Editing Gamma function (section)

=== 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.
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{{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>

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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.

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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>