Beta function

In mathematics, the beta function, also called the Euler integral of the first kind, is a special function defined by

\mathrm{\Beta}(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt \!

for \textrm{Re}(x), \textrm{Re}(y) > 0.\,

The beta function was studied by Euler and Legendre and was given its name by Jacques Binet.

Contents

Properties

The beta function is symmetric, meaning that

\Beta(x,y) = \Beta(y,x). \!

It has many other forms, including:

\Beta(x,y)=\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)} \!
\Beta(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,d\theta, \qquad \textrm{Re}(x)>0,\ \textrm{Re}(y)>0 \!
\Beta(x,y) = \int_0^\infty\dfrac{t^{x-1}}{(1+t)^{x+y}}\,dt, \qquad \textrm{Re}(x)>0,\ \textrm{Re}(y)>0 \!
\Beta(x,y) = \sum_{n=0}^\infty \dfrac{{n-y \choose n}} {x+n}, \!
\Beta(x,y) = \prod_{n=0}^\infty \left( 1+ \dfrac{x y}{n (x+y+n)}\right)^{-1}, \!
\Beta(x,y) \cdot \Beta(x+y,1-y) = \dfrac{\pi}{x \sin(\pi y)}, \!

where \Gamma\, is the gamma function. The second identity shows in particular \Gamma(1/2) = \sqrt \pi.

Just as the gamma function for integers describes factorials, the beta function can define a binomial coefficient after adjusting indices:

{n \choose k} = \frac1{(n+1) \Beta(n-k+1, k+1)}.

The beta function was the first known scattering amplitude in string theory, first conjectured by Gabriele Veneziano. It also occurs in the theory of the preferential attachment process, a type of stochastic urn process.

Relationship between Gamma function and Beta function

To derive the integral representation of the beta function, write the product of two factorials as

\Gamma(x)\Gamma(y) = \int_0^\infty\ e^{-u} u^{x-1}\,du \int_0^\infty\ e^{-v} v^{y-1}\,dv. \!

Now, let u \equiv a^2, v \equiv b^2, so

\begin{align} \Gamma(x)\Gamma(y) & {} = 4\int_0^\infty\ e^{-a^2} a^{2x-1}\mathrm{d}a \int_0^\infty\ e^{-b^2} b^{2y-1}\,db \\ & {} = \int_{-\infty}^\infty\ \int_{-\infty}^\infty\ e^{-(a^2+b^2)} |a|^{2x-1} |b|^{2y-1} \,da \,db. \end{align} \!

Transforming to polar coordinates with a = r\cos\theta, b = r\sin\theta:

\begin{align} \Gamma(x)\Gamma(y) & {} = \int_0^{2\pi}\ \int_0^\infty\ e^{-r^2} |r\cos\theta|^{2x-1} |r\sin\theta|^{2y-1} r \, dr \,d\theta \\ & {} = \int_0^\infty\ e^{-r^2} r^{2x+2y-2} r\, dr \int_0^{2\pi}\ |(\cos\theta)^{2x-1} (\sin\theta)^{2y-1}| \, d\theta \\ & {} = \frac{1}{2}\int_0^\infty\ e^{-r^2} r^{2(x+y-1)} \, d(r^2) 4\int_0^{\pi/2}\ (\cos\theta)^{2x-1} (\sin\theta)^{2y-1} \,d\theta \\ & {} = \Gamma(x+y) 2\int_0^{\pi/2}\ (\cos\theta)^{2x-1} (\sin\theta)^{2y-1} \, d\theta \\ & {} = \Gamma(x+y) \Beta(x,y). \end{align}

Hence, rewrite the arguments with the usual form of Beta function:

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

A somewhat more straightforward derivation:

\begin{align} {\Gamma(x)\Gamma(y)} & {} = \int^\infty_0 t^{x-1} e^{-t} dt \int^\infty_0 s^{y-1} e^{-s} \, ds \\ & {} = \int^\infty_{t=0} \int^\infty_{s=0} \ t^{x-1} s^{y-1} e^{-(t{+}s)} \,ds\, dt. \end{align}

The argument in the exponential inspires us to employ the substitution

\begin{align} \begin{array}{l} \sigma=s{+}t \\ \tau=t \end{array} \quad \Rightarrow \quad |J|=1, \end{align}

where J is the Jacobian of the transformation. Using this transformation we arrive at:

\begin{align} \Gamma(x)\Gamma(y) & =\int^\infty_{\sigma = 0} \int^\sigma_{\tau = 0} \ \tau^{x{-}1} (\sigma{-}\tau)^{y-1} e^{-\sigma} \,d\tau \,d\sigma \\ & {} =\int^\infty_{\sigma = 0} \int^\sigma_{\tau = 0} \ \tau^{x{-}1} \, \sigma^{y{-}1} \, \Big(1{-}\frac{\tau}{\sigma}\Big)^{y{-}1} e^{-\sigma} \,d\tau\,d\sigma. \end{align}

Again, now the comparison to \Beta(x,y)\, leads us to:

\begin{align} r=\displaystyle \frac{\tau}{\sigma}, \ q=\sigma \quad \text{where the Jacobian is now:} \quad |J|=q. \end{align}

This leads to an easy identification with the expected result:

\begin{align} \Gamma(x)\Gamma(y) & = \int^\infty_{q = 0} \int^1_{r = 0} \, q \ (rq)^{x{-}1} \,q^{y{-}1} \, (1{-}r)^{y{-}1} \, e^{-q}\, dr \,dq \\ & = \int^\infty_{q = 0} \int^1_{r = 0} \ r^{x{-}1} \,(1{-}r)^{y{-}1} \, q^{x{+}y{-}1} \, e^{-q} \,dr\, dq \\ & = \int^\infty_0 q^{x{+}y{-}1} \, e^{-q} \, dq \ \int^1_0 r^{x{-}1} (1{-}r)^{y{-}1} \, dr \\ & = \Gamma(x+y) \, \Beta(x, y). \end{align}

Derivatives

The derivatives follow:

{\partial \over \partial x} \mathrm{B}(x, y) = \mathrm{B}(x, y) \left( {\Gamma'(x) \over \Gamma(x)} - {\Gamma'(x + y) \over \Gamma(x + y)} \right) = \mathrm{B}(x, y) (\psi(x) - \psi(x + y))

where \ \psi(x) is the digamma function.

Integrals

The Nörlund-Rice integral is a contour integral involving the beta function.

Approximation

Stirling's approximation gives the asymptotic formula

\Beta(x,y) \sim \sqrt {2\pi } \frac{{x^{x - \frac{1}{2}} y^{y - \frac{1}{2}} }}{{\left( {x + y} \right)^{x + y - \frac{1}{2}} }}

for large x and large y. If on the other hand x is large and y is fixed, then

\Beta(x,y) \sim \Gamma(y)\,x^{-y}.

Incomplete beta function

The incomplete beta function is a generalization of the beta function that replaces the definite integral of the beta function with an indefinite integral. The situation is analogous to the incomplete gamma function being a generalization of the gamma function.

The incomplete beta function is defined as

\Beta(x;\,a,b) = \int_0^x t^{a-1}\,(1-t)^{b-1}\,dt. \!

For x = 1, the incomplete beta function coincides with the complete beta function.

The regularized incomplete beta function (or regularized beta function for short) is defined in terms of the incomplete beta function and the complete beta function:

I_x(a,b) = \dfrac{\Beta(x;\,a,b)}{\Beta(a,b)}. \!

Working out the integral (one can use integration by parts to do that) for integer values of a and b, one finds:

I_x(a,b) = \sum_{j=a}^{a+b-1} {(a+b-1)! \over j!(a+b-1-j)!} x^j (1-x)^{a+b-1-j}.

Properties

I_0(a,b) = 0 \,
I_1(a,b) = 1 \,
I_x(a,b) = 1 - I_{1-x}(b,a) \,

(Many other properties could be listed here.)

See also

References

External links