TAOCP 1.2.6 Exercise 43

Exercise 43. [HM20] Show that $B(1/2,1/2)=\pi$.

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Solution

We are asked to evaluate the beta function

$$ B!\left(\frac{1}{2},\frac{1}{2}\right) = \int_0^1 t^{-1/2} (1-t)^{-1/2}, dt. $$

We perform the substitution $t = \sin^2\theta$, so that $dt = 2\sin\theta\cos\theta, d\theta$. Then

$$ t^{-1/2} = (\sin^2\theta)^{-1/2} = \frac{1}{\sin\theta}, \qquad (1-t)^{-1/2} = (\cos^2\theta)^{-1/2} = \frac{1}{\cos\theta}. $$

Hence the integrand transforms as

$$ t^{-1/2} (1-t)^{-1/2} , dt = \frac{1}{\sin\theta}\cdot \frac{1}{\cos\theta} \cdot 2\sin\theta \cos\theta , d\theta = 2, d\theta. $$

The limits of integration change as $t = 0 \mapsto \theta = 0$ and $t = 1 \mapsto \theta = \pi/2$. Therefore

$$ B!\left(\frac{1}{2},\frac{1}{2}\right) = \int_0^{\pi/2} 2 , d\theta = 2 \cdot \frac{\pi}{2} = \pi. $$

Thus the value of the beta function is

$$ \boxed{\pi}. $$

This completes the proof.

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