnarray}

and the residue at $z=0$ can now be read off as

$$ p'(z_0)-p(z_0)\gamma\;. $$

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  • $\begingroup$ Thanks! This really helps, As to why I was dividing by the Gamma function, I was just fishing and trying to remove the singularity at $\Gamma(z-z_0)$ seemed intuitive. $\endgroup$ Commented Mar 20, 2020 at 3:28

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