The Beta Integral and Beta-Gamma Identity

Why the Beta Integral Appears

The gamma function integrates a power against exponential decay over the positive half-line. The beta integral, by contrast, integrates powers over the finite interval from zero to one. These two integrals are closely connected: a change of variables in a product of two gamma integrals separates a radial coordinate from a proportion coordinate, and the latter produces the beta integral.

The connection is useful both for evaluating integrals and for normalizing distributions built from powers of $t$ and $1-t$. Here the aim is to establish the integral identity itself. We will first check exactly when the beta integral converges, then derive the identity with the Jacobian included.

Definition and Convergence

Definition: For real parameters $a>0$ and $b>0$, the beta integral is $$ B(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1}\,dt. $$ The integral is understood as an improper integral at the endpoints when necessary.

The integrand is positive and continuous on every closed subinterval of $(0,1)$. The only possible convergence issues are therefore at zero and one. Near zero, the factor $(1-t)^{b-1}$ stays bounded while $t^{a-1}$ is integrable precisely when $a>0$. Near one, the corresponding condition is $b>0$. The following proof makes those endpoint checks explicit.

Theorem: The beta integral converges to a finite, strictly positive value for every $a>0$ and $b>0$.

Proof. Split the interval at $1/2$. On $0<t<1/2$, the factor $(1-t)^{b-1}$ is bounded above by $C_b=\max(1,2^{1-b})$. Thus $$ 0\leq \int_0^{1/2} t^{a-1}(1-t)^{b-1}\,dt \leq C_b\int_0^{1/2}t^{a-1}\,dt =\frac{C_b}{a}\left(\frac12\right)^a<\infty, $$ because $a>0$. On $1/2<t<1$, the factor $t^{a-1}$ is bounded above by $C_a=\max(1,2^{1-a})$. Hence $$ 0\leq \int_{1/2}^1 t^{a-1}(1-t)^{b-1}\,dt \leq C_a\int_{1/2}^1(1-t)^{b-1}\,dt =\frac{C_a}{b}\left(\frac12\right)^b<\infty, $$ because $b>0$. The integral is therefore finite. Its integrand is strictly positive throughout $(0,1)$, so its integral is strictly positive as well. $\square$

Worked Example: Checking Convergence Near an Endpoint

Consider $B(1/3,2)$. Its integrand is $t^{-2/3}(1-t)$, which becomes unbounded as $t$ approaches zero. Unboundedness alone does not imply divergence; the relevant question is whether the power is integrable.

Near zero, $1-t$ is bounded, and $$ \int_0^{1/2}t^{-2/3}\,dt =\left[3t^{1/3}\right]_0^{1/2} =3\left(\frac12\right)^{1/3}<\infty. $$ Near one, $t^{-2/3}$ is bounded and $(1-t)^1$ is integrable. The theorem therefore applies: $B(1/3,2)$ is finite.

Deriving the Beta-Gamma Identity

For $s>0$, the gamma function is defined by $$ \Gamma(s)=\int_0^\infty x^{s-1}e^{-x}\,dx. $$ This integral is finite and positive for $s>0$. Its recurrence, established with the gamma function, is $\Gamma(s+1)=s\Gamma(s)$, and $\Gamma(1)=1$. We now use the gamma integral twice, with parameters $a$ and $b$.

Theorem (Beta-Gamma Identity): For $a>0$ and $b>0$, $$ B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}. $$

Proof. Multiply the two gamma integrals. Since their integrands are nonnegative, Tonelli's theorem permits writing the product as a double integral: $$ \Gamma(a)\Gamma(b) =\int_0^\infty\int_0^\infty x^{a-1}y^{b-1}e^{-(x+y)}\,dy\,dx. $$ Make the change of variables $$ x=ru,\qquad y=r(1-u),\qquad r>0,\quad 0<u<1. $$ This is a one-to-one differentiable map from $(0,\infty)\times(0,1)$ onto the positive quadrant. Its Jacobian determinant is $$ \begin{vmatrix} \partial x/\partial r & \partial x/\partial u\\ \partial y/\partial r & \partial y/\partial u \end{vmatrix} = \begin{vmatrix} u&r\\ 1-u&-r \end{vmatrix} =-r, $$ so the absolute Jacobian is $r$. Also $x+y=r$. Substituting gives $$ \Gamma(a)\Gamma(b) =\int_0^\infty\int_0^1 r^{a+b-1}e^{-r}u^{a-1}(1-u)^{b-1}\,du\,dr. $$ The integrand is nonnegative, so Tonelli's theorem allows the two factors to be separated: $$ \Gamma(a)\Gamma(b) =\left(\int_0^\infty r^{a+b-1}e^{-r}\,dr\right) \left(\int_0^1u^{a-1}(1-u)^{b-1}\,du\right) =\Gamma(a+b)B(a,b). $$ Both factors are finite for $a,b>0$, by the gamma integral's convergence and the beta-integral convergence proved above. Since $\Gamma(a+b)>0$, division by it proves the identity. $\square$

The variables $r$ and $u$ have distinct roles in this calculation. The coordinate $r=x+y$ records the total, while $u=x/(x+y)$ records the fraction of that total contributed by $x$. The Jacobian factor $r$ is essential: together with the powers of $x$ and $y$, it produces $r^{a+b-1}$, the power in $\Gamma(a+b)$.

Start with two gamma integrals.
Their product is a nonnegative double integral over $x>0,y>0$.

Use total and fraction coordinates.
Set $r=x+y$ and $u=x/(x+y)$, so $x=ru$ and $y=r(1-u)$.

Include the Jacobian.
The absolute Jacobian is $r$; the transformed integral factors into $\Gamma(a+b)B(a,b)$.

Worked Evaluations

Worked Example: An Integral With One Integer Parameter

Evaluate $B(1/3,2)$ directly and verify the result using the beta-gamma identity.

Directly, $$ B\left(\frac13,2\right) =\int_0^1t^{-2/3}(1-t)\,dt =\int_0^1\left(t^{-2/3}-t^{1/3}\right)\,dt. $$ Both powers are integrable at zero. Evaluating their antiderivatives gives $$ \left[3t^{1/3}-\frac34t^{4/3}\right]_0^1 =3-\frac34=\frac94. $$ For the identity check, $\Gamma(2)=1$, and the recurrence gives $$ \Gamma\left(\frac73\right) =\frac43\Gamma\left(\frac43\right) =\frac43\cdot\frac13\Gamma\left(\frac13\right) =\frac49\Gamma\left(\frac13\right). $$ Therefore $$ \frac{\Gamma(1/3)\Gamma(2)}{\Gamma(7/3)} =\frac{\Gamma(1/3)}{(4/9)\Gamma(1/3)} =\frac94, $$ matching the direct integral.

Worked Example: An Integer-Parameter Check

Evaluate $B(2,3)$ by integration and by the gamma functions. Direct expansion gives $$ B(2,3)=\int_0^1t(1-t)^2\,dt =\int_0^1(t-2t^2+t^3)\,dt =\frac12-\frac23+\frac14 =\frac1{12}. $$ For the second calculation, the recurrence and $\Gamma(1)=1$ give $\Gamma(2)=1$, $\Gamma(3)=2$, and $\Gamma(5)=24$. Thus $$ \frac{\Gamma(2)\Gamma(3)}{\Gamma(5)} =\frac{1\cdot2}{24} =\frac1{12}. $$ The two methods agree. In general, for positive integers $m,n$, repeated use of the recurrence gives $B(m,n)=(m-1)!(n-1)!/(m+n-1)!$.

Worked Example: A Beta Integral With Half Parameters

Evaluate $B(1/2,1/2)$ directly, then check its value through the identity.

In the integral, set $t=\sin^2\theta$, where $0<\theta<\pi/2$. Then $dt=2\sin\theta\cos\theta\,d\theta$, and $\sqrt{t(1-t)}=\sin\theta\cos\theta$ on this interval. Consequently, $$ B\left(\frac12,\frac12\right) =\int_0^1\frac{dt}{\sqrt{t(1-t)}} =2\int_0^{\pi/2}d\theta =\pi. $$ For the gamma-function check, substituting $x=z^2$ in the defining integral yields $$ \Gamma\left(\frac12\right) =2\int_0^\infty e^{-z^2}\,dz =\int_{-\infty}^\infty e^{-z^2}\,dz =\sqrt{\pi}. $$ The final equality is the Gaussian integral: squaring the positive integral over the real line and using polar coordinates gives $$ \left(\int_{-\infty}^\infty e^{-z^2}\,dz\right)^2 =\int_0^{2\pi}\int_0^\infty e^{-r^2}r\,dr\,d\theta =2\pi\cdot\frac12=\pi. $$ Since $\Gamma(1)=1$, the beta-gamma identity gives $$ \frac{\Gamma(1/2)^2}{\Gamma(1)} =\frac{(\sqrt{\pi})^2}{1} =\pi, $$ as required.

Common Pitfall and Why the Identity Matters

A common error is to omit the Jacobian $r$ in the two-dimensional change of variables. Doing so would leave the radial power as $r^{a+b-2}$, suggesting $\Gamma(a+b-1)$ instead of $\Gamma(a+b)$. That is not a harmless change in notation: it gives the wrong integral. The correct determinant has absolute value $r$, and the transformed power is $r^{a-1}r^{b-1}r=r^{a+b-1}$.

Another mistake is to assume that the beta integral converges simply because the interval is finite. The interval's length does not control endpoint singularities. The exponents must still satisfy $a>0$ and $b>0$; for example, $t^{-1}$ is not integrable at zero. In the beta integral, those conditions follow from checking the two endpoints separately.

The identity turns an integral on $(0,1)$ into values of the gamma function, whose recurrence often makes special cases straightforward to evaluate. It also explains the normalizing constant used in beta distributions, which will be developed separately. The derivation's central idea—separating a total from a proportion—appears broadly when a joint positive-variable calculation is reorganized into scale and relative composition.

Key takeaway: For $a,b>0$, both endpoints of $B(a,b)$ are integrable, and changing variables in $\Gamma(a)\Gamma(b)$ gives $\Gamma(a)\Gamma(b)=\Gamma(a+b)B(a,b)$.

Check Your Understanding

Use the endpoint tests and the change-of-variables argument to answer these questions.

For which values of $a$ and $b$ is the beta integral defined as a finite integral?
Which factor in the beta integrand controls convergence near $t=0$, and what condition does it impose?
Under $x=ru,\ y=r(1-u)$, what is the absolute Jacobian, and what is the resulting power of $r$?
Use the beta-gamma identity and the gamma recurrence to evaluate $B(1,3)$.
Why is it invalid to omit the Jacobian when transforming the double gamma integral?