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I'm currently dealing with a problem my professor raised (since I just studied the Mittag-Leffler's Partial Fractions Theorem). The problem is to derive a partial fractions decomposition of the Gamma Function that displays its (simple) poles and principal parts.

So in my textbook, the Gamma Function is defined as the following limit:

$$\lim_{n \to \infty}\frac{n!\ n^z}{z(z+1)...(z+n)}$$

where $z$ is an arbitrary complex number that is not $0, -1, -2,...$

With the residue of each pole $z_v=-v,\ v=0,1,2,...$ as

$$a_{-1}=\frac{(-1)^v}{v!}$$

We have the principal parts of the Gamma Function at each pole $z_v$ as

$$h_v(z)=\frac{(-1)^v}{v!\ (z+v)}$$

And so it is sufficient to show that

$$\sum_{v=1}^{\infty} \left[h_v(z)-g_v(z) \right]$$ is uniformly convergent on $|z|\leq R,\ R>0$ , where $g_v(z)$ is defined to be the first few finite terms of the series expansion of $h_v(z)$ around $z=0$.

By considering $v$ large enough so that $|z_v|=v>2R$ and taking $g_v(z)=0$, we have that for $|z|\leq R$

$$\left|\frac{(-1)^v}{v!\ (z+v)} \right|=\frac{1}{v!\ |z+v|}<\frac{2}{v!\ v}$$

So the series above converges uniformly and hence there exists an entire function $G(z)$ such that

$$\Gamma(z) = G(z)+\frac{1}{z} + \sum_{v=1}^{\infty}\frac{(-1)^v}{v!\ (z+v)}$$

And now I am stuck at finding $G(z)$. Anyone can provide some help?

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    $\begingroup$ You can identify your function from dlmf.nist.gov/5.9#E4, but I have no clue how to give a proof with your lim/product definition of $\Gamma$. With the integral definition you can find a proof in Lebedev's Special functions Ch.1.1 $\endgroup$ Jun 11 '14 at 14:04
  • $\begingroup$ Hmm I just need a clue about the function $G(z)$... $\endgroup$ Jun 11 '14 at 14:27
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    $\begingroup$ Maybe it is a bit late: In your notation the NIST/Lebedev formula is $$\Gamma(z)=\int_{1}^{\infty}t^{z-1}e^{-t}dt+\sum_{v=0}^{\infty}\frac{(-1)^{v}}{(z+v)v!}$$ and $G(z) = \int_{1}^{\infty}t^{z-1}e^{-t}dt$ is the function you seek. According to Lebedev (who refers to Titchmarsh, The Theory of Functions) the function $G(z)$ is entire. $\endgroup$ Jun 12 '14 at 6:52

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