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Gamma and Beta Functions

The gamma function extends factorials from nonnegative integers to complex arguments. The beta function is a two-parameter integral built from gamma functions. Together they are the normalization language behind special functions, radial integrals, angular integrals, Gaussian moments, dimensional formulas, and hypergeometric series.

Use this page as the canonical home for the complete gamma and beta functions. Incomplete gamma and beta functions, regularized distribution functions, and specialized numerical algorithms should be treated on more focused probability or numerical pages.

For Re⁡z>0\operatorname{Re}z>0, the gamma function is defined by Euler’s integral

Γ(z)=∫0∞tz−1e−t dt.\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}\,dt.

This integral converges near t=0t=0 because Re⁡z>0\operatorname{Re}z>0, and it converges at infinity because of the exponential decay. The function is then extended to the complex plane by analytic continuation.

The most important special case is the factorial relation:

Γ(n+1)=n!,n=0,1,2,….\Gamma(n+1)=n!, \qquad n=0,1,2,\ldots.

Equivalently,

Γ(n)=(n−1)!,n=1,2,….\Gamma(n)=(n-1)!, \qquad n=1,2,\ldots.

The shift by one is a common source of normalization errors.

Integration by parts gives, for Re⁡z>0\operatorname{Re}z>0,

Γ(z+1)=∫0∞tze−t dt=z∫0∞tz−1e−t dt=zΓ(z).\begin{aligned} \Gamma(z+1) &= \int_0^\infty t^z e^{-t}\,dt \\ &= z\int_0^\infty t^{z-1}e^{-t}\,dt \\ &= z\Gamma(z). \end{aligned}

This recurrence,

Γ(z+1)=zΓ(z),\Gamma(z+1)=z\Gamma(z),

is the analytic version of n!=n(n−1)!n!=n(n-1)!. Starting from Γ(1)=1\Gamma(1)=1, it gives Γ(n+1)=n!\Gamma(n+1)=n!.

The recurrence also continues the function leftward:

Γ(z)=Γ(z+1)z.\Gamma(z) = \frac{\Gamma(z+1)}{z}.

This reveals simple poles at

z=0,−1,−2,….z=0,-1,-2,\ldots.

The gamma function is meromorphic, not entire. Its reciprocal 1/Γ(z)1/\Gamma(z) is entire and has zeros at those nonpositive integers.

The Gaussian integral gives

Γ ⁣(12)=∫0∞t−1/2e−t dt=π.\Gamma\!\left(\frac12\right) = \int_0^\infty t^{-1/2}e^{-t}\,dt = \sqrt\pi.

Using the recurrence,

Γ ⁣(n+12)=(2n)!4nn!π,n=0,1,2,….\Gamma\!\left(n+\frac12\right) = \frac{(2n)!}{4^n n!}\sqrt\pi, \qquad n=0,1,2,\ldots.

Half-integer values appear in Gaussian moments, harmonic-oscillator normalization, spherical integrals, and Bessel-function coefficients.

For Re⁡x>0\operatorname{Re}x>0 and Re⁡y>0\operatorname{Re}y>0, the beta function is

B(x,y)=∫01tx−1(1−t)y−1 dt.B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1}\,dt.

It is symmetric:

B(x,y)=B(y,x).B(x,y)=B(y,x).

The beta function is connected to gamma functions by

B(x,y)=Γ(x)Γ(y)Γ(x+y).B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.

For positive integers,

B(m,n)=(m−1)!(n−1)!(m+n−1)!.B(m,n) = \frac{(m-1)!(n-1)!}{(m+n-1)!}.

This is why beta integrals often simplify angular and normalization calculations with polynomial powers.

Start from the product

Γ(x)Γ(y)=∫0∞∫0∞sx−1ty−1e−(s+t) ds dt,\Gamma(x)\Gamma(y) = \int_0^\infty \int_0^\infty s^{x-1}t^{y-1}e^{-(s+t)} \,ds\,dt,

valid for Re⁡x>0\operatorname{Re}x>0 and Re⁡y>0\operatorname{Re}y>0. Change variables to

u=s+t,v=ss+t,u=s+t, \qquad v=\frac{s}{s+t},

so that

s=uv,t=u(1−v),ds dt=u du dv.s=uv, \qquad t=u(1-v), \qquad ds\,dt=u\,du\,dv.

The region s,t>0s,t\gt0 becomes u>0u\gt0 and 0<v<10\lt v\lt1. Therefore

Γ(x)Γ(y)=∫0∞∫01ux+y−1vx−1(1−v)y−1e−u dv du=Γ(x+y)B(x,y).\begin{aligned} \Gamma(x)\Gamma(y) &= \int_0^\infty \int_0^1 u^{x+y-1} v^{x-1}(1-v)^{y-1} e^{-u} \,dv\,du \\ &= \Gamma(x+y)B(x,y). \end{aligned}

Dividing by Γ(x+y)\Gamma(x+y) gives the beta-gamma identity.

Beta functions often appear after trigonometric substitutions. A standard form is

∫0π/2sin⁡2x−1θ cos⁡2y−1θ dθ=12B(x,y),\int_0^{\pi/2} \sin^{2x-1}\theta\, \cos^{2y-1}\theta\,d\theta = \frac12 B(x,y),

for Re⁡x>0\operatorname{Re}x>0 and Re⁡y>0\operatorname{Re}y>0.

For example,

∫0π/2sin⁡mθ dθ=π2Γ ⁣(m+12)Γ ⁣(m+22),m>−1.\int_0^{\pi/2} \sin^m\theta\,d\theta = \frac{\sqrt\pi}{2} \frac{ \Gamma\!\left(\frac{m+1}{2}\right) }{ \Gamma\!\left(\frac{m+2}{2}\right) }, \qquad m>-1.

Such formulas are useful for angular normalization, spherical averages, and selection-rule integrals.

The rising factorial, or Pochhammer symbol, is

(a)n=a(a+1)⋯(a+n−1),(a)0=1.(a)_n = a(a+1)\cdots(a+n-1), \qquad (a)_0=1.

In gamma notation,

(a)n=Γ(a+n)Γ(a),(a)_n = \frac{\Gamma(a+n)}{\Gamma(a)},

provided the quotient is interpreted away from poles or by a limiting process. This is the compact coefficient language used in Hypergeometric Functions.

When a numerator parameter is a nonpositive integer, a Pochhammer symbol can vanish after finitely many terms. That is the algebraic mechanism behind polynomial termination in Laguerre, Legendre, and hypergeometric solutions.

The integral definition of Γ(z)\Gamma(z) is initially restricted to Re⁡z>0\operatorname{Re}z>0. The meromorphic continuation satisfies the recurrence throughout the complex plane except at poles. The poles are at the nonpositive integers and are simple.

The reflection formula is

Γ(z)Γ(1−z)=πsin⁡πz.\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin\pi z}.

It encodes both the pole structure and many half-integer identities. It also warns that gamma expressions can change character when parameters cross integers.

For the beta function, analytic continuation follows from

B(x,y)=Γ(x)Γ(y)Γ(x+y).B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.

Thus B(x,y)B(x,y) is meromorphic in its parameters. The integral over 0<t<10\lt t\lt1 is only one initial representation, not the whole analytic object.

The sheet and branch conventions of powers such as tx−1t^{x-1} are part of the same analytic-continuation bookkeeping discussed in Branch Cuts.

For large ∣z∣\lvert z\rvert away from the negative real axis, Stirling’s formula gives the leading behavior

Γ(z)∼2π zz−1/2e−z.\Gamma(z) \sim \sqrt{2\pi}\, z^{z-1/2}e^{-z}.

Taking logarithms is often more stable:

log⁡Γ(z)∼(z−12)log⁡z−z+12log⁡(2π)+⋯ .\log\Gamma(z) \sim \left(z-\frac12\right)\log z -z +\frac12\log(2\pi) +\cdots.

In quantum mechanics, this asymptotic behavior appears when estimating high quantum-number normalization constants, large angular-momentum coefficients, semiclassical densities of states, and factorially growing perturbation coefficients. The general language of asymptotic series is developed in Asymptotic Analysis.

Gamma and beta functions appear whenever normalization integrals reduce to powers times exponentials or powers on a finite interval.

For α>0\alpha>0 and Re⁡ν>−1\operatorname{Re}\nu>-1,

∫0∞rνe−αr dr=Γ(ν+1)αν+1.\int_0^\infty r^\nu e^{-\alpha r}\,dr = \frac{\Gamma(\nu+1)}{\alpha^{\nu+1}}.

This type of integral appears in radial hydrogenic calculations and expectation values.

Gaussian moments reduce to gamma functions:

∫0∞xpe−αx2 dx=12α−(p+1)/2Γ ⁣(p+12),α>0.\int_0^\infty x^{p}e^{-\alpha x^2}\,dx = \frac12 \alpha^{-(p+1)/2} \Gamma\!\left(\frac{p+1}{2}\right), \qquad \alpha>0.

This is the normalization backbone for oscillator wavefunctions, Gaussian wave packets, and many variational trial states.

Angular integrals often reduce to beta functions. For example, after u=cos⁡θu=\cos\theta or t=sin⁡2θt=\sin^2\theta, powers of sine and cosine become beta integrals. This is why gamma ratios appear in spherical averages and in normalization constants for special functions.

In scattering and field-theory bridges, gamma functions appear in dimensional regularization, Coulomb phases, special-function solutions, and asymptotic matching. Those uses require additional physical context; the present page supplies the basic function identities.

  • Writing Γ(n)=n!\Gamma(n)=n! instead of Γ(n)=(n−1)!\Gamma(n)=(n-1)!.
  • Applying the gamma integral outside Re⁡z>0\operatorname{Re}z>0 without analytic continuation.
  • Forgetting the simple poles at 0,−1,−2,…0,-1,-2,\ldots.
  • Treating B(x,y)B(x,y) as only an integral instead of a meromorphic function of its parameters.
  • Canceling gamma functions across poles without checking limits.
  • Confusing complete gamma and beta functions with incomplete or regularized versions.
  • Using Stirling’s formula near the negative real axis without tracking branches.
  • Ignoring normalization conventions when importing special-function formulas.
  • NIST Digital Library of Mathematical Functions, Chapter 5, Gamma Function.
  • F. W. J. Olver, D. W. Lozier, R. F. Boisvert, and C. W. Clark, eds., NIST Handbook of Mathematical Functions, Cambridge University Press, 2010.
  • G. E. Andrews, R. Askey, and R. Roy, Special Functions, Cambridge University Press, 1999.
  • E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, 4th ed., Cambridge University Press, 1927.
  • N. M. Temme, Special Functions: An Introduction to the Classical Functions of Mathematical Physics, Wiley, 1996.
  • G. B. Arfken, H. J. Weber, and F. E. Harris, Mathematical Methods for Physicists, 7th ed., Academic Press, 2013.
  1. Use integration by parts to prove Γ(z+1)=zΓ(z)\Gamma(z+1)=z\Gamma(z) for Re⁡z>0\operatorname{Re}z>0.
Solution

Start with

Γ(z+1)=∫0∞tze−t dt.\Gamma(z+1) = \int_0^\infty t^z e^{-t}\,dt.

Let u=tzu=t^z and dv=e−tdtdv=e^{-t}dt. Then du=ztz−1dtdu=zt^{z-1}dt and v=−e−tv=-e^{-t}. The boundary term −tze−t∣0∞-t^z e^{-t}\rvert_0^\infty vanishes for Re⁡z>0\operatorname{Re}z>0, so

Γ(z+1)=z∫0∞tz−1e−t dt=zΓ(z).\Gamma(z+1) = z\int_0^\infty t^{z-1}e^{-t}\,dt = z\Gamma(z).
  1. Show that Γ(5/2)=3π/4\Gamma(5/2)=3\sqrt\pi/4.
Solution

Using Γ(1/2)=π\Gamma(1/2)=\sqrt\pi and the recurrence,

Γ ⁣(32)=12Γ ⁣(12)=π2.\Gamma\!\left(\frac32\right) = \frac12\Gamma\!\left(\frac12\right) = \frac{\sqrt\pi}{2}.

Then

Γ ⁣(52)=32Γ ⁣(32)=32π2=3π4.\Gamma\!\left(\frac52\right) = \frac32\Gamma\!\left(\frac32\right) = \frac32\frac{\sqrt\pi}{2} = \frac{3\sqrt\pi}{4}.
  1. Compute B(1/2,1/2)B(1/2,1/2).
Solution

Use

B(x,y)=Γ(x)Γ(y)Γ(x+y).B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.

For x=y=1/2x=y=1/2,

B ⁣(12,12)=Γ(1/2)2Γ(1)=π1=π.B\!\left(\frac12,\frac12\right) = \frac{\Gamma(1/2)^2}{\Gamma(1)} = \frac{\pi}{1} = \pi.
  1. Evaluate ∫0∞r4e−αr dr\int_0^\infty r^4e^{-\alpha r}\,dr for α>0\alpha>0.
Solution

Use the substitution u=αru=\alpha r. Then

∫0∞r4e−αr dr=1α5∫0∞u4e−u du=Γ(5)α5.\int_0^\infty r^4e^{-\alpha r}\,dr = \frac{1}{\alpha^5} \int_0^\infty u^4e^{-u}\,du = \frac{\Gamma(5)}{\alpha^5}.

Since Γ(5)=4!=24\Gamma(5)=4!=24,

∫0∞r4e−αr dr=24α5.\int_0^\infty r^4e^{-\alpha r}\,dr = \frac{24}{\alpha^5}.
  1. Why is it unsafe to cancel Γ(a)\Gamma(a) from a formula without checking whether aa is a nonpositive integer?
Solution

The gamma function has poles at a=0,−1,−2,…a=0,-1,-2,\ldots. A quotient such as Γ(a+n)/Γ(a)\Gamma(a+n)/\Gamma(a) may still have a finite limiting value, but that value must be obtained by a limiting argument or by using the Pochhammer symbol. Algebraic cancellation that assumes both factors are finite can miss zeros, poles, or polynomial termination.