General 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 \Z ,which implies\Gamma(z - n) = (-1)^{n-1} \; \frac{\Gamma(-z) \Gamma(1+z)}{\Gamma(n+1-z)}, \qquad n \in \Z and the
Legendre duplication formula \Gamma(z) \Gamma\left(z + \frac12\right) = 2^{1-2z} \, \sqrt{\pi} \, \Gamma(2z).
Proof 1 With Euler's infinite product \Gamma(z) = \frac1z \prod_{n=1}^{\infty} \frac{(1+1/n)^z}{1 + z/n} 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 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+2\pi i and -R+2\pi i where {{tmath|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 {{tmath|1=I_R'=-e^{2\pi i a}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|\le \int_0^{2\pi}\left|\frac{e^{a(R+it)}}{1+e^{R+it}}\right|\, dt\le Ce^{(a-1)R} for some constant C and since , the integral tends to 0 as . Analogously, the integral over the left vertical side of the rectangle tends to 0 as . 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{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} Proving the reflection formula for all z\in (0,1) proves it for all z\in\mathbb{C}\setminus\mathbb{Z} by analytic continuation. The
beta function can be represented as \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. 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 {{tmath|1= t=\frac{1+u}{2} }}: \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 \Beta \left(\frac{1}{2},z\right)=\int_0^1 t^{\frac{1}{2}-1}(1-t)^{z-1} \, dt, \quad t=s^2. Then \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. This implies 2^{2z-1}\Gamma^2(z)=\Gamma (2z)\Beta \left(\frac{1}{2},z\right). Since \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}, 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 Eq. 5.5.6): \prod_{k=0}^{m-1}\Gamma\left(z + \frac{k}{m}\right) = (2 \pi)^{\frac{m-1}{2}} \; m^{\frac12 - 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 , 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{align} \left|\Gamma\left(\tfrac{1}{2}+bi\right)\right|^2 & = \frac{\pi}{\cosh \pi b} \\[6pt] \left|\Gamma\left(1+bi\right)\right|^2 & = \frac{\pi b}{\sinh \pi b} \\[6pt] \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 \\[6pt] \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 \\[6pt] \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} First, consider the reflection formula applied to . \Gamma(bi)\Gamma(1-bi)=\frac{\pi}{\sin \pi bi} Applying the recurrence relation to the second term: -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 {{tmath|1=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(\tfrac12\right)=\sqrt{\pi}, which can be found by setting z = \frac{1}{2} in the reflection formula, by using the relation to the
beta function given below with {{tmath|1=z_1 = z_2 = \frac{1}}}, or simply by making the substitution t = u^2 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{align} \Gamma\left(\frac 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} \\[6pt] \Gamma\left(\frac 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}where the
double factorial . See
Particular values of the gamma function for calculated values. It might be tempting to generalize the result that \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 {{tmath|1=\textstyle r = \frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{2}{3}, \frac{3}{4}, \frac{5}{6} }}. 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, {{tmath|\psi^{(m)} (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 \gamma is the
Euler–Mascheroni constant. For \Re(z) > 0 the nth derivative of the gamma function is:\frac{d^n}{dz^n}\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} (\log t)^n \, dt.(This can be derived by
differentiating the integral form of the gamma function with respect to .) 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 nth
Bell polynomial, we have in particular the
Laurent series expansion of the gamma function \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).
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 x_1 and , and for any , \Gamma(tx_1 + (1 - t)x_2) \le \Gamma(x_1)^t\Gamma(x_2)^{1 - t}. • For any two positive real numbers x_1 and , 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 , \Gamma''(x) \Gamma(x) > \Gamma'(x)^2. The last of these statements is, essentially by definition, the same as the statement that {{tmath|\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 , consists of only positive terms. Logarithmic convexity and
Jensen's inequality together imply, for any positive real numbers and , \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}}. There are also bounds on ratios of gamma functions. The best-known is
Gautschi's inequality, which says that for any positive real number and any , x^{1 - s}
Stirling's formula 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 in the limit . \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] .
Extension to negative, non-integer values Although the main definition of the gamma function—the Euler integral of the second kind—is only valid (on the real axis) for positive arguments, its domain can be extended with
analytic continuation to negative arguments by shifting the negative argument to positive values by using either the Euler's reflection formula,\Gamma(-x) = \frac{1}{\Gamma(x+1)}\frac{\pi}{\sin\big(\pi(x+1)\big)},or the fundamental property,\Gamma(-x):=\frac{\, 1}{-x}\,\Gamma(-x+1) ,when {{tmath|x\not\in\mathbb{Z} }}. For example,\Gamma\!\left(\!-\frac12\right)=-2\,\Gamma\!\left(\frac12\right) .
Residues The behavior for non-positive z is more intricate. Euler's integral does not converge for , 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,\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 . There is in fact no complex number z for which , and hence the
reciprocal gamma function \dfrac {1}{\Gamma (z)} is an
entire function, with zeros at . where it attains the value . The gamma function rises to either side of this minimum. The solution to is and the common value is . The positive solution to is , the
golden ratio, and the common value is . 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. \Gamma (z)=\int_{-\infty}^\infty e^{zt-e^t}\, dtand\Gamma(z) = \int_0^1 \left(\log \frac{1}{t}\right)^{z-1}\,dt, \Gamma(z) = 2c^z\int_{0}^{\infty}t^{2z-1}e^{-ct^{2}}\,dt \,,\; c>0 where the three integrals respectively follow from the substitutions {{tmath|1= t=e^{-x} }}, t=-\log x and t=cx^2 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 z=1/2,\; c=1 we get\Gamma(1/2)=2\int_{0}^{\infty}e^{-t^{2}}\,dt=\sqrt{\pi} \;. Binet's first integral formula for the gamma function states that, when the real part of is positive, then:\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.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{\frac{z}{2\pi}}\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 is positive, then:\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. Let be a
Hankel contour, meaning a path that begins and ends at the point on the
Riemann sphere, whose unit tangent vector converges to at the start of the path and to at the end, which has
winding number 1 around , and which does not cross . 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. If z is not an integer, then Hankel's formula for the gamma function is:\Gamma(z) = -\frac{1}{2i\sin \pi z}\int_C (-t)^{z-1}e^{-t}\,dt,where (-t)^{z-1} is interpreted as . The reflection formula leads to the closely related expression\frac{1}{\Gamma(z)} = \frac{i}{2\pi}\int_C (-t)^{-z}e^{-t}\,dt,which is valid whenever {{tmath|z \not\in \mathbb{Z} }}.
Continued fraction representation The gamma function can also be represented by a sum of two
continued fractions: \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}where .
Fourier series expansion The
logarithm of the gamma function has the following
Fourier series expansion for 0 \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), which was for a long time attributed to
Ernst Kummer, who derived it in 1847. However, Iaroslav Blagouchine discovered that
Carl Johan Malmsten first derived this series in 1842.
Raabe's formula In 1840
Joseph Ludwig Raabe proved that\int_a^{a+1}\log\Gamma(z)\, dz = \tfrac12\log2\pi + a\log a - a,\quad a>0. In particular, if a = 0 then\int_0^1\log\Gamma(z)\, dz = \frac{1}{2} \log (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 An alternative notation introduced by
Gauss is the \Pi-function, a shifted version of the gamma function: \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 . Using the pi function, the reflection formula is: \Pi(z) \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)} using the normalized
sinc function; while the multiplication theorem becomes: \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)\ . The shifted
reciprocal gamma function is sometimes denoted {{tmath|1=\pi(z) = \frac{1}}}, an
entire function. The
volume of an -ellipsoid with radii can be expressed asV_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 • 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:\Gamma(z,x) = \int_x^\infty t^{z-1} e^{-t} dt.There is a similar lower incomplete gamma function. • The gamma function is related to Euler's
beta function by the formula \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)}. • 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, . \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 . The logarithm of the gamma function satisfies the following formula due to Lerch: \operatorname{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).
Particular values Including up to the first 20 digits after the decimal point, some particular values of the gamma function are: \begin{array}{rcccl} \Gamma\left(-\frac{3}{2}\right) &=& \frac{4\sqrt{\pi}}{3} &\approx& +2.36327\,18012\,07354\,70306 \\[6pt] \Gamma\left(-\frac{1}{2}\right) &=& -2\sqrt{\pi} &\approx& -3.54490\,77018\,11032\,05459 \\[6pt] \Gamma\left(\frac{1}{2}\right) &=& \sqrt{\pi} &\approx& +1.77245\,38509\,05516\,02729 \\[6pt] \Gamma(1) &=& 0! &=& +1 \\[6pt] \Gamma\left(\frac{3}{2}\right) &=& \frac{\sqrt{\pi}}{2} &\approx& +0.88622\,69254\,52758\,01364 \\[6pt] \Gamma(2) &=& 1! &=& +1 \\[6pt] \Gamma\left(\frac{5}{2}\right) &=& \frac{3\sqrt{\pi}}{4} &\approx& +1.32934\,03881\,79137\,02047 \\[6pt] \Gamma(3) &=& 2! &=& +2 \\[6pt] \Gamma\left(\frac{7}{2}\right) &=& \tfrac{15\sqrt{\pi}}{8} &\approx& +3.32335\,09704\,47842\,55118 \\[6pt] \Gamma(4) &=& 3! &=& +6 \end{array}(These numbers can be found in the
OEIS. 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 . The
reciprocal gamma function is
well defined and
analytic at these values (and in the
entire complex plane):\frac{1}{\Gamma(-3)} = \frac{1}{\Gamma(-2)} = \frac{1}{\Gamma(-1)} = \frac{1}{\Gamma(0)} = 0. == Log-gamma function ==