Gamma function

The notation ${\displaystyle \Gamma (z)}$ is due to Legendre.[1] If the real part of the complex number z is strictly positive (${\displaystyle \Re (z)>0}$), then the integral

$${\displaystyle \Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}\,dt}$$

converges absolutely, and is known as the Euler integral of the second kind. (Euler's integral of the first kind is the beta function.[1]) Using integration by parts, one sees that:

Plot of the absolute value of the gamma function in complex plane in 3D with color showing the argument of gamma, using 1000 plot points

${\displaystyle {\begin{aligned}\Gamma (z+1)&=\int _{0}^{\infty }t^{z}e^{-t}\,dt\\&={\Bigl [}-t^{z}e^{-t}{\Bigr ]}_{0}^{\infty }+\int _{0}^{\infty }zt^{z-1}e^{-t}\,dt\\&=\lim _{t\to \infty }\left(-t^{z}e^{-t}\right)-\left(-0^{z}e^{-0}\right)+z\int _{0}^{\infty }t^{z-1}e^{-t}\,dt.\end{aligned}}}$

Recognizing that ${\displaystyle -t^{z}e^{-t}\to 0}$ as ${\displaystyle t\to \infty ,}$

$${\displaystyle {\begin{aligned}\Gamma (z+1)&=z\int _{0}^{\infty }t^{z-1}e^{-t}\,dt\\&=z\Gamma (z).\end{aligned}}}$$

We can calculate ${\displaystyle \Gamma (1)}$:

$${\displaystyle {\begin{aligned}\Gamma (1)&=\int _{0}^{\infty }t^{1-1}e^{-t}\,dt\\&=\int _{0}^{\infty }e^{-t}\,dt\\&=1.\end{aligned}}}$$

Thus we can show that ${\displaystyle \Gamma (n)=(n-1)!}$ for any positive integer n by induction. Specifically, the base case is that ${\displaystyle \Gamma (1)=1=0!}$, and the induction step is that ${\displaystyle \Gamma (n+1)=n\Gamma (n)=n(n-1)!=n!}$.

The identity ${\textstyle \Gamma (z)={\frac {\Gamma (z+1)}{z}}}$ can be used (or, yielding the same result, analytic continuation can be used) to uniquely extend the integral formulation for ${\displaystyle \Gamma (z)}$ to a meromorphic function defined for all complex numbers z, except integers less than or equal to zero.[1] It is this extended version that is commonly referred to as the gamma function.[1]

There are many equivalent definitions.

For a fixed integer ${\displaystyle m}$, as the integer ${\displaystyle n}$ increases, we have that [7]

$${\displaystyle \lim _{n\to \infty }{\frac {n!\,\left(n+1\right)^{m}}{(n+m)!}}=1\,.}$$

If ${\displaystyle m}$ is not an integer then it is not possible to say whether this equation is true because we have not yet (in this section) defined the factorial function for non-integers. However, we do get a unique extension of the factorial function to the non-integers by insisting that this equation continue to hold when the arbitrary integer ${\displaystyle m}$ is replaced by an arbitrary complex number ${\displaystyle z}$,

$${\displaystyle \lim _{n\to \infty }{\frac {n!\,\left(n+1\right)^{z}}{(n+z)!}}=1\,.}$$

Multiplying both sides by ${\displaystyle (z-1)!}$ gives

$${\displaystyle {\begin{aligned}\Gamma (z)&=(z-1)!\\[8pt]&={\frac {1}{z}}\lim _{n\to \infty }n!{\frac {z!}{(n+z)!}}(n+1)^{z}\\[8pt]&={\frac {1}{z}}\lim _{n\to \infty }(1\cdots n){\frac {1}{(1+z)\cdots (n+z)}}\left({\frac {2}{1}}\cdot {\frac {3}{2}}\cdots {\frac {n+1}{n}}\right)^{z}\\[8pt]&={\frac {1}{z}}\prod _{n=1}^{\infty }\left[{\frac {1}{1+{\frac {z}{n}}}}\left(1+{\frac {1}{n}}\right)^{z}\right].\end{aligned}}}$$

This infinite product, which is due to Euler,[8] converges for all complex numbers ${\displaystyle z}$ except the non-positive integers, which fail because of a division by zero. Intuitively, this formula indicates that ${\displaystyle \Gamma (z)}$ is approximately the result of computing ${\displaystyle \Gamma (n+1)=n!}$ for some large integer ${\displaystyle n}$, multiplying by ${\displaystyle (n+1)^{z}}$ to approximate ${\displaystyle \Gamma (n+z+1)}$, and using the relationship ${\displaystyle \Gamma (x+1)=x\Gamma (x)}$ backwards ${\displaystyle n+1}$ times to get an approximation for ${\displaystyle \Gamma (z)}$; and furthermore that this approximation becomes exact as ${\displaystyle n}$ increases to infinity.

The definition for the gamma function due to Weierstrass is also valid for all complex numbers z except the non-positive integers:

$${\displaystyle \Gamma (z)={\frac {e^{-\gamma z}}{z}}\prod _{n=1}^{\infty }\left(1+{\frac {z}{n}}\right)^{-1}e^{z/n},}$$

where ${\displaystyle \gamma \approx 0.577216}$ is the Euler–Mascheroni constant.[1] This is the Hadamard product of ${\displaystyle 1/\Gamma (z)}$ in a rewritten form.

Proof of equivalence of the three definitions

Equivalence of the integral definition and Weierstrass definition By the integral definition, the relation ${\displaystyle \Gamma (z+1)=z\Gamma (z)}$ and Hadamard factorization theorem, $${\displaystyle {\frac {1}{\Gamma (z)}}=ze^{c_{1}z+c_{2}}\prod _{n=1}^{\infty }e^{-{\frac {z}{n}}}\left(1+{\frac {z}{n}}\right),\quad z\in \mathbb {C} \setminus \mathbb {Z} _{0}^{-}}$$ for some constants ${\displaystyle c_{1},c_{2}}$ since ${\displaystyle 1/\Gamma }$ is an entire function of order ${\displaystyle 1}$. Since ${\displaystyle z\Gamma (z)\to 1}$ as ${\displaystyle z\to 0}$, ${\displaystyle c_{2}=0}$ (or an integer multiple of ${\displaystyle 2\pi i}$) and since ${\displaystyle \Gamma (1)=1}$, $${\displaystyle {\begin{aligned}e^{-c_{1}}&=\prod _{n=1}^{\infty }e^{-{\frac {1}{n}}}\left(1+{\frac {1}{n}}\right)\\&=\exp \left(\lim _{N\to \infty }\sum _{n=1}^{N}\left(\log \left(1+{\frac {1}{n}}\right)-{\frac {1}{n}}\right)\right)\\&=\exp \left(\lim _{N\to \infty }\left(\log(N+1)-\sum _{n=1}^{N}{\frac {1}{n}}\right)\right)\\&=\exp \left(\lim _{N\to \infty }\left(\log N+\log \left(1+{\frac {1}{N}}\right)-\sum _{n=1}^{N}{\frac {1}{n}}\right)\right)\\&=\exp \left(\lim _{N\to \infty }\left(\log N-\sum _{n=1}^{N}{\frac {1}{n}}\right)\right)\\&=e^{-\gamma }.\end{aligned}}}$$ Whence ${\displaystyle c_{1}=\gamma +2\pi ik}$ for some integer ${\displaystyle k}$. Since ${\displaystyle \Gamma (z)\in \mathbb {R} }$ for ${\displaystyle z\in \mathbb {R} \setminus \mathbb {Z} _{0}^{-}}$, we have ${\displaystyle k=0}$ and $${\displaystyle {\frac {1}{\Gamma (z)}}=ze^{\gamma z}\prod _{n=1}^{\infty }e^{-{\frac {z}{n}}}\left(1+{\frac {z}{n}}\right),\quad z\in \mathbb {C} \setminus \mathbb {Z} _{0}^{-}.}$$ Equivalence of the Weierstrass definition and Euler definition $${\displaystyle {\begin{aligned}\Gamma (z)&={\frac {e^{-\gamma z}}{z}}\prod _{n=1}^{\infty }\left(1+{\frac {z}{n}}\right)^{-1}e^{z/n}\\&={\frac {1}{z}}\lim _{n\to \infty }e^{z\left(\ln n-1-{\frac {1}{2}}-{\frac {1}{3}}-\cdots -{\frac {1}{n}}\right)}{\frac {e^{z\left(1+{\frac {1}{2}}+{\frac {1}{3}}+\cdots +{\frac {1}{n}}\right)}}{\left(1+z\right)\left(1+{\frac {z}{2}}\right)\cdots \left(1+{\frac {z}{n}}\right)}}\\&={\frac {1}{z}}\lim _{n\to \infty }{\frac {1}{\left(1+z\right)\left(1+{\frac {z}{2}}\right)\cdots \left(1+{\frac {z}{n}}\right)}}e^{z\ln \left(n\right)}\\&=\lim _{n\to \infty }{\frac {n!n^{z}}{z(z+1)\cdots (z+n)}},\quad z\in \mathbb {C} \setminus \mathbb {Z} _{0}^{-}\end{aligned}}}$$ Let $${\displaystyle \Gamma _{n}(z)={\frac {n!n^{z}}{z(z+1)\cdots (z+n)}}}$$ and $${\displaystyle G_{n}(z)={\frac {(n-1)!n^{z}}{z(z+1)\cdots (z+n-1)}}.}$$ Then $${\displaystyle \Gamma _{n}(z)={\frac {n}{z+n}}G_{n}(z)}$$ and $${\displaystyle \lim _{n\to \infty }G_{n+1}(z)=\lim _{n\to \infty }G_{n}(z)=\lim _{n\to \infty }\Gamma _{n}(z)=\Gamma (z),}$$ therefore $${\displaystyle \Gamma (z)=\lim _{n\to \infty }{\frac {n!(n+1)^{z}}{z(z+1)\cdots (z+n)}},\quad z\in \mathbb {C} \setminus \mathbb {Z} _{0}^{-}.}$$ Then $${\displaystyle {\frac {n!(n+1)^{z}}{z(z+1)\cdots (z+n)}}={\frac {(2/1)^{z}(3/2)^{z}(4/3)^{z}\cdots ((n+1)/n)^{z}}{z(1+z)(1+z/2)(1+z/3)\cdots (1+z/n)}}={\frac {1}{z}}\prod _{k=1}^{n}{\frac {(1+1/k)^{z}}{1+z/k}},\quad z\in \mathbb {C} \setminus \mathbb {Z} _{0}^{-}}$$ and taking ${\displaystyle n\to \infty }$ gives the desired result.

Besides the fundamental property discussed above:

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

other important functional equations for the gamma function are Euler's reflection formula

$${\displaystyle \Gamma (1-z)\Gamma (z)={\frac {\pi }{\sin \pi z}},\qquad z\not \in \mathbb {Z} }$$

which implies

$${\displaystyle \Gamma (z-n)=(-1)^{n-1}\;{\frac {\Gamma (-z)\Gamma (1+z)}{\Gamma (n+1-z)}},\qquad n\in \mathbb {Z} }$$

and the Legendre duplication formula

$${\displaystyle \Gamma (z)\Gamma \left(z+{\tfrac {1}{2}}\right)=2^{1-2z}\;{\sqrt {\pi }}\;\Gamma (2z).}$$

Derivation of Euler's reflection formula

Proof 1 We can use Euler's infinite product $${\displaystyle \Gamma (z)={\frac {1}{z}}\prod _{n=1}^{\infty }{\frac {(1+1/n)^{z}}{1+z/n}}}$$ to compute $${\displaystyle {\frac {1}{\Gamma (1-z)\Gamma (z)}}={\frac {1}{(-z)\Gamma (-z)\Gamma (z)}}=z\prod _{n=1}^{\infty }{\frac {(1-z/n)(1+z/n)}{(1+1/n)^{-z}(1+1/n)^{z}}}=z\prod _{n=1}^{\infty }\left(1-{\frac {z^{2}}{n^{2}}}\right)={\frac {\sin \pi z}{\pi }}\,,}$$ where the last equality is a known result. A similar derivation begins with Weierstrass's definition. Proof 2 First we prove that $${\displaystyle I=\int _{-\infty }^{\infty }{\frac {e^{ax}}{1+e^{x}}}\,dx=\int _{0}^{\infty }{\frac {v^{a-1}}{1+v}}\,dv={\frac {\pi }{\sin \pi a}},\quad a\in (0,1).}$$ Consider the positively oriented rectangular contour ${\displaystyle C_{R}}$ with vertices at ${\displaystyle R}$, ${\displaystyle -R}$, ${\displaystyle R+2\pi i}$ and ${\displaystyle -R+2\pi i}$ where ${\displaystyle R\in \mathbb {R} ^{+}}$. Then by the residue theorem, $${\displaystyle \int _{C_{R}}{\frac {e^{az}}{1+e^{z}}}\,dz=-2\pi ie^{a\pi i}.}$$ Let $${\displaystyle I_{R}=\int _{-R}^{R}{\frac {e^{ax}}{1+e^{x}}}\,dx}$$ and let ${\displaystyle I_{R}'}$ be the analogous integral over the top side of the rectangle. Then ${\displaystyle I_{R}\to I}$ as ${\displaystyle R\to \infty }$ and ${\displaystyle I_{R}'=-e^{2\pi ia}I_{R}}$. If ${\displaystyle A_{R}}$ denotes the right vertical side of the rectangle, then $${\displaystyle \left|\int _{A_{R}}{\frac {e^{az}}{1+e^{z}}}\,dz\right|\leq \int _{0}^{2\pi }\left|{\frac {e^{a(R+it)}}{1+e^{R+it}}}\right|\,dt\leq Ce^{(a-1)R}}$$ for some constant ${\displaystyle C}$ and since ${\displaystyle a<1}$, the integral tends to ${\displaystyle 0}$ as ${\displaystyle R\to \infty }$. Analogously, the integral over the left vertical side of the rectangle tends to ${\displaystyle 0}$ as ${\displaystyle R\to \infty }$. Therefore $${\displaystyle I-e^{2\pi ia}I=-2\pi ie^{a\pi i},}$$ from which $${\displaystyle I={\frac {\pi }{\sin \pi a}},\quad a\in (0,1).}$$ Then $${\displaystyle \Gamma (1-z)=\int _{0}^{\infty }e^{-u}u^{-z}\,du=t\int _{0}^{\infty }e^{-vt}(vt)^{-z}\,dv,\quad t>0}$$ and $${\displaystyle {\begin{aligned}\Gamma (z)\Gamma (1-z)&=\int _{0}^{\infty }\int _{0}^{\infty }e^{-t(1+v)}v^{-z}\,dv\,dt\\&=\int _{0}^{\infty }{\frac {v^{-z}}{1+v}}\,dv\\&={\frac {\pi }{\sin \pi (1-z)}}\\&={\frac {\pi }{\sin \pi z}},\quad z\in (0,1).\end{aligned}}}$$ Proving the reflection formula for all ${\displaystyle z\in (0,1)}$ proves it for all ${\displaystyle z\in \mathbb {C} \setminus \mathbb {Z} }$ by analytic continuation.

Derivation of the Legendre duplication formula

The beta function can be represented as $${\displaystyle \mathrm {B} (z_{1},z_{2})={\frac {\Gamma (z_{1})\Gamma (z_{2})}{\Gamma (z_{1}+z_{2})}}=\int _{0}^{1}t^{z_{1}-1}(1-t)^{z_{2}-1}\,dt.}$$ Setting ${\displaystyle z_{1}=z_{2}=z}$ yields $${\displaystyle {\frac {\Gamma ^{2}(z)}{\Gamma (2z)}}=\int _{0}^{1}t^{z-1}(1-t)^{z-1}\,dt.}$$ After the substitution ${\displaystyle t={\frac {1+u}{2}}}$ we get $${\displaystyle {\frac {\Gamma ^{2}(z)}{\Gamma (2z)}}={\frac {1}{2^{2z-1}}}\int _{-1}^{1}\left(1-u^{2}\right)^{z-1}\,du.}$$ The function ${\displaystyle (1-u^{2})^{z-1}}$ is even, hence $${\displaystyle 2^{2z-1}\Gamma ^{2}(z)=2\Gamma (2z)\int _{0}^{1}(1-u^{2})^{z-1}\,du.}$$ Now assume $${\displaystyle \mathrm {B} \left({\frac {1}{2}},z\right)=\int _{0}^{1}t^{{\frac {1}{2}}-1}(1-t)^{z-1}\,dt,\quad t=s^{2}.}$$ Then $${\displaystyle \mathrm {B} \left({\frac {1}{2}},z\right)=2\int _{0}^{1}(1-s^{2})^{z-1}\,ds=2\int _{0}^{1}(1-u^{2})^{z-1}\,du.}$$ This implies $${\displaystyle 2^{2z-1}\Gamma ^{2}(z)=\Gamma (2z)\mathrm {B} \left({\frac {1}{2}},z\right).}$$ Since $${\displaystyle \mathrm {B} \left({\frac {1}{2}},z\right)={\frac {\Gamma \left({\frac {1}{2}}\right)\Gamma (z)}{\Gamma \left(z+{\frac {1}{2}}\right)}},\quad \Gamma \left({\frac {1}{2}}\right)={\sqrt {\pi }},}$$ the Legendre duplication formula follows: $${\displaystyle \Gamma (z)\Gamma \left(z+{\frac {1}{2}}\right)=2^{1-2z}{\sqrt {\pi }}\;\Gamma (2z).}$$

The duplication formula is a special case of the multiplication theorem (see [9] Eq. 5.5.6):

$${\displaystyle \prod _{k=0}^{m-1}\Gamma \left(z+{\frac {k}{m}}\right)=(2\pi )^{\frac {m-1}{2}}\;m^{{\frac {1}{2}}-mz}\;\Gamma (mz).}$$

A simple but useful property, which can be seen from the limit definition, is:

$${\displaystyle {\overline {\Gamma (z)}}=\Gamma ({\overline {z}})\;\Rightarrow \;\Gamma (z)\Gamma ({\overline {z}})\in \mathbb {R} .}$$

In particular, with z = a + bi, this product is

$${\displaystyle |\Gamma (a+bi)|^{2}=|\Gamma (a)|^{2}\prod _{k=0}^{\infty }{\frac {1}{1+{\frac {b^{2}}{(a+k)^{2}}}}}}$$

If the real part is an integer or a half-integer, this can be finitely expressed in closed form:

$${\displaystyle {\begin{aligned}|\Gamma (bi)|^{2}&={\frac {\pi }{b\sinh \pi b}}\\[1ex]\left|\Gamma \left({\tfrac {1}{2}}+bi\right)\right|^{2}&={\frac {\pi }{\cosh \pi b}}\\[1ex]\left|\Gamma \left(1+bi\right)\right|^{2}&={\frac {\pi b}{\sinh \pi b}}\\[1ex]\left|\Gamma \left(1+n+bi\right)\right|^{2}&={\frac {\pi b}{\sinh \pi b}}\prod _{k=1}^{n}\left(k^{2}+b^{2}\right),\quad n\in \mathbb {N} \\[1ex]\left|\Gamma \left(-n+bi\right)\right|^{2}&={\frac {\pi }{b\sinh \pi b}}\prod _{k=1}^{n}\left(k^{2}+b^{2}\right)^{-1},\quad n\in \mathbb {N} \\[1ex]\left|\Gamma \left({\tfrac {1}{2}}\pm n+bi\right)\right|^{2}&={\frac {\pi }{\cosh \pi b}}\prod _{k=1}^{n}\left(\left(k-{\tfrac {1}{2}}\right)^{2}+b^{2}\right)^{\pm 1},\quad n\in \mathbb {N} \\[-1ex]&\end{aligned}}}$$

Proof of absolute value formulas for arguments of integer or half-integer real part

First, consider the reflection formula applied to ${\displaystyle z=bi}$. $${\displaystyle \Gamma (bi)\Gamma (1-bi)={\frac {\pi }{\sin \pi bi}}}$$ Applying the recurrence relation to the second term, we have $${\displaystyle -bi\cdot \Gamma (bi)\Gamma (-bi)={\frac {\pi }{\sin \pi bi}}}$$ which with simple rearrangement gives $${\displaystyle \Gamma (bi)\Gamma (-bi)={\frac {\pi }{-bi\sin \pi bi}}={\frac {\pi }{b\sinh \pi b}}}$$ Second, consider the reflection formula applied to ${\displaystyle z={\tfrac {1}{2}}+bi}$. $${\displaystyle \Gamma ({\tfrac {1}{2}}+bi)\Gamma \left(1-({\tfrac {1}{2}}+bi)\right)=\Gamma ({\tfrac {1}{2}}+bi)\Gamma ({\tfrac {1}{2}}-bi)={\frac {\pi }{\sin \pi ({\tfrac {1}{2}}+bi)}}={\frac {\pi }{\cos \pi bi}}={\frac {\pi }{\cosh \pi b}}}$$ Formulas for other values of ${\displaystyle z}$ for which the real part is integer or half-integer quickly follow by induction using the recurrence relation in the positive and negative directions.

Perhaps the best-known value of the gamma function at a non-integer argument is

$${\displaystyle \Gamma \left({\tfrac {1}{2}}\right)={\sqrt {\pi }},}$$

which can be found by setting ${\textstyle z={\frac {1}{2}}}$ in the reflection or duplication formulas, by using the relation to the beta function given below with ${\textstyle z_{1}=z_{2}={\frac {1}{2}}}$, or simply by making the substitution ${\displaystyle u={\sqrt {z}}}$ in the integral definition of the gamma function, resulting in a Gaussian integral. In general, for non-negative integer values of ${\displaystyle n}$ we have:

$${\displaystyle {\begin{aligned}\Gamma \left({\tfrac {1}{2}}+n\right)&={(2n)! \over 4^{n}n!}{\sqrt {\pi }}={\frac {(2n-1)!!}{2^{n}}}{\sqrt {\pi }}={\binom {n-{\frac {1}{2}}}{n}}n!{\sqrt {\pi }}\\[8pt]\Gamma \left({\tfrac {1}{2}}-n\right)&={(-4)^{n}n! \over (2n)!}{\sqrt {\pi }}={\frac {(-2)^{n}}{(2n-1)!!}}{\sqrt {\pi }}={\frac {\sqrt {\pi }}{{\binom {-1/2}{n}}n!}}\end{aligned}}}$$

where the double factorial ${\displaystyle (2n-1)!!=(2n-1)(2n-3)\cdots (3)(1)}$. See Particular values of the gamma function for calculated values.

It might be tempting to generalize the result that ${\textstyle \Gamma \left({\frac {1}{2}}\right)={\sqrt {\pi }}}$ by looking for a formula for other individual values ${\displaystyle \Gamma (r)}$ where ${\displaystyle r}$ is rational, especially because according to Gauss's digamma theorem, it is possible to do so for the closely related digamma function at every rational value. However, these numbers ${\displaystyle \Gamma (r)}$ are not known to be expressible by themselves in terms of elementary functions. It has been proved that ${\displaystyle \Gamma (n+r)}$ is a transcendental number and algebraically independent of ${\displaystyle \pi }$ for any integer ${\displaystyle n}$ and each of the fractions ${\textstyle r={\frac {1}{6}},{\frac {1}{4}},{\frac {1}{3}},{\frac {2}{3}},{\frac {3}{4}},{\frac {5}{6}}}$.[10] In general, when computing values of the gamma function, we must settle for numerical approximations.

The derivatives of the gamma function are described in terms of the polygamma function, ψ⁽⁰⁾(z):

$${\displaystyle \Gamma '(z)=\Gamma (z)\psi ^{(0)}(z).}$$

For a positive integer m the derivative of the gamma function can be calculated as follows:

Plot of gamma function in the complex plane from -2-2i to 6+2i with colors showing the argument of gamma

$${\displaystyle \Gamma '(m+1)=m!\left(-\gamma +\sum _{k=1}^{m}{\frac {1}{k}}\right)=m!\left(-\gamma +H(m)\right)\,,}$$

where H(m) is the mth harmonic number and γ is the Euler–Mascheroni constant.

For ${\displaystyle \Re (z)>0}$ the ${\displaystyle n}$th derivative of the gamma function is:

$${\displaystyle {\frac {d^{n}}{dz^{n}}}\Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}(\ln t)^{n}\,dt.}$$

(This can be derived by differentiating the integral form of the gamma function with respect to ${\displaystyle z}$, and using the technique of differentiation under the integral sign.)

Using the identity

$${\displaystyle \Gamma ^{(n)}(1)=(-1)^{n}n!\sum \limits _{\pi \,\vdash \,n}\,\prod _{i=1}^{r}{\frac {\zeta ^{*}(a_{i})}{k_{i}!\cdot a_{i}}}\qquad \zeta ^{*}(x):={\begin{cases}\zeta (x)&x\neq 1\\\gamma &x=1\end{cases}}}$$

where ${\displaystyle \zeta (z)}$ is the Riemann zeta function, and ${\displaystyle \pi }$ is a partition of ${\displaystyle n}$ given by

$${\displaystyle \pi =\underbrace {a_{1}+\cdots +a_{1}} _{k_{1}{\text{ terms}}}+\cdots +\underbrace {a_{r}+\cdots +a_{r}} _{k_{r}{\text{ terms}}},}$$

we have in particular the Laurent series expansion of the gamma function [11]

$${\displaystyle \Gamma (z)={\frac {1}{z}}-\gamma +{\tfrac {1}{2}}\left(\gamma ^{2}+{\frac {\pi ^{2}}{6}}\right)z-{\tfrac {1}{6}}\left(\gamma ^{3}+{\frac {\gamma \pi ^{2}}{2}}+2\zeta (3)\right)z^{2}+O(z^{3}).}$$

When restricted to the positive real numbers, the gamma function is a strictly logarithmically convex function. This property may be stated in any of the following three equivalent ways:

• For any two positive real numbers ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$, and for any ${\displaystyle t\in [0,1]}$,
  $${\displaystyle \Gamma (tx_{1}+(1-t)x_{2})\leq \Gamma (x_{1})^{t}\Gamma (x_{2})^{1-t}.}$$

  
• For any two positive real numbers ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$, and ${\displaystyle x_{2}}$ > ${\displaystyle x_{1}}$
  $${\displaystyle \left({\frac {\Gamma (x_{2})}{\Gamma (x_{1})}}\right)^{\frac {1}{x_{2}-x_{1}}}>\exp \left({\frac {\Gamma '(x_{1})}{\Gamma (x_{1})}}\right).}$$

  
• For any positive real number ${\displaystyle x}$,
  $${\displaystyle \Gamma ''(x)\Gamma (x)>\Gamma '(x)^{2}.}$$

  

The last of these statements is, essentially by definition, the same as the statement that ${\displaystyle \psi ^{(1)}(x)>0}$, where ${\displaystyle \psi ^{(1)}}$ is the polygamma function of order 1. To prove the logarithmic convexity of the gamma function, it therefore suffices to observe that ${\displaystyle \psi ^{(1)}}$ has a series representation which, for positive real x, consists of only positive terms.

Logarithmic convexity and Jensen's inequality together imply, for any positive real numbers ${\displaystyle x_{1},\ldots ,x_{n}}$ and ${\displaystyle a_{1},\ldots ,a_{n}}$,

$${\displaystyle \Gamma \left({\frac {a_{1}x_{1}+\cdots +a_{n}x_{n}}{a_{1}+\cdots +a_{n}}}\right)\leq {\bigl (}\Gamma (x_{1})^{a_{1}}\cdots \Gamma (x_{n})^{a_{n}}{\bigr )}^{\frac {1}{a_{1}+\cdots +a_{n}}}.}$$

There are also bounds on ratios of gamma functions. The best-known is Gautschi's inequality, which says that for any positive real number x and any s ∈ (0, 1),

$${\displaystyle x^{1-s}<{\frac {\Gamma (x+1)}{\Gamma (x+s)}}<\left(x+1\right)^{1-s}.}$$

Representation of the gamma function in the complex plane. Each point ${\displaystyle z}$ is colored according to the argument of ${\displaystyle \Gamma (z)}$. The contour plot of the modulus ${\displaystyle |\Gamma (z)|}$ is also displayed.

3-dimensional plot of the absolute value of the complex gamma function

The behavior of ${\displaystyle \Gamma (x)}$ for an increasing positive real variable is given by Stirling's formula

$${\displaystyle \Gamma (x+1)\sim {\sqrt {2\pi x}}\left({\frac {x}{e}}\right)^{x},}$$

where the symbol ${\displaystyle \sim }$ means asymptotic convergence: the ratio of the two sides converges to 1 in the limit ${\textstyle x\to +\infty }$.[1] This growth is faster than exponential, ${\displaystyle \exp(\beta x)}$, for any fixed value of ${\displaystyle \beta }$.

Another useful limit for asymptotic approximations for ${\displaystyle x\to +\infty }$ is:

$${\displaystyle {\Gamma (x+\alpha )}\sim {\Gamma (x)x^{\alpha }},\qquad \alpha \in \mathbb {C} .}$$

The behavior for non-positive ${\displaystyle z}$ is more intricate. Euler's integral does not converge for ${\displaystyle \Re (z)\leq 0}$, but the function it defines in the positive complex half-plane has a unique analytic continuation to the negative half-plane. One way to find that analytic continuation is to use Euler's integral for positive arguments and extend the domain to negative numbers by repeated application of the recurrence formula,[1]

$${\displaystyle \Gamma (z)={\frac {\Gamma (z+n+1)}{z(z+1)\cdots (z+n)}},}$$

choosing ${\displaystyle n}$ such that ${\displaystyle z+n}$ is positive. The product in the denominator is zero when ${\displaystyle z}$ equals any of the integers ${\displaystyle 0,-1,-2,\ldots }$. Thus, the gamma function must be undefined at those points to avoid division by zero; it is a meromorphic function with simple poles at the non-positive integers.[1]

For a function ${\displaystyle f}$ of a complex variable ${\displaystyle z}$, at a simple pole ${\displaystyle c}$, the residue of ${\displaystyle f}$ is given by:

$${\displaystyle \operatorname {Res} (f,c)=\lim _{z\to c}(z-c)f(z).}$$

For the simple pole ${\displaystyle z=-n,}$ we rewrite recurrence formula as:

$${\displaystyle (z+n)\Gamma (z)={\frac {\Gamma (z+n+1)}{z(z+1)\cdots (z+n-1)}}.}$$

The numerator at ${\displaystyle z=-n,}$ is

$${\displaystyle \Gamma (z+n+1)=\Gamma (1)=1}$$

and the denominator

$${\displaystyle z(z+1)\cdots (z+n-1)=-n(1-n)\cdots (n-1-n)=(-1)^{n}n!.}$$

So the residues of the gamma function at those points are:[12]

$${\displaystyle \operatorname {Res} (\Gamma ,-n)={\frac {(-1)^{n}}{n!}}.}$$

The gamma function is non-zero everywhere along the real line, although it comes arbitrarily close to zero as z → −∞. There is in fact no complex number ${\displaystyle z}$ for which ${\displaystyle \Gamma (z)=0}$, and hence the reciprocal gamma function ${\textstyle {\frac {1}{\Gamma (z)}}}$ is an entire function, with zeros at ${\displaystyle z=0,-1,-2,\ldots }$.[1]

On the real line, the gamma function has a local minimum at z_min ≈ +1.46163214496836234126[13] where it attains the value Γ(z_min) ≈ +0.88560319441088870027.[14] The gamma function rises to either side of this minimum. The solution to Γ(z − 0.5) = Γ(z + 0.5) is z = +1.5 and the common value is Γ(1) = Γ(2) = +1. The positive solution to Γ(z − 1) = Γ(z + 1) is z = φ ≈ +1.618, the golden ratio, and the common value is Γ(φ − 1) = Γ(φ + 1) = φ! ≈ +1.44922960226989660037.[15]

The gamma function must alternate sign between its poles at the non-positive integers because the product in the forward recurrence contains an odd number of negative factors if the number of poles between ${\displaystyle z}$ and ${\displaystyle z+n}$ is odd, and an even number if the number of poles is even.[12] The extrema values of the Gamma function between the non-positive integers are Γ(-0.50408300826454938526...[16]) = -3.54464361115500508912..., Γ(-1.57349847316239045877...[17]) = 2.30240725833968013582..., Γ(-2.61072086844414465000...[18]) = -0.88813635840124192009..., Γ(-3.63529336643690109783...[19]) = 0.24512753983436625043..., Γ(-4.65323776174314244171...[20]) = -0.05277963958731940076..., etc.

There are many formulas, besides the Euler integral of the second kind, that express the gamma function as an integral. For instance, when the real part of z is positive,[21]

$${\displaystyle \Gamma (z)=\int _{-\infty }^{\infty }e^{zt-e^{t}}\,dt}$$

and[22]

$${\displaystyle \Gamma (z)=\int _{0}^{1}\left(\log {\frac {1}{t}}\right)^{z-1}\,dt,}$$

$${\displaystyle \Gamma (z)=2\int _{0}^{\infty }t^{2z-1}e^{-t^{2}}\,dt}$$

where the three integrals respectively follow from the substitutions ${\displaystyle t=e^{-x}}$, ${\displaystyle t=-\log x}$ [23] and ${\displaystyle t=x^{2}}$[24] in Euler's second integral. The last integral in particular makes clear the connection between the gamma function at half integer arguments and the Gaussian integral: if we let ${\displaystyle z=1/2}$ we get ${\textstyle \Gamma (1/2)={\sqrt {\pi }}=2\int _{0}^{\infty }e^{-t^{2}}\,dt}$.

Binet's first integral formula for the gamma function states that, when the real part of z is positive, then:[25]

$${\displaystyle \log \Gamma (z)=\left(z-{\frac {1}{2}}\right)\log z-z+{\frac {1}{2}}\log(2\pi )+\int _{0}^{\infty }\left({\frac {1}{2}}-{\frac {1}{t}}+{\frac {1}{e^{t}-1}}\right){\frac {e^{-tz}}{t}}\,dt.}$$

The integral on the right-hand side may be interpreted as a Laplace transform. That is,

$${\displaystyle \log \left(\Gamma (z)\left({\frac {e}{z}}\right)^{z}{\sqrt {2\pi z}}\right)={\mathcal {L}}\left({\frac {1}{2t}}-{\frac {1}{t^{2}}}+{\frac {1}{t(e^{t}-1)}}\right)(z).}$$

Binet's second integral formula states that, again when the real part of z is positive, then:[26]

$${\displaystyle \log \Gamma (z)=\left(z-{\frac {1}{2}}\right)\log z-z+{\frac {1}{2}}\ln(2\pi )+2\int _{0}^{\infty }{\frac {\arctan(t/z)}{e^{2\pi t}-1}}\,dt.}$$

Let C be a Hankel contour, meaning a path that begins and ends at the point ∞ on the Riemann sphere, whose unit tangent vector converges to −1 at the start of the path and to 1 at the end, which has winding number 1 around 0, and which does not cross [0, ∞). Fix a branch of ${\displaystyle \log(-t)}$ by taking a branch cut along [0, ∞) and by taking ${\displaystyle \log(-t)}$ to be real when t is on the negative real axis. Assume z is not an integer. Then Hankel's formula for the gamma function is:[27]

$${\displaystyle \Gamma (z)=-{\frac {1}{2i\sin \pi z}}\int _{C}(-t)^{z-1}e^{-t}\,dt,}$$

where ${\displaystyle (-t)^{z-1}}$ is interpreted as ${\displaystyle \exp((z-1)\log(-t))}$. The reflection formula leads to the closely related expression

$${\displaystyle {\frac {1}{\Gamma (z)}}={\frac {i}{2\pi }}\int _{C}(-t)^{-z}e^{-t}\,dt,}$$

again valid whenever z is not an integer.

The gamma function can also be represented by a sum of two continued fractions:[28][29]