Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 16

In the integral for $B(q,p)$, put $u=1-x$, so $dx=-du$ and the endpoints are interchanged. ThenThus exchanging the positive parameters leaves the Euler beta function unchanged. Positivity ensures convergence at both endpoints of this improper integral.

The increasing substitution $x={\sin }^2\theta$ maps $0\le \theta \le \pi /2$ onto $0\le x\le 1$, with $dx=2\sin \theta \cos \theta d\theta$. Since sine and cosine are nonnegative there, powers combine without a sign ambiguity:This is the trigonometric form of the beta function. The powers at the endpoints are integrable because $p,q>0$.

Write $B(n,n)={\int }_0^1[x(1-x){]}^{n-1}dx$ and set $v=2x-1$. Since $x(1-x)=(1-v^2)/4$ and $dx=dv/2$,In the last integral put $u=v^2$, giving $dv=du/(2\sqrt{u})$. This yields $B(n,n)=2^{1-2n}B(1/2,n)$, and part (a) interchanges the arguments. HenceThe equal-argument beta integral identity holds for every positive real $n$, not only positive integers.

Multiply the two defining Gamma integrals. For $p,q>0$ the nonnegative product integrand allows Tonelli's theorem:Use $r=x+y$, $u=x/(x+y)$, so $x=ru$, $y=r(1-u)$ and the positive quadrant becomes $r>0$, $0<u<1$. The Jacobian determinant has absolute valueAfter including this factor, the transformed integral separates:This proves the beta-gamma identity,All changes of variables are made inside the open positive domains; their boundaries do not contribute to the integrals.