How to Integrate the Incomplete Beta Function

The phrase “integral of the incomplete beta function” can mean an indefinite integral with respect to its argument, or a definite integral over a range. For the usual lower incomplete beta function, both have concise formulas. The key is to keep track of the parameter shift: integrating with respect to $x$ changes $a$ to $a+1$ in one term.

Definition and convention

The lower incomplete beta function is

$$B_x(a,b)=\int_0^x t^{a-1}(1-t)^{b-1}\,dt.$$

For real parameters, this integral is directly defined when $a,b>0$ and $0\leq x\leq1$. The formulas below also apply more broadly by analytic continuation where the functions are defined. Differentiating with respect to the upper limit gives

$$\frac{d}{dx}B_x(a,b)=x^{a-1}(1-x)^{b-1}.$$

There is also an upper incomplete beta function, often written $\int_x^1 t^{a-1}(1-t)^{b-1}\,dt$. Its antiderivative has a different sign; see the section on conventions below.

Indefinite integral of the lower function

Use integration by parts with $u=B_x(a,b)$ and $dv=dx$. Then $du=x^{a-1}(1-x)^{b-1}dx$ and $v=x$, so

$$\begin{aligned} \int B_x(a,b)\,dx &=xB_x(a,b)-\int x\,x^{a-1}(1-x)^{b-1}\,dx\\ &=xB_x(a,b)-\int x^a(1-x)^{b-1}\,dx\\ &=xB_x(a,b)-B_x(a+1,b)+C. \end{aligned}$$

Thus,

$$\boxed{\int B_x(a,b)\,dx=xB_x(a,b)-B_x(a+1,b)+C.}$$

To check the result, differentiate the right-hand side. The product rule gives $B_x(a,b)+xB_x'(a,b)$, and the derivative of $B_x(a+1,b)$ is $x^a(1-x)^{b-1}$. Since $xB_x'(a,b)=x^a(1-x)^{b-1}$, those terms cancel, leaving $B_x(a,b)$.

Worked example

Take $a=2$ and $b=3$. Directly expanding the integrand gives

$$B_x(2,3)=\int_0^x t(1-t)^2\,dt=\frac{x^2}{2}-\frac{2x^3}{3}+\frac{x^4}{4}.$$

The general formula says

$$\int B_x(2,3)\,dx=xB_x(2,3)-B_x(3,3)+C.$$

Here,

$$B_x(3,3)=\int_0^x t^2(1-t)^2\,dt=\frac{x^3}{3}-\frac{x^4}{2}+\frac{x^5}{5}.$$

Substituting and collecting terms gives

$$\int B_x(2,3)\,dx=\frac{x^3}{6}-\frac{x^4}{6}+\frac{x^5}{20}+C.$$

Differentiating this polynomial recovers $B_x(2,3)$, as required.

Definite integral from 0 to 1

Integrating the lower incomplete beta function over its full argument range gives

$$\int_0^1 B_x(a,b)\,dx=B(a,b+1),$$

where $B(a,b+1)$ is the complete beta function. One way to see this is to reverse the order of integration:

$$\begin{aligned} \int_0^1 B_x(a,b)\,dx &=\int_0^1\int_0^x t^{a-1}(1-t)^{b-1}\,dt\,dx\\ &=\int_0^1 (1-t)t^{a-1}(1-t)^{b-1}\,dt\\ &=\int_0^1 t^{a-1}(1-t)^b\,dt\\ &=B(a,b+1). \end{aligned}$$

The region of integration is $0\leq t\leq x\leq1$, so for each fixed $t$, the remaining range for $x$ has length $1-t$. For example, the worked example yields $\int_0^1 B_x(2,3)\,dx=B(2,4)=1/20$.

More generally, for $0\leq c\leq d\leq1$,

$$\int_c^d B_x(a,b)\,dx= \left[xB_x(a,b)-B_x(a+1,b)\right]_c^d.$$

Upper incomplete beta and common mistakes

If your convention is the upper tail

$$U_x(a,b)=\int_x^1 t^{a-1}(1-t)^{b-1}\,dt,$$

then $U_x'(a,b)=-x^{a-1}(1-x)^{b-1}$. Integration by parts therefore gives

$$\int U_x(a,b)\,dx=xU_x(a,b)+B_x(a+1,b)+C.$$

The sign is positive in the second term, unlike the lower-function formula. Check the definition before applying either result: authors do not always use the same notation for lower and upper incomplete beta functions.

Another common error is shifting the wrong parameter. Multiplication of the defining integrand by $x$ changes $x^{a-1}$ into $x^a$, so the shift is $a\mapsto a+1$, not $b\mapsto b+1$. The definite integral over $[0,1]$ does involve $b+1$, because reversing the order of integration contributes a factor of $1-t$.