# Thread: gamma function question for x < 0

1. ## gamma function question for x < 0

so I know the gamma function defined:
T(x) = [integral from 0 to infinity] t^(x-1)e^(-t)dt

converges for x > 0

I also know that T(x+1) = xT(x)

does this make it possible to extend the T(x) function for x <0? (except the negative integers), and if so, why is it okay to apply this identity even though T(x) is not defined (divergent) for x <= 0?

This is more of a conceptual question, so if anyone understands this I would love to hear an explanation. Thanks!

2. This arises from the principle of Analytic Continuation. The Gamma function exists independently of the integral representation you alluded to. It's just that integral representation is equal to the Gamma function when $\textbf{Re}(s)>0$. That may sound confussing. Consider the canonial example: The series representation for $f(x)=\frac{1}{1-x}=\sum_{k=0}^{\infty} x^k$. This series represents'' $f(x)$ only when $-1 but the function $f(x)$ exists for all values of $x\ne 1$. It's a similar situation for the Gamma function: The Gamma function is a complex-analytic function defined throughout the complex plane except for poles (singularities) at zero and the negative integers with the common integral representation of it valid only when $\textbf{Re}(s)>0$:

$\Gamma(s)=-\frac{i}{2}\csc(\pi s)\mathop\int\limits_{-H}e^{w} w^{s-1}dw\overset{\textbf{Re}(s)>0}{=}\int_0^{\infty}t^ {s-1}e^{-t}dt$

where $-H$ is a reverse-Hankel contour in the complex plane (there are other ways to define the Gamma function).

Even though the expression $\Gamma(1+s)=s\Gamma(s)$ can be derived from the common integral representation of Gamma which is only valid when $\textbf{Re}(s)>0$, because of the Principle of Analytic Continuation, the expression $\Gamma(1+s)=s\Gamma(s)$ is still valid for all $s$ and not just s>0.

3. Originally Posted by shawsend
Even though the expression $\Gamma(1+s)=s\Gamma(s)$ can be derived from the common integral representation of Gamma which is only valid when $\textbf{Re}(s)>0$, because of the Principle of Analytic Continuation, the expression $\Gamma(1+s)=s\Gamma(s)$ is still valid for all $s$ and not just s>0.
Perhaps I am missing something shawsend but what you did all above is not necessary.

The necesscity here is to note $\Gamma (s+1) = s\Gamma( s)$ for $\Re (s) > 0$ therefore $\Gamma (s) = \tfrac{1}{s}\cdot \Gamma(s+1)$.
Now define for $-1< \Re(s) < 0$, $\Gamma ( s ) = \tfrac{1}{s}\cdot \Gamma (s+1)$ (notice that $0 < \Re(s+1) < 1$).
Now define for $-2 < \Re(s) < -1$, $\Gamma ( s ) = \tfrac{1}{s}\cdot \Gamma ( s+ 1) = \tfrac{1}{s} \cdot \tfrac{1}{s+1} \cdot \Gamma (s+2)$ (notice that $0 < \Re (s+2) < 1$).

Thus in general we define for $- (n+1) < \Re (s) < -n$, $\Gamma (s) = \left( \prod_{k=0}^n \tfrac{1}{s+k} \right) \Gamma (s+n+1)$.

This defines an analytic function $\Gamma$ on $\mathbb{C} - (\mathbb{Z}^- \cup \{ 0 \} )$.
Of course, it remains to prove this function cannot be further extended.
And it follows that this function is the analytically extended Gamma function.

4. Ok, that's cleaner and simpler. I struggle with analytic continuation. Also should it not be defined on $\mathbb{C}\backslash \{0,\mathbb{Z}^{-}\}$ (C minus the negative integers and zero)?

5. Originally Posted by shawsend
Also should it not be defined on $\mathbb{C}\backslash \{0,\mathbb{Z}^{-}\}$ (C minus the negative integers and zero)?
Yes! Thank you, comrade. I shall fix it now.