What Is the Limiting Value of the Beta Function?

The Beta function has two arguments, so “the limiting value” is not defined until you say which argument changes and what happens to the other one. A common endpoint question is to let one argument approach zero while holding the other positive and fixed.

For that interpretation, the Beta function itself has no finite limit: it grows without bound. Its leading behavior is $B(x,y)\sim 1/x$, so the scaled limit is $\lim_{x\to0^+}xB(x,y)=1$. Other limits, such as letting an argument tend to infinity, have different answers.

Definition and the key identity

For $x,y>0$, the Beta function is

$$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\,dt =\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.$$

The integral definition shows that $B(x,y)$ is finite when both arguments are positive. To study the limit as $x$ approaches zero, the Gamma-function identity is especially convenient. The Gamma function has a pole at zero, with $\Gamma(x)\sim 1/x$ as $x\to0^+$, while $\Gamma(x+y)$ tends to $\Gamma(y)$ for fixed $y>0$.

Limit as one argument approaches zero

Starting from the identity, multiply by $x$:

$$\begin{aligned} xB(x,y) &=x\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\\ &=\frac{\Gamma(1+x)\Gamma(y)}{\Gamma(x+y)}. \end{aligned}$$

Here we used $\Gamma(1+x)=x\Gamma(x)$. As $x\to0^+$, continuity of $\Gamma$ at the positive number $y$ gives $\Gamma(1+x)\to1$ and $\Gamma(x+y)\to\Gamma(y)$. Therefore

$$\boxed{\lim_{x\to0^+}xB(x,y)=1\qquad(y>0\text{ fixed}).}$$

Since $x$ tends to zero through positive values, this also says

$$B(x,y)\sim\frac{1}{x},$$

so $B(x,y)\to+\infty$. The symmetry $B(x,y)=B(y,x)$ gives the corresponding result when $y\to0^+$ with $x>0$ fixed.

Worked example: fixing the second argument at 3

For $y=3$, the Gamma identity simplifies to an elementary expression:

$$B(x,3)=\frac{\Gamma(x)\Gamma(3)}{\Gamma(x+3)} =\frac{2}{x(x+1)(x+2)}.$$

Thus

$$xB(x,3)=\frac{2}{(x+1)(x+2)}\longrightarrow\frac{2}{1\cdot2}=1.$$

At the same time, $B(x,3)$ itself is approximately $1/x$ for small positive $x$, so it diverges. The scaled expression has a finite limit; the unscaled function does not.

Other common limiting cases

The answer changes if you vary the arguments differently. For fixed $y>0$ and $x\to\infty$, the Gamma-ratio asymptotic gives

$$B(x,y)\sim\Gamma(y)x^{-y},$$

so $B(x,y)\to0$. By symmetry, the analogous result holds when $y\to\infty$ and $x$ is fixed positive.

If both arguments approach zero, there is no single path-independent rate. Using $\Gamma(z)\sim1/z$ near zero gives

$$B(x,y)\sim\frac{x+y}{xy}=\frac1x+\frac1y \qquad(x,y\to0^+).$$

This expression still diverges, but its size depends on how $x$ and $y$ approach zero. For example, if $y=x$, then $B(x,x)\sim2/x$; if $y$ is much smaller than $x$, the term $1/y$ dominates.

Common mistakes

• Reporting a limit without specifying the path. A function of two variables can have different limiting behavior depending on which arguments vary.
• Confusing $B(x,y)$ with $xB(x,y)$. For fixed $y>0$, the first tends to infinity as $x\to0^+$; the second tends to 1.
• Using the endpoint integral without checking convergence. At $x=0$, the integrand contains $t^{-1}$ near zero, which is not integrable. The Beta integral is defined for positive arguments, and the limit is taken from within that domain.
• Treating $x\to0$ as a two-sided real limit. The standard Beta integral requires positive arguments, so this endpoint limit is $x\to0^+$.