# Prove: $B(x,x) = 2^{(1-x)}B(x,\frac{1}{2})$, for $B(x,y) = \int_{0}^{1}t^{x-1}(1-t)^{y-1}$

Prove: $B(x,x) = 2^{(1-x)}B(x,\frac{1}{2})$, for $B(x,y) = \int_{0}^{1}t^{x-1}(1-t)^{y-1}$

I'm not really sure how to approach this. I'm pretty sure integrating by parts doesn't work, I can't think of a good substitution and I'm not sure I can evaluate the integral.

Any help would be appreciated.

We use the identity (letting $t=\sin^2\theta$) \begin{align}\text{B}(x,y)&=\int_0^1t^{x-1}(1-t)^{y-1}dt\\ &=\int_0^{\frac{\pi}2}\sin^{2x-2}\theta\left(1-\cos^2\theta\right)^{y-1}\cdot2\sin\theta\cos\theta\,d\theta\\ &=2\int_0^{\frac{\pi}2}\sin^{2x-1}\theta\cos^{2y-1}\theta\,d\theta\end{align} So \begin{align}\text{B}(x,x)&=2\int_0^{\frac{\pi}2}\sin^{2x-1}\theta\cos^{2y-1}\theta\,d\theta\\ &=\frac2{2^{2x-1}}\int_0^{\frac{\pi}2}\left(2\sin\theta\cos\theta\right)^{2x-1}d\theta\\ &=\frac2{2^{2x-1}}\int_0^{\frac{\pi}2}\sin^{2x-1}2\theta\,d\theta\\ &=\frac2{2^{2x-1}}\int_0^{\pi}\sin^{2x-1}\phi\left(\frac{d\phi}2\right)\\ &=\frac2{2^{2x-1}}\int_0^{\frac{\pi}2}\sin^{2x-1}\phi\,d\phi\\ &=\frac2{2^{2x-1}}\int_0^{\frac{\pi}2}\sin^{2x-1}\phi\cos^{2\left(\frac12\right)-1}\phi\,d\phi\\ &=\frac1{2^{2x-1}}\text{B}\left(x,\frac12\right)\end{align} So there was a typo in your formula, please correct it. This is a famous theorem called the duplication formula for the Gamma function.