How to Prove the Beta–Gamma Function Identity

The beta–gamma identity connects two standard improper integrals. For positive real numbers $x$ and $y$, it is

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

A direct proof comes from multiplying the integral definitions of the gamma function and changing variables in the resulting double integral. The condition $x,y>0$ ensures the defining integrals converge.

Start with the definitions

For $x>0$, the gamma function is

$$\Gamma(x)=\int_0^\infty u^{x-1}e^{-u}\,du.$$

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

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

Multiplying the two gamma integrals gives

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

The region of integration is the positive quadrant: $u>0$ and $v>0$. The integrand is nonnegative, so the iterated integrals can be combined as a double integral.

Change variables in the double integral

Use the variables

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

Solving for the old variables gives

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

As $u,v$ range over the positive quadrant, $s$ ranges over $(0,\infty)$ and $t$ ranges over $(0,1)$. The Jacobian determinant is

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

Substitute these expressions into the double integral. The power terms become $u^{x-1}v^{y-1}=s^{x+y-2}t^{x-1}(1-t)^{y-1}$, the exponential becomes $e^{-s}$, and the area element contributes the additional factor $s$. Therefore,

$$\begin{aligned} \Gamma(x)\Gamma(y) &=\int_0^\infty\int_0^1 s^{x+y-1}e^{-s}t^{x-1}(1-t)^{y-1}\,dt\,ds\\ &=\left(\int_0^\infty s^{x+y-1}e^{-s}\,ds\right) \left(\int_0^1 t^{x-1}(1-t)^{y-1}\,dt\right)\\ &=\Gamma(x+y)B(x,y). \end{aligned}$$

The first factor in the product is the defining integral for $\Gamma(x+y)$, and the second is the defining integral for $B(x,y)$. Since $x+y>0$, the gamma factor is finite and nonzero. Dividing by it proves

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

Worked example: evaluate $B(2,3)$

The identity and the recurrence $\Gamma(n+1)=n\Gamma(n)$ give

$$B(2,3)=\frac{\Gamma(2)\Gamma(3)}{\Gamma(5)} =\frac{1!\,2!}{4!}=\frac{2}{24}=\frac{1}{12}.$$

You can check this directly from the beta integral:

$$\begin{aligned} B(2,3)&=\int_0^1 t(1-t)^2\,dt\\ &=\int_0^1(t-2t^2+t^3)\,dt\\ &=\frac12-\frac23+\frac14=\frac{1}{12}. \end{aligned}$$

The agreement illustrates how the identity turns a beta integral into gamma values, which are especially simple at positive integers.

Common mistakes

• Missing the Jacobian. The change of variables contributes $s$. Without it, the power of $s$ is wrong and the gamma factor would not be $\Gamma(x+y)$.
• Reversing the integration ranges. The ratio $t=u/(u+v)$ lies between $0$ and $1$, while the sum $s=u+v$ is positive and unbounded.
• Ignoring the parameter conditions. The defining integrals converge for $x>0$ and $y>0$. Extensions to other parameter values require analytic continuation and are not justified by this integral proof alone.
• Confusing the identity with a product formula. The proof first gives $\Gamma(x)\Gamma(y)=\Gamma(x+y)B(x,y)$. The displayed quotient follows by dividing by $\Gamma(x+y)$.

A useful consequence

The beta integral is symmetric: replacing $t$ by $1-t$ shows $B(x,y)=B(y,x)$. The gamma quotient gives the same result immediately, since its numerator and denominator are unchanged when $x$ and $y$ are swapped. For positive parameters, both viewpoints are valid and can be useful: use the integral to study the shape of the integrand, and the gamma formula to compute values or derive identities.