Beta function ($B$)

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