For \(\operatorname{Re}a>0\) and \(\operatorname{Re}b>0\), prove that \[ B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}. \]
Hint: fix \(b\) and show that \[ F(a)=\Gamma(a+b)B(a,b) \] satisfies \(F(a+1)=aF(a)\), using \[ x^a(1-x)^{b+1}=(x^a-x^{a+1})(1-x)^b. \]
Solution
Fix \(b\) with \(\operatorname{Re}b>0\) and set \[ F(a)=\Gamma(a+b)B(a,b), \qquad \operatorname{Re}a>0. \] We show that \(F(a)/\Gamma(b)\) satisfies Wielandt’s characterization of the Gamma function.
1. The beta integral satisfies\[ (a+b)B(a+1,b)=aB(a,b). \]
From \[ 1=x+(1-x) \] inside the beta integral, \[ B(a,b)=B(a+1,b)+B(a,b+1). \] Also integration by parts gives \[ \begin{aligned} B(a+1,b) &=\int_0^1x^a(1-x)^{b-1}\,dx\\ &=\frac{a}{b}\int_0^1x^{a-1}(1-x)^b\,dx\\ &=\frac{a}{b}B(a,b+1). \end{aligned} \] The boundary term vanishes because \(\operatorname{Re}a,\operatorname{Re}b>0\). Hence \[ bB(a+1,b)=aB(a,b+1) =a\bigl(B(a,b)-B(a+1,b)\bigr), \] which rearranges to \[ (a+b)B(a+1,b)=aB(a,b). \] This is the integrated form of the identity suggested in the hint.
2. The function \(F\) satisfies the Gamma functional equation\[ F(a+1)=aF(a). \]
Using 1 and \(\Gamma(z+1)=z\Gamma(z)\), \[ \begin{aligned} F(a+1) &=\Gamma(a+b+1)B(a+1,b)\\ &=(a+b)\Gamma(a+b)B(a+1,b)\\ &=a\Gamma(a+b)B(a,b)\\ &=aF(a). \end{aligned} \]
3. The normalized function\[ G(a)=\frac{F(a)}{\Gamma(b)} \]satisfies \(G(1)=1\) and is bounded on the strip \(1\le\operatorname{Re}a\le2\).
Since \[ B(1,b)=\int_0^1(1-x)^{b-1}\,dx=\frac1b, \] the Gamma functional equation gives \[ G(1) =\frac{\Gamma(b+1)B(1,b)}{\Gamma(b)} =\frac{b\Gamma(b)}{b\Gamma(b)} =1. \]
For \(1\le\operatorname{Re}a\le2\), \[ |B(a,b)| \le\int_0^1x^{\operatorname{Re}a-1}(1-x)^{\operatorname{Re}b-1}\,dx \le\int_0^1(1-x)^{\operatorname{Re}b-1}\,dx =\frac1{\operatorname{Re}b}. \] The estimate \[ |\Gamma(z)|\le\Gamma(\operatorname{Re}z) \qquad(\operatorname{Re}z>0), \] which follows from \(|t^{z-1}|=t^{\operatorname{Re}z-1}\) in Euler’s integral, shows that \(\Gamma(a+b)\) is uniformly bounded on this strip, because \[ 1+\operatorname{Re}b \le\operatorname{Re}(a+b) \le2+\operatorname{Re}b. \] Thus \(G\) is bounded on \(1\le\operatorname{Re}a\le2\).
4. Wielandt’s uniqueness theorem identifies \(G\) with \(\Gamma\).
By 2, \[ G(a+1)=aG(a). \] By 3, \(G(1)=1\) and \(G\) is bounded on the fundamental strip. Wielandt’s characterization therefore gives \[ G(a)=\Gamma(a) \qquad(\operatorname{Re}a>0). \] Thus \[ \Gamma(a+b)B(a,b)=\Gamma(a)\Gamma(b), \] and hence \[ \boxed{ B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} }. \]