0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.