Gamma function

For the gamma function of ordinals, see Veblen function. For the gamma distribution in statistics, see Gamma distribution. For the function used in video and image color representations, see Gamma correction.

In mathematics, the gamma function (represented by Γ, the capital letter gamma from the Greek alphabet) is one commonly used extension of the factorial function to complex numbers. The gamma function is defined for all complex numbers except the non-positive integers. For every positive integer n,

Gamma
The gamma function along part of the real axis
General information
General definition	${\displaystyle \Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}\,dt}$
Fields of application	Calculus, mathematical analysis, statistics, physics

$${\displaystyle \Gamma (n)=(n-1)!\,.}$$

Derived by Daniel Bernoulli, for complex numbers with a positive real part, the gamma function is defined via a convergent improper integral:

$${\displaystyle \Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}{\text{ d}}t,\ \qquad \Re (z)>0\,.}$$

The gamma function then is defined as the analytic continuation of this integral function to a meromorphic function that is holomorphic in the whole complex plane except zero and the negative integers, where the function has simple poles.

The gamma function has no zeros, so the reciprocal gamma function 1/Γ(z) is an entire function. In fact, the gamma function corresponds to the Mellin transform of the negative exponential function:

$${\displaystyle \Gamma (z)={\mathcal {M}}\{e^{-x}\}(z)\,.}$$

Other extensions of the factorial function do exist, but the gamma function is the most popular and useful. It is a component in various probability-distribution functions, and as such it is applicable in the fields of probability and statistics, as well as combinatorics.

Motivation

 

${\displaystyle \Gamma (x+1)}$  interpolates the factorial function to non-integer values.

The gamma function can be seen as a solution to the following interpolation problem:

"Find a smooth curve that connects the points (x, y) given by y = (x − 1)! at the positive integer values for x."

A plot of the first few factorials suggests that such a curve can be drawn, but it would be preferable to have a formula that precisely describes the curve, in which the number of operations does not depend on the size of x. The simple formula for the factorial, x! = 1 × 2 × ⋯ × x, cannot be used directly for non-integer values of x since it is only valid when x is a natural number (or positive integer). There are, relatively speaking, no such simple solutions for factorials; no finite combination of sums, products, powers, exponential functions, or logarithms will suffice to express x!; but it is possible to find a general formula for factorials using tools such as integrals and limits from calculus. A good solution to this is the gamma function.^[1]

There are infinitely many continuous extensions of the factorial to non-integers: infinitely many curves can be drawn through any set of isolated points. The gamma function is the most useful solution in practice, being analytic (except at the non-positive integers), and it can be defined in several equivalent ways. However, it is not the only analytic function that extends the factorial, as adding to it any analytic function that is zero on the positive integers, such as k sin mπx for an integer m, will give another function with that property.^[1] Such a function is known as a pseudogamma function, the most famous being the Hadamard function.^[2]

 

The gamma function, Γ(z) in blue, plotted along with Γ(z) + sin(πz) in green. Notice the intersection at positive integers. Both are valid analytic continuations of the factorials to the non-integers.

A more restrictive property than satisfying the above interpolation is to satisfy the recurrence relation defining a translated version of the factorial function,^[3][4]

$${\displaystyle f(1)=1,}$$

 

$${\displaystyle f(x+1)=xf(x),}$$

 

for any positive real number x. But this would allow for multiplication by any function g(x) satisfying both g(x) = g(x+1) for all real numbers x and g(0) = 1, such as the function g(x) = e ^{k sin 2mπx}. One of several ways to resolve the ambiguity comes from the Bohr–Mollerup theorem. It states that when the condition that f be logarithmically convex (or "super-convex",^[5] meaning that ${\displaystyle \ln \circ f}$  is convex) is added, it uniquely determines f for positive, real inputs. From there, the gamma function can be extended to all real and complex values (except the negative integers and zero) by using the unique analytic continuation of f.^[6]

Definition

Main definition

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 and legend and 1000 plot points created with Mathematica

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

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]

Alternative definitions

There are many equivalent definitions.

Euler's definition as an infinite product

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.

Weierstrass's definition

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.

Properties

General

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 created in Mathematica

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

 

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]

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

 

Inequalities

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),

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

 

Stirling's formula

Main article: Stirling's approximation

 

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:

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

 

Residues

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]

Minima and maxima

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.

Integral representations

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.

Continued fraction representation

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

$${\displaystyle \Gamma (z)={\cfrac {e^{-1}}{2+0-z+1{\cfrac {z-1}{2+2-z+2{\cfrac {z-2}{2+4-z+3{\cfrac {z-3}{2+6-z+4{\cfrac {z-4}{2+8-z+5{\cfrac {z-5}{2+10-z+\ddots }}}}}}}}}}}}+{\cfrac {e^{-1}}{z+0-{\cfrac {z+0}{z+1+{\cfrac {1}{z+2-{\cfrac {z+1}{z+3+{\cfrac {2}{z+4-{\cfrac {z+2}{z+5+{\cfrac {3}{z+6-\ddots }}}}}}}}}}}}}}}$$

 

where ${\displaystyle z\in \mathbb {C} }$ .

Fourier series expansion

The logarithm of the gamma function has the following Fourier series expansion for ${\displaystyle 0<z<1:}$ 

$${\displaystyle \ln \Gamma (z)=\left({\frac {1}{2}}-z\right)(\gamma +\ln 2)+(1-z)\ln \pi -{\frac {1}{2}}\ln \sin(\pi z)+{\frac {1}{\pi }}\sum _{n=1}^{\infty }{\frac {\ln n}{n}}\sin(2\pi nz),}$$

 

which was for a long time attributed to Ernst Kummer, who derived it in 1847.^[30][31] However, Iaroslav Blagouchine discovered that Carl Johan Malmsten first derived this series in 1842.^[32][33]

Raabe's formula

In 1840 Joseph Ludwig Raabe proved that

$${\displaystyle \int _{a}^{a+1}\ln \Gamma (z)\,dz={\tfrac {1}{2}}\ln 2\pi +a\ln a-a,\quad a>0.}$$

October 18, 2023