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Is there any relation between the limiting behaviour of $\Gamma({\epsilon})$ and $\Gamma(-1+{\epsilon})$? I have seen the relation such as $\Gamma(-1+{\epsilon})$ $=$ $\Gamma({\epsilon})/(-1+{\epsilon})$. I think it is basically wrong? But does there exist such a similar relation?

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In a small treatize on the Eulerian numbers I tried to make sense to the gamma-function at zero and negative integers including the aspect of epsilon-range deviations around the integer arguments at which the singularities occur. Perhaps this is giving some ideas to you.... see pg 8 ff – Gottfried Helms Dec 22 '12 at 15:52
You may appreciate the discussion in this 'Limits defined for negative factorials' thread. – Raymond Manzoni Dec 22 '12 at 16:28

The relation $\Gamma(-1+\epsilon) = \Gamma({\epsilon})/(-1+{\epsilon})$ is true so long as $\epsilon$ is not a negative integer (so that $-1+\epsilon$ will then also not be a negative integer) since the gamma function is extended to the complex plane minus the negative integers by using the relation $\Gamma(z)=\Gamma(z+1)/z$ or by using analytic continuation.

Thus, you can say something about the limiting behaviour of $\Gamma(\epsilon)$ and $\Gamma(-1+\epsilon)$, in that you can say that

$$\lim_{\epsilon\to 0} \frac{\Gamma(-1+\epsilon)}{\Gamma(\epsilon)} = \lim_{\epsilon\to 0} \frac{1}{-1+\epsilon} = -1.$$

Note that the fact that $\Gamma(z)$ is not defined at $-1$ does not affect this, since for the limit, we are only interested in the values of the function close to $-1$.

In other words, $|\Gamma(z)\vert$ tends to infinity "at the same rate" as $z\to 0$ or as $z\to -1$, and similar results could be proved at any negative integer.

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Well put. (+1) Computations of this sort often show up in dimensional regularization. – user26872 Dec 22 '12 at 21:22

Your relation /would/ hold if $\Gamma$ were continuous at $-1$. It is not, however: intuitively, we cannot take the factorial of negative integers.

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I think there are other continuations of the factorial function which are defined for negative integers. The Gamma function is a particular one which is not. I'll try and find the link to the various extensions of $n!$ that are out there. – Pedro Tamaroff Dec 22 '12 at 20:53
@Peter, were you thinking of this? – J. M. Apr 3 '13 at 7:30
@JM Yes!! Thank you. – Pedro Tamaroff Apr 3 '13 at 13:33

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