Gamma function of negative argument

Question

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?

Answer 1

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.

Answer 2

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