$$B(p,q)=\int_0^1t^{p-1}(1-t)^{q-1}dt\;(p>0,q>0)$$ I want to find the value of $B(\frac1 2,\frac1 2)$. And I tried to make $t=\cos ^2\theta$ so that the integral is equal to $$\begin{align*}&-\int_{-\infty}^{\infty}\left(\frac{\cos2\theta+1} 2\right)^{-\frac1 2}\left(\frac{1-\cos2\theta} 2\right)^{-\frac1 2}\sin 2\theta \;d\theta\\ &=-2\int_{-\infty}^{\infty}\frac{\sin 2\theta}{\sqrt{(1-\cos 2\theta)(1+\cos 2\theta)}}d\theta\\ &=-2\theta\bigg|_{-\infty}^{\infty}\\&=-\infty\end{align*}$$ Since the negative infinity is not likely to be the correct result, could someone tell me where I went wrong?
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2$\begingroup$ Doesn't $\theta$ vary between $0$ and $\pi/2$ $\endgroup$– Angina SengJul 10, 2020 at 11:03
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$\begingroup$ So the interval should go from $\pi/2$ to $0$ right? Then the result is $\pi$. It makes sense. Sorry for the stupid mistake. $\endgroup$– B.W. ZhangJul 10, 2020 at 11:41
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1 Answer
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Recall that for $x,y>0$, the following holds:
\begin{equation} \mathrm{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \end{equation}
Then:
\begin{equation} \mathrm{B}\left(\frac{1}{2},\frac{1}{2}\right)=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(1)} =\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{0!} = \Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right) \end{equation}
It is known that $\Gamma(1/2)=\sqrt{\pi}$, therefore $\mathrm{B}\left(\frac{1}{2},\frac{1}{2}\right) = \pi$
