Definitions and Properties of the Gamma and Beta Functions

The factorial $n!$ is defined for nonnegative integers. Extending it to noninteger arguments leads to the Gamma function. Another special function, the Beta function, arises naturally when evaluating integrals involving $t^{a-1}(1-t)^{b-1}$.

Although their defining integrals cover an infinite interval and a finite interval, respectively, the two functions are closely related:

$$\mathrm{B}(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}, \qquad a,b>0.$$

This identity reduces many integrals to evaluations of the Gamma function. In probability, it also gives the normalizing constant and moments of the Beta distribution. We will derive it from the integral definitions. Throughout, we consider real parameters and state the convergence conditions for each integral representation.

The Gamma function

Definition and convergence

For $x>0$, define

$$\boxed{\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}\,\mathrm{d}t.}$$

This is an improper integral, so we must examine both endpoints. As $t\to0^+$, we have $e^{-t}\to1$, and the integrand is asymptotic to $t^{x-1}$. Moreover,

$$\int_0^1 t^{x-1}\,\mathrm{d}t=\frac1x, \qquad x>0.$$

As $t\to\infty$, the exponential factor $e^{-t}$ decays faster than any fixed power can grow, ensuring convergence at infinity. For real parameters, the defining integral converges exactly when $x>0$, and $\Gamma(x)>0$. This is Euler’s integral definition.

Recurrence and factorials

Integrating $\Gamma(x+1)$ by parts with $u=t^x$ and $\mathrm{d}v=e^{-t}\mathrm{d}t$ gives

$$\begin{aligned} \Gamma(x+1) &=\int_0^\infty t^xe^{-t}\,\mathrm{d}t\\ &=\left[-t^xe^{-t}\right]_0^\infty +x\int_0^\infty t^{x-1}e^{-t}\,\mathrm{d}t\\ &=x\Gamma(x). \end{aligned}$$

Since $x>0$, the boundary term vanishes as both $t\to0^+$ and $t\to\infty$. Therefore,

$$\boxed{\Gamma(x+1)=x\Gamma(x),\qquad x>0.}$$

Also,

$$\Gamma(1)=\int_0^\infty e^{-t}\,\mathrm{d}t=1,$$

so

$$\Gamma(2)=1,\quad \Gamma(3)=2,\quad \Gamma(4)=6,\quad \Gamma(5)=24.$$

Repeated application of the recurrence gives, for every positive integer $n$,

$$\boxed{\Gamma(n)=(n-1)!.}$$

Equivalently, $\Gamma(n+1)=n!$, including $n=0$. Keep this shift in the argument in mind: $n!$ corresponds to $\Gamma(n+1)$.

Values at half-integers

A useful starting point for evaluating the Gamma function at noninteger arguments is $x=1/2$:

$$\Gamma\left(\frac12\right) =\int_0^\infty t^{-1/2}e^{-t}\,\mathrm{d}t.$$

Let $t=u^2$, so that $\mathrm{d}t=2u\,\mathrm{d}u$. Then

$$\begin{aligned} \Gamma\left(\frac12\right) &=2\int_0^\infty e^{-u^2}\,\mathrm{d}u\\ &=\int_{-\infty}^\infty e^{-u^2}\,\mathrm{d}u. \end{aligned}$$

The last step uses the fact that $e^{-u^2}$ is even. This Gaussian integrand has no elementary antiderivative, but we can evaluate the integral by squaring it and switching to polar coordinates.

Write

$$I=\int_{-\infty}^\infty e^{-u^2}\,\mathrm{d}u.$$

Since the integrand is nonnegative, Tonelli’s theorem allows us to express the product of two integrals as a double integral over the plane:

$$\begin{aligned} I^2 &=\left(\int_{-\infty}^\infty e^{-x^2}\,\mathrm{d}x\right) \left(\int_{-\infty}^\infty e^{-y^2}\,\mathrm{d}y\right)\\ &=\iint_{\mathbb{R}^2}e^{-(x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y. \end{aligned}$$

Set $x=r\cos\theta$ and $y=r\sin\theta$. The area element becomes $\mathrm{d}x\,\mathrm{d}y=r\,\mathrm{d}r\,\mathrm{d}\theta$, giving

$$\begin{aligned} I^2 &=\int_0^{2\pi}\int_0^\infty e^{-r^2}r\,\mathrm{d}r\,\mathrm{d}\theta\\ &=2\pi\int_0^\infty re^{-r^2}\,\mathrm{d}r\\ &=-\pi\left[e^{-r^2}\right]_0^\infty\\ &=\pi. \end{aligned}$$

Because $I>0$, we obtain

$$\boxed{\Gamma\left(\frac12\right)=\sqrt\pi.}$$

Applying the recurrence then gives

$$\Gamma\left(\frac32\right)=\frac{\sqrt\pi}{2},\qquad \Gamma\left(\frac52\right)=\frac{3\sqrt\pi}{4},\qquad \Gamma\left(\frac72\right)=\frac{15\sqrt\pi}{8}.$$

More generally, for every nonnegative integer $n$,

$$\boxed{ \Gamma\left(n+\frac12\right) =\frac{(2n)!}{4^n n!}\sqrt\pi. }$$

For $n\geq1$, this comes from multiplying $n-\tfrac12,n-\tfrac32,\ldots,\tfrac12$ and then multiplying by $\sqrt\pi$. The formula also holds for $n=0$.

The Beta function

Definition and convergence

For $a>0$ and $b>0$, define

$$\boxed{ \mathrm{B}(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1}\,\mathrm{d}t. }$$

As $t\to0^+$, the integrand is asymptotic to $t^{a-1}$; as $t\to1^-$, it is asymptotic to $(1-t)^{b-1}$. For real parameters, the integral therefore converges exactly when $a,b>0$.

If $0<a<1$ or $0<b<1$, the integrand diverges at the corresponding endpoint, but the integral still converges. For example, when $a=b=1/2$, both endpoints have integrable singularities.

Symmetry

Substituting $u=1-t$ gives

$$\begin{aligned} \mathrm{B}(a,b) &=\int_0^1(1-u)^{a-1}u^{b-1}\,\mathrm{d}u\\ &=\mathrm{B}(b,a). \end{aligned}$$

Thus, interchanging the two parameters leaves the value unchanged:

$$\boxed{\mathrm{B}(a,b)=\mathrm{B}(b,a).}$$

The Beta–Gamma relation

Start with the product of two Gamma functions:

$$\Gamma(a)\Gamma(b) =\int_0^\infty\int_0^\infty u^{a-1}v^{b-1}e^{-(u+v)}\,\mathrm{d}u\,\mathrm{d}v.$$

Again, nonnegativity allows us to write the product as a double integral. We now replace $u,v$ with their sum and proportion:

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

Conversely,

$$u=st,\qquad v=s(1-t).$$

The first quadrant $u,v>0$ corresponds to $s>0$ and $0<t<1$. The absolute value of the Jacobian determinant is

$$\left|\frac{\partial(u,v)}{\partial(s,t)}\right| =\left|\begin{vmatrix}t&s\\1-t&-s\end{vmatrix}\right| =s.$$

Hence the area element is $\mathrm{d}u\,\mathrm{d}v=s\,\mathrm{d}s\,\mathrm{d}t$. Substituting into the integral,

$$\begin{aligned} \Gamma(a)\Gamma(b) &=\int_0^\infty\int_0^1 (st)^{a-1}[s(1-t)]^{b-1}e^{-s}s\,\mathrm{d}t\,\mathrm{d}s\\ &=\left(\int_0^\infty s^{a+b-1}e^{-s}\,\mathrm{d}s\right) \left(\int_0^1 t^{a-1}(1-t)^{b-1}\,\mathrm{d}t\right)\\ &=\Gamma(a+b)\mathrm{B}(a,b). \end{aligned}$$

Since $\Gamma(a+b)>0$, we obtain

$$\boxed{ \mathrm{B}(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}, \qquad a,b>0. }$$

The substitution works by separating $u,v$ into a total $s$ and a proportion $t$, allowing the integral to factor into two independent parts. The result is Euler’s Beta integral formula.

Properties of the Beta function

Throughout this section, assume $a,b>0$.

Recurrence relations

Using the Beta–Gamma relation and the Gamma recurrence,

$$\begin{aligned} \mathrm{B}(a+1,b) &=\frac{\Gamma(a+1)\Gamma(b)}{\Gamma(a+b+1)}\\ &=\frac{a}{a+b}\mathrm{B}(a,b). \end{aligned}$$

By symmetry, we also have the corresponding recurrence in the second parameter:

$$\boxed{ \begin{aligned} \mathrm{B}(a+1,b)&=\frac{a}{a+b}\mathrm{B}(a,b),\\ \mathrm{B}(a,b+1)&=\frac{b}{a+b}\mathrm{B}(a,b). \end{aligned} }$$

Adding these identities gives

$$\boxed{\mathrm{B}(a+1,b)+\mathrm{B}(a,b+1)=\mathrm{B}(a,b).}$$

This addition identity also follows directly from the definition: adding the two integrands on the left produces a common factor of $t+(1-t)=1$.

Special values

When one parameter equals one,

$$\mathrm{B}(a,1)=\int_0^1t^{a-1}\,\mathrm{d}t=\frac1a, \qquad \mathrm{B}(1,b)=\frac1b.$$

For positive integers $m,n$,

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

For example,

$$\int_0^1t^2(1-t)^3\,\mathrm{d}t =\mathrm{B}(3,4) =\frac{2!\,3!}{6!} =\frac1{60}.$$

For half-integer parameters, using $\Gamma(1/2)=\sqrt\pi$ gives

$$\mathrm{B}\left(\frac12,\frac12\right)=\pi, \qquad \mathrm{B}\left(\frac12,\frac32\right)=\frac\pi2, \qquad \mathrm{B}\left(\frac32,\frac32\right)=\frac\pi8.$$

In particular, the first identity evaluates an integral that converges despite having singularities at both endpoints:

$$\int_0^1\frac{\mathrm{d}t}{\sqrt{t(1-t)}}=\pi.$$

Positivity and monotonicity

The defining integrand is positive on $(0,1)$, so $\mathrm{B}(a,b)>0$.

Fix $b>0$ and let $a_2>a_1>0$. Because $0<t<1$, we have $t^{a_2-1}<t^{a_1-1}$. Multiplying both sides by the positive factor $(1-t)^{b-1}$ and integrating gives

$$\mathrm{B}(a_2,b)<\mathrm{B}(a_1,b).$$

Thus, with either positive parameter fixed, the Beta function is strictly decreasing in the other parameter. The result for the second parameter follows by symmetry.

Other integral representations of the Beta function

Trigonometric integrals

In the definition, substitute $t=\sin^2\theta$, so that $\mathrm{d}t=2\sin\theta\cos\theta\,\mathrm{d}\theta$. This gives

$$\mathrm{B}(a,b) =2\int_0^{\pi/2} \sin^{2a-1}\theta\cos^{2b-1}\theta\,\mathrm{d}\theta.$$

Therefore, for $m,n>-1$,

$$\boxed{ \int_0^{\pi/2}\sin^m\theta\cos^n\theta\,\mathrm{d}\theta =\frac12\mathrm{B}\left(\frac{m+1}{2},\frac{n+1}{2}\right). }$$

The conditions $m,n>-1$ ensure integrability at the two endpoints. This is the trigonometric form of the Beta integral. For example,

$$\begin{aligned} \int_0^{\pi/2}\sin^4\theta\cos^2\theta\,\mathrm{d}\theta &=\frac12\mathrm{B}\left(\frac52,\frac32\right)\\ &=\frac{\Gamma(5/2)\Gamma(3/2)}{2\Gamma(4)}\\ &=\frac\pi{32}. \end{aligned}$$

Integrals over an infinite interval

In the definition, set

$$t=\frac{u}{1+u},\qquad 1-t=\frac1{1+u},\qquad \mathrm{d}t=\frac{\mathrm{d}u}{(1+u)^2},$$

to obtain the Beta integral over an infinite interval:

$$\boxed{ \mathrm{B}(a,b) =\int_0^\infty\frac{u^{a-1}}{(1+u)^{a+b}}\,\mathrm{d}u, \qquad a,b>0. }$$

In practice, an integral of the form

$$\int_0^\infty\frac{u^{p-1}}{(1+u)^q}\,\mathrm{d}u$$

can be evaluated by setting $a=p$ and $b=q-p$. The convergence condition is $q>p>0$, and the result is

$$\frac{\Gamma(p)\Gamma(q-p)}{\Gamma(q)}.$$

For example,

$$\int_0^\infty\frac{\sqrt u}{(1+u)^3}\,\mathrm{d}u =\mathrm{B}\left(\frac32,\frac32\right) =\frac\pi8.$$

Deriving the Gamma duplication formula using the Beta function

The Gamma function helps us evaluate the Beta function, and the Beta function can in turn help us derive identities for the Gamma function.

Let $x>0$. In the integral for $\mathrm{B}(x,x)$, substitute $t=(1+u)/2$ to obtain

$$\begin{aligned} \mathrm{B}(x,x) &=2^{1-2x}\int_{-1}^1(1-u^2)^{x-1}\,\mathrm{d}u\\ &=2^{2-2x}\int_0^1(1-u^2)^{x-1}\,\mathrm{d}u. \end{aligned}$$

The second step uses the symmetry of the even integrand. Next, let $v=u^2$, giving

$$\mathrm{B}(x,x) =2^{1-2x}\int_0^1v^{-1/2}(1-v)^{x-1}\,\mathrm{d}v =2^{1-2x}\mathrm{B}\left(\frac12,x\right).$$

Expressing both sides in terms of Gamma functions gives

$$\frac{\Gamma(x)^2}{\Gamma(2x)} =2^{1-2x}\frac{\sqrt\pi\,\Gamma(x)}{\Gamma(x+1/2)}.$$

Canceling the positive factor $\Gamma(x)$ and rearranging yields Legendre’s duplication formula:

$$\boxed{ \Gamma(x)\Gamma\left(x+\frac12\right) =2^{1-2x}\sqrt\pi\,\Gamma(2x),\qquad x>0. }$$

For example, taking $x=n+1/2$, where $n$ is a nonnegative integer, recovers the formula for the Gamma function at half-integers derived earlier.

Application to the Beta distribution

In probability, a direct use of the Beta function is to normalize a positive function into a probability density. If a random variable $X$ has a Beta distribution with parameters $a,b>0$, written $X\sim\operatorname{Beta}(a,b)$, its density is

$$f_X(t)=\frac{t^{a-1}(1-t)^{b-1}}{\mathrm{B}(a,b)}, \qquad 0<t<1,$$

and zero outside this interval. The denominator is exactly the integral of the numerator, so the total probability is one.

We can reuse the Beta function to calculate moments. For any nonnegative integer $k$,

$$\begin{aligned} \mathbb{E}[X^k] &=\frac1{\mathrm{B}(a,b)}\int_0^1t^{a+k-1}(1-t)^{b-1}\,\mathrm{d}t\\ &=\frac{\mathrm{B}(a+k,b)}{\mathrm{B}(a,b)}. \end{aligned}$$

Taking $k=1$ and $k=2$ and applying the recurrence relations gives

$$\mathbb{E}[X]=\frac{a}{a+b}, \qquad \mathbb{E}[X^2]=\frac{a(a+1)}{(a+b)(a+b+1)}.$$

The variance is therefore

$$\begin{aligned} \operatorname{Var}(X) &=\mathbb{E}[X^2]-\bigl(\mathbb{E}[X]\bigr)^2\\ &=\frac{a(a+1)}{(a+b)(a+b+1)}-\frac{a^2}{(a+b)^2}\\ &=\frac{ab}{(a+b)^2(a+b+1)}. \end{aligned}$$

For example, when $a=b=1$, we have $\mathrm{B}(1,1)=1$, and the density is constant at one on $(0,1)$. This Beta distribution is the uniform distribution on that interval, with mean $1/2$ and variance $1/12$.

This calculation illustrates a useful role of the Beta function: multiplying the density by $t^k$ simply increases one parameter by $k$, turning the moment integral into a ratio of two Beta functions.