Compute Residue of $\Gamma(z)$ at Negative Integers & 0

[Dustinsfl]

Compute the residue of $\Gamma(z)$ at each of its poles.

So the poles are at the negative integers and 0. I suspect there must be a formula than since this is an infinite set.

$\Gamma(z) = \dfrac{e^{-\gamma z}}{z}\prod\limits_{n=1}^{\infty}\left(1+\dfrac{z}{n}\right)^{-1}e^{z/n}$

Should I start by logarithmically differentiating?

 

[Jose27]

Just a thought (I don't know if it'll work), but maybe using the $\Gamma$ function's integral representation (remember we have to split the integral several times in order to extend it past a certain integer) and using Fubini's theorem. Something along these lines.

 

[Dustinsfl]

Just a thought (I don't know if it'll work), but maybe using the $\Gamma$ function's integral representation (remember we have to split the integral several times in order to extend it past a certain integer) and using Fubini's theorem. Something along these lines.

$\displaystyle\Gamma(z) =\int_0^{\infty}e^{-t}t^z\dfrac{dt}{t}$

What should I do with this?

---------- Post added at 21:29 ---------- Previous post was at 20:37 ----------

The $\text{Res} = \lim\limits_{z\to -m}(z+m)\Gamma(z)$ and we can write $\Gamma(z) = \dfrac{\Gamma(z+m)}{(z+m-1)(z+m-2)\cdots (z+1)z}$ for the $\text{Re} z> -m$.

By substitution, we have
$$
\text{Res} = \lim_{z\to -m}(z+m)\Gamma(z) = \lim_{z\to -m}(z+m)\frac{\Gamma(z+m)}{(z+m-1)(z+m-2)\cdots (z+1)z}.
$$

By definition, we have $(z+m)\Gamma(z+m) = \Gamma(z+m+1)$.

So
$$
\text{Res} = \lim_{z\to -m}\frac{\Gamma(z+m+1)}{(z+m-1)(z+m-2)\cdots (z+1)z} = \frac{\Gamma(1) = 1}{(-1)(-2)\cdots (-m)} = \frac{(-1)^m}{m!}.
$$

 

[Opalg]

$$
\text{Res} = \ldots = \frac{(-1)^m}{m!}.
$$

That is correct. The functional equation $\Gamma(z+1) = z\Gamma(z)$ is the key to the calculation.

 

[blue_raver22]

Yes, starting with logarithmically differentiating would be a good approach. Let's begin by writing the logarithmically differentiated form of $\Gamma(z)$:

$\dfrac{\Gamma'(z)}{\Gamma(z)} = -\gamma + \dfrac{1}{z} - \sum\limits_{n=1}^{\infty}\left(\dfrac{1}{z+n}+\dfrac{1}{n}\right)$

Next, we can use the fact that the residue of a function at a pole $z_0$ is given by the coefficient of $(z-z_0)^{-1}$ in the Laurent series expansion of the function around that pole. In this case, we can see that the only poles of $\Gamma(z)$ are at the negative integers and 0, so we can expand the above expression around each of these poles to find their respective residues.

For the pole at $z=0$, we have:

$\dfrac{\Gamma'(z)}{\Gamma(z)} = -\gamma + \dfrac{1}{z} - \sum\limits_{n=1}^{\infty}\left(\dfrac{1}{z+n}+\dfrac{1}{n}\right) = -\gamma + \dfrac{1}{z} + \sum\limits_{n=1}^{\infty}\left(\dfrac{1}{n}-\dfrac{1}{z+n}\right)$

We can see that the coefficient of $(z-0)^{-1}$ is simply 1, so the residue at $z=0$ is 1.

For the poles at the negative integers, we can use the fact that the Laurent series expansion of $\Gamma(z)$ around a pole $z=-n$ is given by:

$\Gamma(z) = \dfrac{(-1)^n}{n!}\left(\dfrac{1}{z+n}+\dfrac{1}{n}\right)^{-1}e^{z/n}$

Therefore, the residue at $z=-n$ is given by the coefficient of $(z+n)^{-1}$ in this expansion, which is simply $(-1)^n$. So the residues at the negative integers are $(-1)^n$.

In summary, the residue of $\Gamma(z)$ at the pole $z=-n$ is $(-1)^n$, and the residue at