Gamma function ($\Gamma$)

The Gamma function extends factorials to a continuous argument. Its defining Gamma integral iswhere $t^{z-1}=\exp ((z-1)\log t)$ uses the real natural logarithm for $t>0$. The Gamma function recurrence gives its meromorphic continuation to the complex plane, with simple poles at $0,-1,-2,\dots$ and no zeros. See Gamma function residue at a nonpositive integer and Gamma function has no zeros for proof outlines.^[ref]

For nonnegative integers $n$, the factorial identity has a shift of one:Thus $\Gamma (n)=(n-1)!$ for positive integers $n$. Half-integer arguments supply a different familiar constant:The first value follows from the Gaussian integral; the second follows from the Gamma function recurrence.^[ref]

The plot shows only positive real arguments; the gamma function also accepts complex numbers away from its poles. Matching the factorials alone does not select a unique interpolation, even if the Gamma function recurrence is imposed: $f(x)=\Gamma (x)\exp (\sin (2\pi x))$ has the same positive-integer values and the same recurrence relation. The Bohr-Mollerup theorem supplies the extra condition that selects the gamma function.

The useful idea is to replace a discrete product of numbers by an integral whose exponent varies continuously. The Gamma integral then evaluates whole families of integrals at once, provides the normalizing constant for the Gamma distribution in probability theory, and appears in physics, for example in the Bose integral. Its large-positive-argument growth reconnects this continuous description to factorials through Stirling's formula.^[ref]

Historically, Leonhard Euler studied the interpolation of factorials in his paper E19, written in 1729 and published in 1738. His integral over $(0,1)$ becomes the Gamma integral after a change of variables, with the same one-step shift in the argument.^[ref]

For positive real $a$, differentiating the gamma function integral under the integral sign gives the displayed formula; successive derivatives insert higher powers of $\log t$. Dominated differentiation on compact positive-$a$ intervals is justified by exponential decay at infinity and integrability of $t^{a-1}|\log t{|}^n$ at zero. At $a=1$, these derivatives give moments used in logarithmic endpoint asymptotic expansions.

For $t$ in a compact positive interval $[a,b]$, split the integration region at one. Near zero the smallest exponent gives the displayed bound, whose integral is $1/a^2$. Above one the largest exponent gives the bound, integrable by exponential decay. A single integrable function therefore dominates all parameter derivatives locally. Dominated differentiation under the integral sign gives ${\Gamma }^{\prime }(t)={\int }_0^{\infty }{e}^{-x}{x}^{t-1}\log xdx$. The same argument with higher powers of $|\log x|$ gives all higher derivatives.

The beta-gamma identity gives $\Gamma (z/2{)}^2=B(z/2,z/2)\Gamma (z)$ for $\mathrm{Re}z>0$. A zero of $\Gamma (z)$ would therefore force zeros at every $z/2^m$. But $w\Gamma (w)=\Gamma (1+w)\to 1$ as $w\to 0$ in the right half-plane, a contradiction. The Gamma function recurrence then extends nonvanishing to every point where the meromorphic gamma function is finite. Nonpositive integers are poles, not zeros.

Rotating the integral defining the Gamma function toward the positive imaginary axis introduces the factor $i{e}^{-\pi n/2}$. After scaling by $n$, the remaining phase is $\log s-s$, whose unique stationary point is $s=1$ with second derivative $-1$. The stationary phase method gives $e^{-in-i\pi /4}\sqrt{2\pi /n}$ for this scaled integral, producing the displayed asymptotic. The boundary contour integral must be Abel-regularized; its off-saddle tails are smaller by integration by parts.

The Gamma function recurrence and $\Gamma (z)\sim 1/z$ at zero imply a simple pole at every $z=-n$, $n\ge 0$, with residue $(-1{)}^n/n!$.

For $z$ away from the poles, the gamma function satisfiesIts finite product is $m!(m+1{)}^z/[z(z+1)\cdots (z+m)]$. Splitting even and odd factors gives the gamma duplication formula.

The Gamma function recurrence gives $\Gamma (s)=\Gamma (s+j+1)/(s(s+1)\cdots (s+j))$. Its numerator equals one at $s=-j$, so the residue there is $(-1{)}^j/j!$ for every nonnegative integer $j$.

For real $a>0$ and $\mathrm{Re}s>0$, the change of variables $u=at$ in the defining improper integral for the Gamma function givesThe improper integral converges absolutely: near zero its absolute integrand behaves like $t^{\mathrm{Re}s-1}$, and at infinity the exponential function overcomes every fixed power.^[ref]

For real $s=1,2,4$ and $a=1$, the integrands are $e^{-t}$, $t{e}^{-t}$, and $t^3{e}^{-t}$, with total areas $\Gamma (1)=1$, $\Gamma (2)=1$, and $\Gamma (4)=6$. The third integrand is taller and has six times the area of the first, although the exponential function makes each tail decay.

For real $s>0$, the kernel $t^{s-1}{e}^{-t}$ becomes the probability density function of a Gamma distribution with shape $s$ and unit rate only after division by its normalizing constant $\Gamma (s)$. In particular, the shape-four probability density function is $t^3{e}^{-t}/6$. More generally, the Gamma integral verifies that $a^s{t}^{s-1}{e}^{-at}/\Gamma (s)$ integrates to one for shape $s>0$ and rate $a>0$.

For the half-integer evaluation, the change of variables $t=u^2$ reduces the Gamma integral to the Gaussian integral:Consequently, for real $u>0$,

For $t>0$, contour rotation through angle $-\pi /6$ givesTaking real parts yields

The Gamma function obeys the recurrence relationInitially, for $\mathrm{Re}z>0$, integration by parts in the Gamma integral gives the short proofThe boundary term vanishes at zero because $|t^z|=t^{\mathrm{Re}z}\to 0$, and at infinity because of exponential decay. Together with $\Gamma (1)={\int }_0^{\infty }{e}^{-t}dt=1$, the Gamma function recurrence immediately gives $\Gamma (n+1)=n!$ for every nonnegative integer $n$.^[ref]

Read backwards, the Gamma function recurrence defineswhere $m$ is a positive integer. These overlapping formulas agree by the Gamma function recurrence, giving the meromorphic continuation. Since $\Gamma (1)=1$, they give a simple pole at zero and then at every negative integer; Gamma function residue at a nonpositive integer records the residues. The recurrence relation therefore holds throughout the continuation as an identity of meromorphic functions.

On $(0,\infty )$, the Gamma function is the unique positive function $f$ satisfying $f(1)=1$, $f(x+1)=xf(x)$, and the condition that $\log f$ is a convex function. This last condition concerns the natural logarithm of $f$, rather than $f$ itself, and removes the interpolation freedom left by the Gamma function recurrence.^[ref]

For $\mathrm{Re}s>1$, expansion of $(e^x-1{)}^{-1}$ as a geometric series and termwise integration give

The Euler--Mascheroni constant is

For the harmonic number $H_N$, the upper sequence $u_N=H_N-\ln N$ decreases and the lower sequence $v_N=H_N-\ln (N+1)$ increases. The inequalities $\ln (1+1/N)>1/(N+1)$ and $\ln (1+1/(N+1))<1/(N+1)$ follow by integrating the decreasing function $1/x$ over a unit interval. They give the two monotonicities, while $u_N-v_N=\ln (1+1/N)\to 0$. The monotone convergence theorem for sequences therefore gives the common limit Euler's constant $\gamma$. Strict monotonicity makes each finite-index bound strict. At $N=1$, this gives $0<1-\ln 2<\gamma <1$, since $\ln 2<1$.

The reciprocal gamma function has the entire-product representation

This is a particular application of the Weierstrass factorization theorem.

The digamma function is the logarithmic derivativeLogarithmically differentiating the Weierstrass product gives

The trigamma function is ${\psi }^{\prime }(z)$. For real $z>0$,

The digamma function is strictly increasing on the positive real axis, whileIt therefore has exactly one positive zero, and that zero lies in $(1,2)$.

It follows by making the logarithm of the quotient entire and periodic; growth forces its periodic remainder to be constant.

For $\mathrm{Re}p,\mathrm{Re}q>0$, Euler's beta function is

For positive real numbers $a,b$, the beta integral represents the beta function by $B(a,b)={\int }_0^1{x}^{a-1}(1-x{)}^{b-1}dx$. In particular, for nonnegative integers $m,n$, it gives ${\int }_0^1{x}^m(1-x{)}^{n}dx=m!n!/(m+n+1)!$, connecting integration to binomial coefficients.

For $p,q>0$, express the product of the two Gamma integrals as the nonnegative double integralUse $r=x+y$ and $u=x/(x+y)$, with inverse $x=ru$, $y=r(1-u)$ and Jacobian determinant of absolute value $r$. The new domain of integration is $r>0$, $0<u<1$, and the integrand with this factor becomes $r^{p+q-1}{e}^{-r}{u}^{p-1}(1-u{)}^{q-1}$. Tonelli's theorem and separation of the two integrals therefore give $\Gamma (p+q)B(p,q)$. The positive real parameter hypothesis ensures that all the original integrals are finite.

For complex parameters with positive real parts, the same proof first gives the undivided identity $\Gamma (p+q)B(p,q)=\Gamma (p)\Gamma (q)$. Absolute integrability follows by taking real parts in the exponents, so Fubini's theorem applies. The nonvanishing of the Gamma function then permits division by $\Gamma (p+q)$.

For $s,t>0$, put $r=s+t$ and $u=t/(s+t)$. Then $(s,t)=(r(1-u),ru)$ and $dsdt=rdrdu$, separating radial gamma and ratio beta integrals.

In $B(n,n)={\int }_0^1[x(1-x){]}^{n-1}dx$, put $v=2x-1$. Then $x(1-x)=(1-v^2)/4$, and evenness givesThe further substitution $u=v^2$ turns this into $2^{1-2n}B(1/2,n)$. The symmetry of the Euler beta function yields the title identity. This proof works for every positive real $n$, without requiring $n$ to be an integer or using the Gamma function.

For positive real $p,q$, substitute $x={\sin }^2\theta$ in the defining Euler beta function integral. The increasing substitution maps $[0,\pi /2]$ onto $[0,1]$ and has $dx=2\sin \theta \cos \theta d\theta$. Combining this factor with $x^{p-1}(1-x{)}^{q-1}$ gives the displayed identity. Positivity of both parameters ensures convergence at the two endpoints, even when the integrand is unbounded there.

For $p>0$ and a nonnegative integer $n$, integration by parts gives, when $n>0$,The endpoint term $[s(1-s^p{)}^n{]}_0^1$ vanishes. ThereforeRepeated application yields $J_n(p)={\prod }_{j=1}^{n}pj/(pj+1)$. The integration by substitution $w=s^p$ also identifies $J_n(p)=B(1/p,n+1)/p$ with an Euler beta function. A related weighted integral follows immediately from the same substitution:

For real $p,q>0$ and $0\le z\le 1$, the displayed integral is the incomplete beta function. The regularized version is $I_z(p,q)=B_z(p,q)/B_1(p,q)$, the cumulative distribution function of the beta distribution. It has endpoint values zero and one and derivative $z^{p-1}(1-z{)}^{q-1}/B_1(p,q)$. This is a useful explicit scale function of a one-dimensional diffusion when the scale density has two power-law endpoint singularities.

For $c=1-a-b$ with positive real parts $\mathrm{Re}a,\mathrm{Re}b,\mathrm{Re}c$, the integral converges at its two finite singularities and at infinity, and equalsTo prove it, use Schwinger parameterization with exponents $1-a$ and $1-b$. Integrating the resulting planar Gaussian integral gives $\pi /(\lambda +\rho )$ times $\exp [-\lambda \rho /(\lambda +\rho )]$. Set $q=\lambda +\rho$ and $x=\lambda /q$; the $q$ integral is a gamma function and the $x$ integral is $B(a,b)$, giving the stated ratio. This domain justifies the interchanges; elsewhere the answer is interpreted by analytic continuation. The string measure $d^{2}z=2dxdy$ multiplies the answer by two.

The Beta function obeysThese identities follow immediately from the beta-gamma identity and the Gamma function recurrence.

For $0<a<2$,Differentiating at $a=1$ and expanding the secant gives

The functional equation and $\Gamma (1)=1$ give $\Gamma (z)\sim 1/z$ as $z\to 0$ away from the negative real axis.

The beta--gamma identity gives $B(z,z)=\Gamma (z{)}^2/\Gamma (2z)\sim 2/z$ as $z\to 0$ through the right half-plane.