# Domain of the Gamma function

I need to find the domain of the Gamma function, that is to say all $z \in \mathbb{C}$, for which the integral:

$$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} \mathrm dt$$

converges. I started by splitting up the integral into an integral running from $0$ to $1$ and another one from $1$ to $\infty$. I first tried to figure out for what $z \in \mathbb{C}$ the integral from $0$ to $1$ converges and I came to the conclusion, that $\Re(z) > 0$ is the condition.

The other integral, I believe, converges for every $z$, as the exponential function dominates the monomial eventually. So I concluded:

$$\exists \Gamma(z) \iff \Re(z) > 0$$

However, I just learned that this is wrong. I found out that the integral only diverges for non-positive integers. What did I do wrong or what is a better way to find the domain of the Gamma function?

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Your problem is that the two issues "I need to find the domain of the Gamma function" and "that is to say all $z\in\mathbb{C}$ such that the integral converges" are different. They are different problems with different answers. The integral is only used to define the Gamma function on $\Re[z] > 0$, but the Gamma function itself extends to all of $\mathbb{C}$ other than simple poles at the nonnegative integers. – George Lowther Dec 11 '10 at 21:11

The integral does converge iff $\Re z > 0$. However, it defines a function that can be analytically extended to the whole complex plane except the non-positive integers.

There are analogous integral representations for $\Gamma(z)$ which hold true for $\Re z < 0$. For instance, it is not difficult to show that for any $k\in\mathbb N$ $$\Gamma(z)=\int_{0}^{\infty} t^{z-1}\left(e^{-t}-1+t-\frac{t^2}{2!}+\dots+ (-1)^{k+1}\frac{t^k}{k!}\right)dt,$$ where $-k < \Re z < - k+1$ (the Cauchy–Saalschütz integral).

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The problem to me is, that the integral also exists for e.g. $z=i$, that is to say $Re(z) = 0$. Also, a problem is, that the integral exists for $z = -1.5$ and other negative non-integers. So I found that $Re(z) > 0 \implies \exists \Gamma(z)$ but that's not specific enough... – Huy Dec 11 '10 at 20:52
$t^{z-1}e^{-t}$ has a non-integrable singularity at $t=0$ when $\Re z\leq 0$. – Andrey Rekalo Dec 11 '10 at 21:06
@Huy: Dear Huy, Just to repeat what Andrey has said, what you wrote here in your comment is wrong, and what you wrote in the original post is correct: the given integral converges precisely when $z$ has positive imaginary part. Although the Gamma function can be analytically continued to other values of $z$ (e.g. via the formula that Andrey wrote in his answer), it cannot be computed by the integral you wrote down when $z$ has non-positive real part. – Matt E Dec 11 '10 at 21:23

There are some similar questions (Q1 and Q2, e.g.) related to the convergence/divergence of the standard integral rep of $\Gamma (s)$ on this site, so I thought it might be helpful for those familiar with basic complex analysis to look at an extended integral rep for $\Gamma (s)$ that is related to the function's singularities, i.e., the Cauchy–Saalschütz integral, which Andrey highlights.

In understanding the convergence of real Taylor series, you need to look at the complex domain and the singularities (hopefully, only simple poles at most) of the function it represents. The same can apply to integrals over the real line, so first do a partial fraction expansion of the Euler/Gauss rep (Eqn. 6.1.2 in Abramowitz and Stegun, pg. 255; also EOM article) of $\Gamma (s)$ in the complex plane and note the simple poles at $s=0, -1, -2, ...$ , consistent with the identity $\frac{1}{s!(-s)!}=\frac{\sin(\pi s)}{\pi s}$. Then if you are comfortable with the Mellin transform, you can easily write down a Hadamard finite part integral representation for sections between the poles:

For $-n<\Re(s)=\sigma<-(n+1)$, the inverse Mellin transform gives

$$\frac{1}{2\pi i}\int_{\sigma -i\infty }^{\sigma +i\infty } \Gamma(s) x^{-s}ds=\frac{1}{2\pi i}\int_{\sigma -i\infty }^{\sigma +i\infty }\frac{\pi }{\sin \left ( \pi s \right )}\frac{x^{-s}}{(-s)!}ds$$

$$=\exp(-x)-\left(1-x+\frac{x^2}{2!}+ ... + \frac{(-x)^n}{n!}\right),$$

and, therefore, the associated Mellin transform gives

$$\Gamma (s)=\mathrm{FP}\int_{0}^{\infty }x^{s-1}\exp(-x)dx = \int_{0}^{\infty }x^{s-1}\left[\exp(-x)-\left(1-x+\frac{x^2}{2!}+ ... + \frac{(-x)^n}{n!}\right)\right]dx.$$

Roughly speaking, the singularities at the lower limit $x=0$ of $\frac{x^{s+m}}{s+m}$ for $m=0, -1, ..., -n$ are being subtracted out.

For $\Re(s)=\sigma>0$, you obtain the standard integral rep.

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