Editing Gamma function (section)

== Properties ==

=== General ===

Besides the fundamental property discussed above:
<math display="block">\Gamma(z+1) = z\ \Gamma(z)</math>
other important functional equations for the gamma function are [[reflection formula|Euler's reflection formula]]
<math display="block">\Gamma(1-z) \Gamma(z) = \frac{\pi}{\sin \pi z}, \qquad z \not\in \Z</math>
which implies
<math display="block">\Gamma(z - n) = (-1)^{n-1} \; \frac{\Gamma(-z) \Gamma(1+z)}{\Gamma(n+1-z)}, \qquad n \in \Z</math>
and the [[Multiplication theorem#Gamma function–Legendre formula|Legendre duplication formula]]
<math display="block">\Gamma(z) \Gamma\left(z + \tfrac12\right) = 2^{1-2z} \; \sqrt{\pi} \; \Gamma(2z).</math>

{{Collapse top|title=Derivation of Euler's reflection formula}}
'''Proof 1'''

With Euler's infinite product
<math display=block>\Gamma(z) = \frac1z \prod_{n=1}^{\infty} \frac{(1+1/n)^z}{1 + z/n}</math> compute
<math display=block>\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}\,,</math>
where the last equality is a [[Sine#Partial fraction and product expansions of complex sine|known result]].  A similar derivation begins with Weierstrass's definition.

'''Proof 2'''

First prove that
<math display="block">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).</math>
Consider the positively oriented rectangular contour <math>C_R</math> with vertices at <math>R</math>, <math>-R</math>, <math>R+2\pi i</math> and <math>-R+2\pi i</math> where <math>R\in\mathbb{R}^+</math>. Then by the [[residue theorem]],
<math display="block">\int_{C_R}\frac{e^{az}}{1+e^z}\, dz=-2\pi ie^{a\pi i}.</math>
Let
<math display="block">I_R=\int_{-R}^R \frac{e^{ax}}{1+e^x}\, dx</math>
and let <math>I_R'</math> be the analogous integral over the top side of the rectangle. Then <math>I_R\to I</math> as <math>R\to\infty</math> and <math>I_R'=-e^{2\pi i a}I_R</math>. If <math>A_R</math> denotes the right vertical side of the rectangle, then
<math display="block">\left|\int_{A_R} \frac{e^{az}}{1+e^z}\, dz\right|\le \int_0^{2\pi}\left|\frac{e^{a(R+it)}}{1+e^{R+it}}\right|\, dt\le Ce^{(a-1)R}</math>
for some constant <math>C</math> and since <math>a<1</math>, the integral tends to <math>0</math> as <math>R\to\infty</math>. Analogously, the integral over the left vertical side of the rectangle tends to <math>0</math> as <math>R\to\infty</math>. Therefore
<math display="block">I-e^{2\pi ia}I=-2\pi ie^{a\pi i},</math>
from which
<math display="block">I=\frac{\pi}{\sin \pi a},\quad a\in (0,1).</math>
Then
<math display="block">\Gamma (1-z)=\int_0^\infty e^{-u}u^{-z}\, du=t\int_0^\infty e^{-vt}(vt)^{-z}\, dv,\quad t>0</math>
and
<math display="block">\begin{align}\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{align}</math>
Proving the reflection formula for all <math>z\in (0,1)</math> proves it for all <math>z\in\mathbb{C}\setminus\mathbb{Z}</math> by analytic continuation.
{{Collapse bottom}}

{{Collapse top|title=Derivation of the Legendre duplication formula}}

The [[beta function]] can be represented as
<math display="block">\Beta (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.</math>

Setting <math>z_1=z_2=z</math> yields
<math display="block">\frac{\Gamma^2(z)}{\Gamma (2z)}=\int_0^1 t^{z-1}(1-t)^{z-1} \, dt.</math>

After the substitution <math>t=\frac{1+u}{2}</math>:
<math display="block">\frac{\Gamma^2(z)}{\Gamma (2z)}=\frac{1}{2^{2z-1}}\int_{-1}^1 \left(1-u^{2}\right)^{z-1} \, du.</math>

The function <math>(1-u^2)^{z-1}</math> is even, hence
<math display="block">2^{2z-1}\Gamma^2(z)=2\Gamma (2z)\int_0^1 (1-u^2)^{z-1} \, du.</math>

Now
<math display="block">\Beta \left(\frac{1}{2},z\right)=\int_0^1 t^{\frac{1}{2}-1}(1-t)^{z-1} \, dt, \quad t=s^2.</math>

Then
<math display="block">\Beta \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.</math>

This implies
<math display="block">2^{2z-1}\Gamma^2(z)=\Gamma (2z)\Beta \left(\frac{1}{2},z\right).</math>

Since
<math display="block">\Beta \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},</math>
the Legendre duplication formula follows:
<math display="block">\Gamma (z)\Gamma \left(z+\frac{1}{2}\right)=2^{1-2z}\sqrt{\pi} \; \Gamma (2z).</math>

{{Collapse bottom}}

The duplication formula is a special case of the [[multiplication theorem]] (see&nbsp;<ref name="ReferenceA">{{dlmf|authorlink=Richard Askey|first=R. A.|last=Askey|first2=R.|last2=Roy|id=8.7|title=Series Expansions|ref=none}}</ref> Eq.&nbsp;5.5.6):
<math display="block">\prod_{k=0}^{m-1}\Gamma\left(z + \frac{k}{m}\right) = (2 \pi)^{\frac{m-1}{2}} \; m^{\frac12 - mz} \; \Gamma(mz).</math>

A simple but useful property, which can be seen from the limit definition, is:
<math display="block">\overline{\Gamma(z)} = \Gamma(\overline{z}) \; \Rightarrow \; \Gamma(z)\Gamma(\overline{z}) \in \mathbb{R} .</math>

In particular, with {{math|1=''z'' = ''a'' + ''bi''}}, this product is
<math display="block">|\Gamma(a+bi)|^2 = |\Gamma(a)|^2 \prod_{k=0}^\infty \frac{1}{1+\frac{b^2}{(a+k)^2}}</math>

If the real part is an integer or a half-integer, this can be finitely expressed in [[Closed-form expression|closed form]]:
<math display="block">
\begin{align}
|\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 \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 \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 \N
\\[-1ex]&
\end{align}
</math>

{{Collapse top|title=Proof of absolute value formulas for arguments of integer or half-integer real part}}

First, consider the reflection formula applied to <math>z=bi</math>.
<math display="block">\Gamma(bi)\Gamma(1-bi)=\frac{\pi}{\sin \pi bi}</math>
Applying the recurrence relation to the second term:
<math display="block">-bi \cdot \Gamma(bi)\Gamma(-bi)=\frac{\pi}{\sin \pi bi}</math>
which with simple rearrangement gives
<math display="block">\Gamma(bi)\Gamma(-bi)=\frac{\pi}{-bi\sin \pi bi}=\frac{\pi}{b\sinh \pi b}</math>

Second, consider the reflection formula applied to <math>z=\tfrac{1}{2}+bi</math>.
<math display="block">\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}</math>

Formulas for other values of <math>z</math> for which the real part is integer or half-integer quickly follow by [[mathematical induction|induction]] using the recurrence relation in the positive and negative directions.

{{Collapse bottom}}

Perhaps the best-known value of the gamma function at a non-integer argument is
<math display="block">\Gamma\left(\tfrac12\right)=\sqrt{\pi},</math>
which can be found by setting <math display="inline">z = \frac{1}{2}</math> in the reflection formula, by using the relation to the [[beta function]] given below with <math display="inline">z_1 = z_2 = \frac{1}{2}</math>, or simply by making the substitution <math>t = u^2</math> in the integral definition of the gamma function, resulting in a [[Gaussian integral]]. In general, for non-negative integer values of <math>n</math> we have:
<math display="block">\begin{align}
\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{align}</math>
where the [[double factorial]] <math>(2n-1)!! = (2n-1)(2n-3)\cdots(3)(1)</math>. See [[Particular values of the gamma function]] for calculated values.

It might be tempting to generalize the result that <math display="inline">\Gamma \left( \frac{1}{2} \right) = \sqrt\pi</math> by looking for a formula for other individual values <math>\Gamma(r)</math> where <math>r</math> is rational, especially because according to [[Digamma function#Gauss's digamma theorem|Gauss's digamma theorem]], it is possible to do so for the closely related [[digamma function]] at every rational value. However, these numbers <math>\Gamma(r)</math> are not known to be expressible by themselves in terms of elementary functions. It has been proved that <math>\Gamma (n + r)</math> is a [[transcendental number]] and [[algebraic independence|algebraically independent]] of <math>\pi</math> for any integer <math>n</math> and each of the fractions <math display="inline">r = \frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{2}{3}, \frac{3}{4}, \frac{5}{6}</math>.<ref>{{cite journal|last=Waldschmidt |first=M. |date=2006 |url=http://www.math.jussieu.fr/~miw/articles/pdf/TranscendencePeriods.pdf |archive-url=https://web.archive.org/web/20060506050646/http://www.math.jussieu.fr/~miw/articles/pdf/TranscendencePeriods.pdf |archive-date=2006-05-06 |url-status=live |title=Transcendence of Periods: The State of the Art |journal=Pure Appl. Math. Quart. |volume=2 |issue=2 |pages=435–463 |doi=10.4310/pamq.2006.v2.n2.a3|doi-access=free}}</ref> 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]],&nbsp;{{math|''ψ''{{isup|(0)}}(''z'')}}:
<math display="block">\Gamma'(z)=\Gamma(z)\psi^{(0)}(z).</math>
For a positive integer&nbsp;{{mvar|m}} the derivative of the gamma function can be calculated as follows:
[[File:Plot of gamma function in the complex plane from -2-i to 6+2i with colors created in Mathematica.svg|alt=Gamma function in the complex plane with colors showing its argument|thumb|Colors showing the argument of the gamma function in the complex plane from {{math|−2 − 2''i''}} to {{math|6 + 2''i''}}]]
<math display="block">\Gamma'(m+1) = m! \left( - \gamma + \sum_{k=1}^m\frac{1}{k} \right)= m! \left( - \gamma + H(m) \right)\,,</math>
where H(m) is the mth [[harmonic number]] and {{math|''γ''}} is the [[Euler–Mascheroni constant]].

For <math>\Re(z) > 0</math> the <math>n</math>th derivative of the gamma function is:
<math display="block">\frac{d^n}{dz^n}\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} (\log t)^n \, dt.</math>
(This can be derived by differentiating the integral form of the gamma function with respect to <math>z</math>, and using the technique of [[differentiation under the integral sign]].)

Using the identity
<math display="block">\Gamma^{(n)}(1)=(-1)^n B_n(\gamma, 1! \zeta(2), \ldots, (n-1)! \zeta(n))</math>
where <math>\zeta(z)</math> is the [[Riemann zeta function]], and <math>B_n</math> is the <math>n</math>-th [[Bell polynomials|Bell polynomial]], we have in particular the [[Laurent series]] expansion of the gamma function <ref>{{Cite web |title=How to obtain the Laurent expansion of gamma function around $z=0$? |url=https://math.stackexchange.com/q/1287555 |access-date=2022-08-17 |website=Mathematics Stack Exchange |language=en}}</ref>
<math display="block">\Gamma(z) = \frac1z - \gamma + \frac12\left(\gamma^2 + \frac{\pi^2}6\right)z - \frac16\left(\gamma^3 + \frac{\gamma\pi^2}2 + 2 \zeta(3)\right)z^2 + O(z^3).</math>

=== 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 <math>x_1</math> and <math>x_2</math>, and for any <math>t \in [0, 1]</math>, <math display="block">\Gamma(tx_1 + (1 - t)x_2) \le \Gamma(x_1)^t\Gamma(x_2)^{1 - t}.</math>
* For any two positive real numbers <math>x_1</math> and <math>x_2</math>, and <math>x_2</math> > <math>x_1</math><math display="block"> \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).</math>
* For any positive real number <math>x</math>, <math display="block"> \Gamma''(x) \Gamma(x) > \Gamma'(x)^2.</math>
The last of these statements is, essentially by definition, the same as the statement that <math>\psi^{(1)}(x) > 0</math>, where <math>\psi^{(1)}</math> is the [[polygamma function]] of order 1. To prove the logarithmic convexity of the gamma function, it therefore suffices to observe that <math>\psi^{(1)}</math> has a series representation which, for positive real {{mvar|x}}, consists of only positive terms.

Logarithmic convexity and [[Jensen's inequality]] together imply, for any positive real numbers <math>x_1, \ldots, x_n</math> and <math>a_1, \ldots, a_n</math>,
<math display="block">\Gamma\left(\frac{a_1x_1 + \cdots + a_nx_n}{a_1 + \cdots + a_n}\right) \le \bigl(\Gamma(x_1)^{a_1} \cdots \Gamma(x_n)^{a_n}\bigr)^{\frac{1}{a_1 + \cdots + a_n}}.</math>

There are also bounds on ratios of gamma functions. The best-known is [[Gautschi's inequality]], which says that for any positive real number {{mvar|x}} and any {{math|''s'' ∈ (0, 1)}},
<math display="block">x^{1 - s} < \frac{\Gamma(x + 1)}{\Gamma(x + s)} < \left(x + 1\right)^{1 - s}.</math>

=== Stirling's formula ===
{{Main|Stirling's approximation}}

[[File:Gamma cplot.svg|thumb|Representation of the gamma function in the complex plane. Each point <math>z</math> is colored according to the argument of {{nowrap|<math>\Gamma(z)</math>.}} The contour plot of the modulus <math>|\Gamma(z)|</math> is also displayed.]]

[[File:Gamma abs 3D.png|thumb|3-dimensional plot of the absolute value of the complex gamma function]]
The behavior of <math>\Gamma(x)</math> for an increasing positive real variable is given by [[Stirling's formula]]
<math display="block">\Gamma(x+1)\sim\sqrt{2\pi x}\left(\frac{x}{e}\right)^x,</math>
where the symbol <math>\sim</math> means asymptotic convergence: the ratio of the two sides converges to 1 in the limit {{nowrap|<math display="inline">x \to + \infty</math>.<ref name="Davis"/>}}  This growth is faster than exponential, <math>\exp(\beta x)</math>, for any fixed value of <math>\beta</math>.

Another useful limit for asymptotic approximations for <math>x \to + \infty</math> is:
<math display="block">  {\Gamma(x+\alpha)}\sim{\Gamma(x)x^\alpha}, \qquad \alpha \in \Complex. </math>

When writing the error term as an infinite product, Stirling's formula can be used to define the gamma function: <ref>{{cite book |last1=Artin |first1=Emil |title=The Gamma Function |date=2015 |page = 24|publisher=Dover }}</ref>
<math display="block"> \Gamma(x) = \sqrt{\frac{2\pi}{x}} \left(\frac{x}{e}\right)^x  \prod_{n=0}^{\infty} \left[\frac{1}{e}\left(1+\frac{1}{x+n}\right)^{x+n+\frac{1}{2}} \right]</math>

=== Extension to negative, non-integer values ===
Although the main definition of the gamma function&mdash;the Euler integral of the second kind&mdash;is only valid (on the real axis) for positive arguments, its domain can be extended with [[analytic continuation]]<ref>{{cite book |last1=Oldham |first1=Keith |last2=Myland |first2=Jan |last3=Spanier |first3=Jerome |title=An Atlas of Functions |date=2010 |publisher=Springer Science & Business Media |location=Ch 43 |isbn=9780387488073 |edition=2}}</ref> to negative arguments by shifting the negative argument to positive values by using either the Euler's reflection formula,
<math display="block">
\Gamma(-x) = \frac{1}{\Gamma(x+1)}\frac{\pi}{\sin\big(\pi(x+1)\big)},
</math>
or the fundamental property,
<math display="block">
\Gamma(-x):=\frac1{-x}\Gamma(-x+1) ,
</math>
when <math>x\not\in\mathbb{Z}</math>. 
For example,
<math display="block">
\Gamma\left(-\frac12\right)=-2\Gamma\left(\frac12\right) .
</math>

=== Residues ===
The behavior for non-positive <math>z</math> is more intricate. Euler's integral does not converge for {{nowrap|<math>\Re(z) \le 0</math>,}} 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,<ref name="Davis" />
<math display="block">\Gamma(z)=\frac{\Gamma(z+n+1)}{z(z+1)\cdots(z+n)},</math>
choosing <math>n</math> such that <math>z + n</math> is positive. The product in the denominator is zero when <math>z</math> equals any of the integers <math>0, -1, -2, \ldots</math>. Thus, the gamma function must be undefined at those points to avoid [[division by zero]]; it is a [[meromorphic function]] with [[simple pole]]s at the non-positive integers.<ref name="Davis" />

For a function <math>f</math> of a complex variable <math>z</math>, at a [[simple pole]] <math>c</math>, the [[Residue (complex analysis)|residue]] of <math>f</math> is given by:
<math display="block">\operatorname{Res}(f,c)=\lim_{z\to c}(z-c)f(z).</math>

For the simple pole <math>z = -n</math>, the recurrence formula can be rewritten as:
<math display="block">(z+n) \Gamma(z)=\frac{\Gamma(z+n+1)}{z(z+1)\cdots(z+n-1)}.</math>
The numerator at <math>z = -n,</math> is
<math display="block">\Gamma(z+n+1) = \Gamma(1) = 1</math>
and the denominator 
<math display="block">z(z+1)\cdots(z+n-1) = -n(1-n)\cdots(n-1-n) = (-1)^n n!.</math>
So the residues of the gamma function at those points are:<ref name="Mathworld">{{MathWorld|urlname=GammaFunction |title=Gamma Function}}</ref>
<math display="block">\operatorname{Res}(\Gamma,-n)=\frac{(-1)^n}{n!}.</math>The gamma function is non-zero everywhere along the real line, although it comes arbitrarily close to zero as {{math|''z'' → −∞}}. There is in fact no complex number <math>z</math> for which <math>\Gamma (z) = 0</math>, and hence the [[reciprocal gamma function]] <math display="inline">\frac {1}{\Gamma (z)}</math> is an [[entire function]], with zeros at <math>z = 0, -1, -2, \ldots</math>.<ref name="Davis" />

=== Minima and maxima ===
On the real line, the gamma function has a local minimum at {{math|''z''<sub>min</sub> ≈ {{gaps|+1.46163|21449|68362|34126}}}}<ref>{{Cite OEIS|A030169|2=Decimal expansion of real number x such that y = Gamma(x) is a minimum}}</ref> where it attains the value {{math|Γ(''z''<sub>min</sub>) ≈ {{gaps|+0.88560|31944|10888|70027}}}}.<ref>{{Cite OEIS|A030171|2=Decimal expansion of real number y such that y = Gamma(x) is a minimum}}</ref> The gamma function rises to either side of this minimum. The solution to {{math|1=Γ(''z'' − 0.5) = Γ(''z'' + 0.5)}} is {{math|1=''z'' = +1.5}} and the common value is {{math|1=Γ(1) = Γ(2) = +1}}.  The positive solution to {{math|1=Γ(''z'' − 1) = Γ(''z'' + 1)}} is {{math|1=''z'' = ''φ'' ≈ +1.618}}, the [[golden ratio]], and the common value is {{math|1=Γ(''φ'' − 1) = Γ(''φ'' + 1) = ''φ''! ≈ {{gaps|+1.44922|96022|69896|60037}}}}.<ref>{{Cite OEIS|A178840|Decimal expansion of the factorial of Golden Ratio}}</ref>

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 <math>z</math> and <math>z + n</math> is odd, and an even number if the number of poles is even.<ref name="Mathworld" />  The values at the local extrema of the gamma function along the real axis between the non-positive integers are:
: {{math|1=Γ({{gaps|−0.50408|30082|64455|40925...}}<ref>{{Cite OEIS|A175472|Decimal expansion of the absolute value of the abscissa of the local maximum of the Gamma function in the interval [ -1,0]}}</ref>) = {{gaps|−3.54464|36111|55005|08912...}}}},
: {{math|1=Γ({{gaps|−1.57349|84731|62390|45877...}}<ref>{{Cite OEIS|A175473|Decimal expansion of the absolute value of the abscissa of the local minimum of the Gamma function in the interval [ -2,-1]}}</ref>) =  {{gaps|2.30240|72583|39680|13582...}}}},
: {{math|1=Γ({{gaps|−2.61072|08684|44144|65000...}}<ref>{{Cite OEIS|A175474|Decimal expansion of the absolute value of the abscissa of the local maximum of the Gamma function in the interval [ -3,-2]}}</ref>) = {{gaps|−0.88813|63584|01241|92009...}}}},
: {{math|1=Γ({{gaps|−3.63529|33664|36901|09783...}}<ref>{{Cite OEIS|A256681|Decimal expansion of the [negated] abscissa of the Gamma function local minimum in the interval [-4,-3]}}</ref>) =  {{gaps|0.24512|75398|34366|25043...}}}},
: {{math|1=Γ({{gaps|−4.65323|77617|43142|44171...}}<ref>{{Cite OEIS|A256682|Decimal expansion of the [negated] abscissa of the Gamma function local maximum in the interval [-5,-4]}}</ref>) = {{gaps|−0.05277|96395|87319|40076...}}}}, 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 {{mvar|z}} is positive,<ref>{{Cite book |last1=Gradshteyn |first1=I. S.|last2=Ryzhik |first2=I. M. |title=Table of Integrals, Series, and Products |edition=Seventh |publisher=Academic Press |year=2007 |isbn=978-0-12-373637-6|page=893}}</ref>
<math display="block">\Gamma (z)=\int_{-\infty}^\infty e^{zt-e^t}\, dt</math>
and<ref>Whittaker and Watson, 12.2 example 1.</ref>
<math display="block">\Gamma(z) = \int_0^1 \left(\log \frac{1}{t}\right)^{z-1}\,dt,</math>
<math display="block">\Gamma(z) = 2c^z\int_{0}^{\infty}t^{2z-1}e^{-ct^{2}}\,dt \,,\; c>0</math>
where the three integrals respectively follow from the substitutions <math>t=e^{-x}</math>, <math>t=-\log x</math> <ref>{{Cite journal |last=Detlef |first=Gronau |title=Why is the gamma function so as it is? |url=https://imsc.uni-graz.at/gronau/TMCS_1_2003.pdf |journal=Imsc.uni-graz.at}}</ref> and <math>t=cx^2</math><ref>{{cite journal |last1=Pascal Sebah |first1=Xavier Gourdon |title=Introduction to the Gamma Function |journal=Numbers Computation |url=https://www.csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf |access-date=30 January 2023 |archive-date=30 January 2023 |archive-url=https://web.archive.org/web/20230130155521/https://www.csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf |url-status=dead }}</ref> 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 <math>z=1/2,\; c=1</math> we get
<math display="block">
\Gamma(1/2)=2\int_{0}^{\infty}e^{-t^{2}}\,dt=\sqrt{\pi} \;.
</math>

Binet's first integral formula for the gamma function states that, when the real part of {{mvar|z}} is positive, then:<ref>Whittaker and Watson, 12.31.</ref>
<math display="block">\operatorname{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.</math>
The integral on the right-hand side may be interpreted as a [[Laplace transform]].  That is,
<math display="block">\log\left(\Gamma(z)\left(\frac{e}{z}\right)^z\sqrt{\frac{z}{2\pi}}\right) = \mathcal{L}\left(\frac{1}{2t} - \frac{1}{t^2} + \frac{1}{t(e^t - 1)}\right)(z).</math>

Binet's second integral formula states that, again when the real part of {{mvar|z}} is positive, then:<ref>Whittaker and Watson, 12.32.</ref>
<math display="block">\operatorname{log\Gamma}(z) = \left(z - \frac{1}{2}\right)\log z - z + \frac{1}{2}\log(2\pi) + 2\int_0^\infty \frac{\arctan(t/z)}{e^{2\pi t} - 1}\,dt.</math>

Let {{math|''C''}} be a [[Hankel contour]], meaning a path that begins and ends at the point {{math|∞}} on the [[Riemann sphere]], whose unit tangent vector converges to {{math|−1}} at the start of the path and to {{math|1}} at the end, which has [[winding number]] 1 around {{math|0}}, and which does not cross {{closed-open|0, ∞}}.  Fix a branch of <math>\log(-t)</math> by taking a branch cut along {{closed-open|0, ∞}} and by taking <math>\log(-t)</math> to be real when {{math|t}} is on the negative real axis.  Assume {{mvar|z}} is not an integer.  Then Hankel's formula for the gamma function is:<ref>Whittaker and Watson, 12.22.</ref>
<math display="block">\Gamma(z) = -\frac{1}{2i\sin \pi z}\int_C (-t)^{z-1}e^{-t}\,dt,</math>
where <math>(-t)^{z-1}</math> is interpreted as <math>\exp((z-1)\log(-t))</math>.  The reflection formula leads to the closely related expression 
<math display="block">\frac{1}{\Gamma(z)} = \frac{i}{2\pi}\int_C (-t)^{-z}e^{-t}\,dt,</math>
again valid whenever {{math|''z''}} is not an integer.

=== Continued fraction representation ===
The gamma function can also be represented by a sum of two [[continued fraction]]s:<ref>{{cite web | url=https://functions.wolfram.com/GammaBetaErf/ExpIntegralE/10/0005/ | title=Exponential integral E: Continued fraction representations (Formula 06.34.10.0005) }}</ref><ref>{{cite web | url=https://functions.wolfram.com/GammaBetaErf/ExpIntegralE/10/0003/ | title=Exponential integral E: Continued fraction representations (Formula 06.34.10.0003) }}</ref>
<math display="block">\begin{aligned}
    \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
                            }
                        }
                    }
                }
            }
        }
    }
\end{aligned}</math>
where <math>z\in\mathbb{C}</math>.

=== Fourier series expansion ===
The [[#Log-gamma function|logarithm of the gamma function]] has the following [[Fourier series]] expansion for <math>0 < z < 1:</math>
<math display="block">\operatorname{log\Gamma}(z) = \left(\frac{1}{2} - z\right)(\gamma + \log 2) + (1 - z)\log\pi - \frac{1}{2}\log\sin(\pi z) + \frac{1}{\pi}\sum_{n=1}^\infty \frac{\log n}{n} \sin (2\pi n z),</math>
which was for a long time attributed to [[Ernst Kummer]], who derived it in 1847.<ref>{{cite book|first1=Harry |last1=Bateman |first2=Arthur |last2=Erdélyi |title=Higher Transcendental Functions |publisher=McGraw-Hill |date=1955 |oclc=627135 }}</ref><ref>{{cite book|first1=H. M. |last1=Srivastava |first2=J. |last2=Choi |title=Series Associated with the Zeta and Related Functions |publisher=Kluwer Academic |location=The Netherlands |date=2001 |isbn=0-7923-7054-6 }}</ref> However, [[Iaroslav Blagouchine]] discovered that [[Carl Johan Malmsten]] first derived this series in 1842.<ref name="iaroslav_06">{{cite journal|doi=10.1007/s11139-013-9528-5 |first=Iaroslav V. |last=Blagouchine |title=Rediscovery of Malmsten's integrals, their evaluation by contour integration methods and some related results |journal=Ramanujan J. |volume=35 |issue=1 |pages=21–110 |date=2014 |s2cid=120943474 |url=https://www.researchgate.net/publication/257381156}}</ref><ref name="iaroslav_06bis">{{cite journal|doi=10.1007/s11139-015-9763-z |first=Iaroslav V. |last=Blagouchine |title=Erratum and Addendum to "Rediscovery of Malmsten's integrals, their evaluation by contour integration methods and some related results" |journal=Ramanujan J. |volume=42 |issue=3 |pages=777–781 |date=2016 |s2cid=125198685 }}</ref>

=== Raabe's formula ===
In 1840 [[Joseph Ludwig Raabe]] proved that
<math display="block">\int_a^{a+1}\log\Gamma(z)\, dz = \tfrac12\log2\pi + a\log a - a,\quad a>0.</math>
In particular, if <math>a = 0</math> then
<math display="block">\int_0^1\log\Gamma(z)\, dz = \tfrac12\log2\pi.</math>

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 <math>a \to \infty</math> gives the formula.

=== Pi function ===
An alternative notation introduced by [[Carl Friedrich Gauss|Gauss]] is the <math>\Pi</math>-function, a shifted version of the gamma function:
<math display="block">\Pi(z) = \Gamma(z+1) = z \Gamma(z) = \int_0^\infty e^{-t} t^z\, dt,</math>
so that <math>\Pi(n) = n!</math> for every non-negative integer <math>n</math>.

Using the pi function, the reflection formula is:
<math display="block">\Pi(z) \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}</math>
using the normalized [[sinc function]]; while the multiplication theorem becomes:
<math display="block">\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-\frac12} \Pi(z)\ .</math>

The shifted [[reciprocal gamma function]] is sometimes denoted <math display="inline">\pi(z) = \frac{1}{\Pi(z)}\ ,</math>
an [[entire function]].

The [[volume of an n-ball|volume of an {{math|''n''}}-ellipsoid]] with radii {{math|''r''{{sub|1}}, …, ''r''{{sub|''n''}}}} can be expressed as
<math display="block">V_n(r_1,\dotsc,r_n)=\frac{\pi^{\frac{n}{2}}}{\Pi\left(\frac{n}{2}\right)} \prod_{k=1}^n r_k.</math>

=== Relation to other functions ===
* In the first integral defining the gamma function, the limits of integration are fixed. The upper [[incomplete gamma function]] is obtained by allowing the lower limit of integration to vary:<math display="block">\Gamma(z,x) = \int_x^\infty t^{z-1} e^{-t} dt.</math>There is a similar lower incomplete gamma function.
* The gamma function is related to Euler's [[beta function]] by the formula <math display="block">\Beta(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)}.</math>
* The [[logarithmic derivative]] of the gamma function is called the [[digamma function]]; higher derivatives are the [[polygamma function]]s.
* The analog of the gamma function over a [[finite field]] or a [[finite ring]] is the [[Gaussian sum]]s, 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]], <math>\zeta (z)</math>. <math display="block">\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).</math> It also appears in the following formula: <math display="block">\zeta(z) \Gamma(z) = \int_0^\infty \frac{u^{z}}{e^u - 1} \, \frac{du}{u},</math> which is valid only for <math>\Re (z) > 1</math>.{{pb}} The logarithm of the gamma function satisfies the following formula due to Lerch: <math display="block">\operatorname{log\Gamma}(z) = \zeta_H'(0,z) - \zeta'(0),</math> where <math>\zeta_H</math> is the [[Hurwitz zeta function]], <math>\zeta</math> is the Riemann zeta function and the prime ({{math|′}}) denotes differentiation in the first variable.
* The gamma function is related to the [[stretched exponential function]]. For instance, the moments of that function are <math display="block">\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).</math>

=== Particular values ===
{{Main|Particular values of the gamma function}}
Including up to the first 20 digits after the decimal point, some particular values of the gamma function are:
<math display="block">\begin{array}{rcccl}
\Gamma\left(-\tfrac{3}{2}\right) &=& \tfrac{4\sqrt{\pi}}{3} &\approx& +2.36327\,18012\,07354\,70306 \\
\Gamma\left(-\tfrac{1}{2}\right) &=& -2\sqrt{\pi} &\approx& -3.54490\,77018\,11032\,05459 \\
\Gamma\left(\tfrac{1}{2}\right) &=& \sqrt{\pi} &\approx& +1.77245\,38509\,05516\,02729 \\
\Gamma(1) &=& 0! &=& +1 \\
\Gamma\left(\tfrac{3}{2}\right) &=& \tfrac{\sqrt{\pi}}{2} &\approx& +0.88622\,69254\,52758\,01364 \\
\Gamma(2) &=& 1! &=& +1 \\
\Gamma\left(\tfrac{5}{2}\right) &=& \tfrac{3\sqrt{\pi}}{4} &\approx& +1.32934\,03881\,79137\,02047 \\
\Gamma(3) &=& 2! &=& +2 \\
\Gamma\left(\tfrac{7}{2}\right) &=& \tfrac{15\sqrt{\pi}}{8} &\approx& +3.32335\,09704\,47842\,55118 \\
\Gamma(4) &=& 3! &=& +6
\end{array}</math>
(These numbers can be found in the [[On-Line Encyclopedia of Integer Sequences|OEIS]].<ref>{{Cite OEIS|A245886|Decimal expansion of Gamma(-3/2), where Gamma is Euler's gamma function}}</ref><ref>{{Cite OEIS|A019707|Decimal expansion of sqrt(Pi)/5}}</ref><ref>{{Cite OEIS|A002161|Decimal expansion of square root of Pi}}</ref><ref>{{Cite OEIS|A019704|Decimal expansion of sqrt(Pi)/2}}</ref><ref>{{Cite OEIS|A245884|Decimal expansion of Gamma(5/2), where Gamma is Euler's gamma function}}</ref><ref>{{Cite OEIS|A245885|Decimal expansion of Gamma(7/2), where Gamma is Euler's gamma function}}</ref> The values presented here are truncated rather than rounded.)
The complex-valued gamma function is undefined for non-positive integers, but in these cases the value can be defined in the [[Riemann sphere]] as {{math|∞}}. The [[reciprocal gamma function]] is [[well defined]] and [[analytic function|analytic]] at these values (and in the [[entire function|entire complex plane]]):
<math display="block">\frac{1}{\Gamma(-3)} = \frac{1}{\Gamma(-2)} = \frac{1}{\Gamma(-1)} = \frac{1}{\Gamma(0)} = 0.</math>