How to prove $\frac{\mathcal{B}(\frac{1}{2},\frac{1}{6})}{3} = -\mathcal{B}(-\frac{1}{2},\frac{7}{6})$

Question

Recently I solved an integral with answer $\frac{\mathcal{B}(\frac{1}{2},\frac{1}{6})}{3}$ and WolframAlpah solve the same problem with answer $-\mathcal{B}(-\frac{1}{2},\frac{7}{6})$. Both are same value. How to prove the are same value using Beta function identities?

Answer 1

Try using $$ B(a,b) = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)}$$ and the properties of the gamma function

Answer 2

$$LHS=\frac{1}{3}B(1/2,1/6)=\frac{\Gamma(1/2) \Gamma(1/6)}{3\Gamma(2/3)}$$ Use $\Gamma(1+z)=z\Gamma(z)$, then $$RHS=-B(-1/2,7/6)=-\frac{\Gamma(-1/2) \Gamma(7/6)}{\Gamma(2/3)}=2\frac{(-1/2)\Gamma(-1/2)(1/6)\Gamma(1/6)}{\Gamma(2/3)}=LHS $$