# What is the Gamma Function for negative complex numbers?

I know that:

$\Gamma(n) \quad=\quad (n-1)!\quad\forall n\in\mathbb N$

And that:

$\Gamma(t) \quad=\quad \int_{0}^\infty {x^{t-1}\over e^x}dx,\quad\Re(t)>0$

During some further research, (please correct me if I'm wrong) I learned that the Gamma Function can be defined for a negative value in such manner:

$\Gamma(s)\quad=\quad \Gamma(s+1)/s\quad\forall s\in\mathbb R^*$

Therefore (e.g.):

$\Gamma(-{1\over 2})\quad= \quad\Gamma({1\over 2})/(-{1\over 2})\quad=\quad -2\sqrt\pi$

Does this method also work for complex values of s with Real part less than $0$?

Or is there another method?

$\Gamma(z)$ is not going to be defined for all $z$ in $\mathbb{R}^*$. We can show this by doing the following manipulations, and in the process we can also show what values it will be defined for.

\begin{align*} \Gamma(z)&=\int_0^{\infty}e^{-t}t^{z-1}dt \\ &= \int_0^{1}e^{-t}t^{z-1}dt+\int_1^{\infty}e^{-t}t^{z-1}dt\\ &= \int_0^{1}\sum_{n=0}^{\infty}\frac{(-t)^n}{n!} t^{z-1}dt+\int_1^{\infty}e^{-t}t^{z-1}dt\\ &= \sum_{n=0}^{\infty}\int_0^{1}\frac{(-t)^n}{n!} t^{z-1}dt+\int_1^{\infty}e^{-t}t^{z-1}dt\\ &= \sum_{n=0}^{\infty}\frac{(-1)^n}{n!(z+n)}+\int_1^{\infty}e^{-t}t^{z-1}dt\;\;\;\;\;\;(*) \end{align*}

Note that the integral in line $(*)$ is defined $\forall z\in\mathbb{C}$, but the series runs over $n\in\left \{ \;0,\;1,\;2,\;3,\;\cdots\right \}$ and contains $(z+n)$ in the denominator. From this we can see that $\Gamma(z)$ is defined $\forall z\in\mathbb{C}\setminus\left \{ \;0,\;-1,\;-2,\;-3,\;\cdots\right \}$.

So we conclude that the statement $\quad\forall s\in\mathbb {R}^*,\;\;\Gamma(s)= \Gamma(s+1)/s$ is false, as $\Gamma(s)$ is undefined for some $s$ in $\mathbb {R}^*$.

The relation

$$\Gamma(z)=\frac{\Gamma(z+1)}{z}$$

will hold $\forall z\in\mathbb{C}\setminus\left \{ \;0,\;-1,\;-2,\;-3,\;\cdots\right \}$. We see this from the above section.

I am also assuming you are looking for a general way to create closed form representations of $\Gamma(z)$ for $z=iy-x$ where $x\in\mathbb{R}^+\setminus\mathbb{N}$ and $y\in\mathbb{R}$.

As far as I am aware there is no way to further simplify the following any more than shown:

\begin{align*} \Gamma(iy-x)&=\frac{\Gamma(iy-x+1)}{iy-x} \\ &= \frac{\Gamma(iy-x+2)}{(iy-x)(iy-x+1)}\\ &= \frac{\Gamma(iy-\left \{ x\right \})}{(iy-x)(iy-x+1)\cdots(iy-\left \{ x \right \}-1)} \end{align*}

as there is no other 'nicer' way to represent $\Gamma(iy-\left \{ x\right \})$ where $\left \{ x\right \}$ is the fractional part of $x$. One could also argue that what is shown directly above isn't much of a simplification, but this does show that the relation holds for $z$ such that $\Re(z)<0$.

• Thank you so much. This helped me more than you can imagine. – Max Echendu Nov 18 '15 at 0:11
• You are very much welcome. – dxdydz Nov 18 '15 at 1:29