Gamma function

Gamma
Gamma
	The gamma function along part of the real axis
General information
General definition
Fields of application	Calculus, mathematical analysis, statistics, physics

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 (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:

\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.^{[clarification needed]}

The gamma function has no zeros, so the reciprocal gamma function $.mw-parser-output .sfrac{white-space:nowrap}.mw-parser-output .sfrac.tion,.mw-parser-output .sfrac .tion{display:inline-block;vertical-align:-0.5em;font-size:85%;text-align:center}.mw-parser-output .sfrac .num{display:block;line-height:1em;margin:0.0em 0.1em;border-bottom:1px solid}.mw-parser-output .sfrac .den{display:block;line-height:1em;margin:0.1em 0.1em}.mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);clip-path:polygon(0px 0px,0px 0px,0px 0px);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px}1/Γ(z)$ is an entire function. In fact, the gamma function corresponds to the Mellin transform of the negative exponential function:

\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[edit]

The gamma function can be seen as a solution to the interpolation problem of finding a smooth curve $y=f(x)$ that connects the points of the factorial sequence: $(x,y)=(n,n!)$ for all positive integer values of $n$ . The simple formula for the factorial, $x! = 1 \times 2 \times \dots \times x$ is only valid when $x$ is a positive integer, and no elementary function has this property, but a good solution is the gamma function $f(x)=\Gamma (x+1)$ .^[1]

The gamma function is not only smooth but analytic (except at the non-positive integers), and it can be defined in several explicit ways. However, it is not the only analytic function that extends the factorial, as one may add any analytic function that is zero on the positive integers, such as $k\sin(m\pi x)$ for an integer $m$ .^[1] Such a function is known as a pseudogamma function, the most famous being the Hadamard function.^[2]

A more restrictive requirement is the functional equation which interpolates the shifted factorial $f(n)=(n{-}1)!$ :^[3]^[4]

f(x+1)=xf(x)\ {\text{ for any }}x>0,\qquad f(1)=1.

But this still does not give a unique solution, since it allows for multiplication by any periodic function $g(x)$ with $g(x)=g(x+1)$ and $g(0)=1$ , such as $g(x)=e^{k\sin(m\pi x)}$ . One way to resolve the ambiguity is the Bohr–Mollerup theorem, which shows that $f(x)=\Gamma (x)$ is the unique interpolating function over the positive reals which is logarithmically convex (super-convex),^[5] meaning that $y=\ln f(x)$ is convex.^[6]

Definition[edit]

Main definition[edit]

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

\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:

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

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

We can calculate $\Gamma (1)$ :

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

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 $\Gamma (z)$ 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[edit]

There are many equivalent definitions.

Euler's definition as an infinite product[edit]

For a fixed integer $m$ , as the integer $n$ increases, we have that ^[7]

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

If $m$ 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 $m$ is replaced by an arbitrary complex number $z$ ,

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

Multiplying both sides by

(z-1)!

gives

{\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

z

except the non-positive integers, which fail because of a division by zero. Intuitively, this formula indicates that

\Gamma (z)

is approximately the result of computing

\Gamma (n+1)=n!

for some large integer

n

, multiplying by

(n+1)^{z}

to approximate

\Gamma (n+z+1)

, and using the relationship

\Gamma (x+1)=x\Gamma (x)

backwards

n+1

times to get an approximation for

\Gamma (z)

; and furthermore that this approximation becomes exact as

n

increases to infinity.

Weierstrass's definition[edit]

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

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

where

\gamma \approx 0.577216

is the Euler–Mascheroni constant.^[1] This is the Hadamard product of

1/\Gamma (z)

in a rewritten form. The utility of this definition cannot be overstated as it appears in a certain identity involving pi.^{[citation needed]}

Proof of equivalence of the three definitions

Equivalence of the integral definition and Weierstrass definition

By the integral definition, the relation $\Gamma (z+1)=z\Gamma (z)$ and Hadamard factorization theorem,

{\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

c_{1},c_{2}

since

1/\Gamma

is an entire function of order

1

. Since

z\Gamma (z)\to 1

as

z\to 0

,

c_{2}=0

(or an integer multiple of

2\pi i

) and since

\Gamma (1)=1

,

{\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

c_{1}=\gamma +2\pi ik

for some integer

k

. Since

\Gamma (z)\in \mathbb {R}

for

z\in \mathbb {R} \setminus \mathbb {Z} _{0}^{-}

, we have

k=0

and

{\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

{\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

\Gamma _{n}(z)={\frac {n!n^{z}}{z(z+1)\cdots (z+n)}}

and

G_{n}(z)={\frac {(n-1)!n^{z}}{z(z+1)\cdots (z+n-1)}}.

Then

\Gamma _{n}(z)={\frac {n}{z+n}}G_{n}(z)

and

\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

\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

{\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

n\to \infty

gives the desired result.

Properties[edit]

General[edit]

Besides the fundamental property discussed above:

\Gamma (z+1)=z\ \Gamma (z)

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

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

which implies

\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

\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

\Gamma (z)={\frac {1}{z}}\prod _{n=1}^{\infty }{\frac {(1+1/n)^{z}}{1+z/n}}

to compute

{\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

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

C_{R}

with vertices at

R

,

-R

,

R+2\pi i

and

-R+2\pi i

where

R\in \mathbb {R} ^{+}

. Then by the residue theorem,

\int _{C_{R}}{\frac {e^{az}}{1+e^{z}}}\,dz=-2\pi ie^{a\pi i}.

Let

I_{R}=\int _{-R}^{R}{\frac {e^{ax}}{1+e^{x}}}\,dx

and let

I_{R}'

be the analogous integral over the top side of the rectangle. Then

I_{R}\to I

as

R\to \infty

and

I_{R}'=-e^{2\pi ia}I_{R}

. If

A_{R}

denotes the right vertical side of the rectangle, then

\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

C

and since

a<1

, the integral tends to

0

as

R\to \infty

. Analogously, the integral over the left vertical side of the rectangle tends to

0

as

R\to \infty

. Therefore

I-e^{2\pi ia}I=-2\pi ie^{a\pi i},

from which

I={\frac {\pi }{\sin \pi a}},\quad a\in (0,1).

Then

\Gamma (1-z)=\int _{0}^{\infty }e^{-u}u^{-z}\,du=t\int _{0}^{\infty }e^{-vt}(vt)^{-z}\,dv,\quad t>0

and

{\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

z\in (0,1)

proves it for all

z\in \mathbb {C} \setminus \mathbb {Z}

by analytic continuation.

Derivation of the Legendre duplication formula

The beta function can be represented as

\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 $z_{1}=z_{2}=z$ yields

{\frac {\Gamma ^{2}(z)}{\Gamma (2z)}}=\int _{0}^{1}t^{z-1}(1-t)^{z-1}\,dt.

After the substitution $t={\frac {1+u}{2}}$ we get

{\frac {\Gamma ^{2}(z)}{\Gamma (2z)}}={\frac {1}{2^{2z-1}}}\int _{-1}^{1}\left(1-u^{2}\right)^{z-1}\,du.

The function $(1-u^{2})^{z-1}$ is even, hence

2^{2z-1}\Gamma ^{2}(z)=2\Gamma (2z)\int _{0}^{1}(1-u^{2})^{z-1}\,du.

Now assume

\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

\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

2^{2z-1}\Gamma ^{2}(z)=\Gamma (2z)\mathrm {B} \left({\frac {1}{2}},z\right).

Since

\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:

\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):

\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:

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

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

|\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:

{\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 $z=bi$ .

\Gamma (bi)\Gamma (1-bi)={\frac {\pi }{\sin \pi bi}}

Applying the recurrence relation to the second term, we have

-bi\cdot \Gamma (bi)\Gamma (-bi)={\frac {\pi }{\sin \pi bi}}

which with simple rearrangement gives

\Gamma (bi)\Gamma (-bi)={\frac {\pi }{-bi\sin \pi bi}}={\frac {\pi }{b\sinh \pi b}}

Second, consider the reflection formula applied to $z={\tfrac {1}{2}}+bi$ .

\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 $z$ 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

\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

u={\sqrt {z}}

in the integral definition of the gamma function, resulting in a Gaussian integral. In general, for non-negative integer values of

n

we have:

{\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

(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 $\Gamma (r)$ where $r$ 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 $\Gamma (r)$ are not known to be expressible by themselves in terms of elementary functions. It has been proved that $\Gamma (n+r)$ is a transcendental number and algebraically independent of $\pi$ for any integer $n$ 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, $ψ (0) (z)$ :

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

For a positive integer

m

the derivative of the gamma function can be calculated as follows:

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

{\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

z

, and using the technique of differentiation under the integral sign.)

Using the identity

\Gamma ^{(n)}(1)=(-1)^{n}B_{n}(\gamma ,1!\zeta (2),\ldots ,(n-1)!\zeta (n))

where

\zeta (z)

is the Riemann zeta function, and

B_{n}

is the

n

-th Bell polynomial, we have in particular the Laurent series expansion of the gamma function ^[11]

\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[edit]

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 $x_{1}$ and $x_{2}$ , and for any $t\in [0,1]$ , $\Gamma (tx_{1}+(1-t)x_{2})\leq \Gamma (x_{1})^{t}\Gamma (x_{2})^{1-t}.$
For any two positive real numbers $x_{1}$ and $x_{2}$ , and $x_{2}$ > $x_{1}$ $\left({\frac {\Gamma (x_{2})}{\Gamma (x_{1})}}\right)^{\frac {1}{x_{2}-x_{1}}}>\exp \left({\frac {\Gamma '(x_{1})}{\Gamma (x_{1})}}\right).$
For any positive real number $x$ , $\Gamma ''(x)\Gamma (x)>\Gamma '(x)^{2}.$

The last of these statements is, essentially by definition, the same as the statement that $\psi ^{(1)}(x)>0$ , where $\psi ^{(1)}$ is the polygamma function of order 1. To prove the logarithmic convexity of the gamma function, it therefore suffices to observe that $\psi ^{(1)}$ 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 $x_{1},\ldots ,x_{n}$ and $a_{1},\ldots ,a_{n}$ ,

\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 $s \in (0, 1)$ ,

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

Stirling's formula[edit]

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

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

where the symbol

\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,

\exp(\beta x)

, for any fixed value of

\beta

.

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

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

When writing the error term as an infinite product, Stirling's formula can be used to define the gamma function: ^[12]

\Gamma \left(x\right)={\sqrt {\frac {2\pi }{x}}}{\frac {x}{e}}^{x}\prod _{n=0}^{\infty }e^{-1}\left(1+{\frac {1}{x+n}}\right)^{{\frac {1}{2}}+x+n}

Residues[edit]

The behavior for non-positive $z$ is more intricate. Euler's integral does not converge for $\Re (z)\leq 0$ , 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]

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

choosing

n

such that

z+n

is positive. The product in the denominator is zero when

z

equals any of the integers

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 $f$ of a complex variable $z$ , at a simple pole $c$ , the residue of $f$ is given by:

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

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

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

The numerator at

z=-n,

is

\Gamma (z+n+1)=\Gamma (1)=1

and the denominator

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:^[13]

\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 \to -\infty

. There is in fact no complex number

z

for which

\Gamma (z)=0

, and hence the reciprocal gamma function

{\textstyle {\frac {1}{\Gamma (z)}}}

is an entire function, with zeros at

z=0,-1,-2,\ldots

.^[1]

Minima and maxima[edit]

On the real line, the gamma function has a local minimum at $z min \approx +1.46163 214496836234126$ ^[14] where it attains the value $Γ(z min) \approx +0.88560 319441088870027$ .^[15] 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 $z = φ \approx +1.618$ , the golden ratio, and the common value is $Γ(φ - 1) = Γ(φ + 1) = φ! \approx +1.44922 960226989660037$ .^[16]

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 $z$ and $z+n$ is odd, and an even number if the number of poles is even.^[13] The values at the local extrema of the gamma function along the real axis between the non-positive integers are:

Γ(-0.50408 3008264455 40925... [17]) = -3.54464 3611155005 08912...

,

Γ(-1.57349 8473162390 45877... [18]) = 2.30240 7258339680 13582...

,

Γ(-2.61072 0868444144 65000... [19]) = -0.88813 6358401241 92009...

,

Γ(-3.63529 3366436901 09783... [20]) = 0.24512 7539834366 25043...

,

Γ(-4.65323 7761743142 44171... [21]) = -0.05277 9639587319 40076...

, etc.

Integral representations[edit]

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,^[22]

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

and^[23]

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

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

where the three integrals respectively follow from the substitutions

t=e^{-x}

,

t=-\log x

^[24] and

t=x^{2}

^[25] 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

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:^[26]

\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,

\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:^[27]

\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 $\infty$ 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 $[0, \infty)$ . Fix a branch of $\log(-t)$ by taking a branch cut along $[0, \infty)$ and by taking $\log(-t)$ 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:^[28]

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

where

(-t)^{z-1}

is interpreted as

\exp((z-1)\log(-t))

. The reflection formula leads to the closely related expression

{\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[edit]

The gamma function can also be represented by a sum of two continued fractions:^[29]^[30]

\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

z\in \mathbb {C}

.

Fourier series expansion[edit]

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

\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.^[31]^[32] However, Iaroslav Blagouchine discovered that Carl Johan Malmsten first derived this series in 1842.^[33]^[34]

Raabe's formula[edit]

In 1840 Joseph Ludwig Raabe proved that

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

In particular, if

a=0

then

\int _{0}^{1}\ln \Gamma (z)\,dz={\tfrac {1}{2}}\ln 2\pi .

The latter can be derived taking the logarithm in the above multiplication formula, which gives an expression for the Riemann sum of the integrand. Taking the limit for $a\to \infty$ gives the formula.

Pi function[edit]

An alternative notation that was originally introduced by Gauss is the $\Pi$ -function, which, in terms of the gamma function, is

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

so that

\Pi (n)=n!

for every non-negative integer

n

.

Using the pi function the reflection formula takes on the form

\Pi (z)\Pi (-z)={\frac {\pi z}{\sin(\pi z)}}={\frac {1}{\operatorname {sinc} (z)}}

where

sinc

is the normalized sinc function, while the multiplication theorem takes on the form

\Pi \left({\frac {z}{m}}\right)\,\Pi \left({\frac {z-1}{m}}\right)\cdots \Pi \left({\frac {z-m+1}{m}}\right)=(2\pi )^{\frac {m-1}{2}}m^{-z-{\frac {1}{2}}}\Pi (z)\ .

We also sometimes find

\pi (z)={\frac {1}{\Pi (z)}}\ ,

which is an entire function, defined for every complex number, just like the reciprocal gamma function. That

\pi (z)

is entire entails it has no poles, so

\Pi \left(z\right)

, like

\Gamma \left(z\right)

, has no zeros.

The volume of an $n$ -ellipsoid with radii $r 1, \dots, r n$ can be expressed as

V_{n}(r_{1},\dotsc ,r_{n})={\frac {\pi ^{\frac {n}{2}}}{\Pi \left({\frac {n}{2}}\right)}}\prod _{k=1}^{n}r_{k}.

Relation to other functions[edit]

In the first integral above, which defines the gamma function, the limits of integration are fixed. The upper and lower incomplete gamma functions are the functions obtained by allowing the lower or upper (respectively) limit of integration to vary.
The gamma function is related to the beta function by the formula $\mathrm {B} (z_{1},z_{2})=\int _{0}^{1}t^{z_{1}-1}(1-t)^{z_{2}-1}\,dt={\frac {\Gamma (z_{1})\,\Gamma (z_{2})}{\Gamma (z_{1}+z_{2})}}.$
The logarithmic derivative of the gamma function is called the digamma function; higher derivatives are the polygamma functions.
The analog of the gamma function over a finite field or a finite ring is the Gaussian sums, a type of exponential sum.
The reciprocal gamma function is an entire function and has been studied as a specific topic.
The gamma function also shows up in an important relation with the Riemann zeta function, $\zeta (z)$ . $\pi ^{-{\frac {z}{2}}}\;\Gamma \left({\frac {z}{2}}\right)\zeta (z)=\pi ^{-{\frac {1-z}{2}}}\;\Gamma \left({\frac {1-z}{2}}\right)\;\zeta (1-z).$ It also appears in the following formula: $\zeta (z)\Gamma (z)=\int _{0}^{\infty }{\frac {u^{z}}{e^{u}-1}}\,{\frac {du}{u}},$ which is valid only for $\Re (z)>1$ .
The logarithm of the gamma function satisfies the following formula due to Lerch: $\log \Gamma (z)=\zeta _{H}'(0,z)-\zeta '(0),$ where $\zeta _{H}$ is the Hurwitz zeta function, $\zeta$ is the Riemann zeta function and the prime ( $'$ ) denotes differentiation in the first variable.
The gamma function is related to the stretched exponential function. For instance, the moments of that function are $\langle \tau ^{n}\rangle \equiv \int _{0}^{\infty }t^{n-1}\,e^{-\left({\frac {t}{\tau }}\right)^{\beta }}\,\mathrm {d} t={\frac {\tau ^{n}}{\beta }}\Gamma \left({n \over \beta }\right).$

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Gamma function

Motivation[edit]

Definition[edit]

Main definition[edit]

Alternative definitions[edit]

Euler's definition as an infinite product[edit]

Weierstrass's definition[edit]

Properties[edit]

General[edit]

Inequalities[edit]

Stirling's formula[edit]

Residues[edit]

Minima and maxima[edit]

Integral representations[edit]

Continued fraction representation[edit]

Fourier series expansion[edit]

Raabe's formula[edit]

Pi function[edit]

Relation to other functions[edit]

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