Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 18 / b

For $m=2$, evaluate the improper integral directly:For $m\ge 3$, use integration by parts with $u=x^{m-2}$ and $dv=(1+x{)}^{-(m+n-2)}dx$. Then $v=-(1+x{)}^{-(m+n-3)}/(m+n-3)$ andThe boundary term vanishes: at zero $m-2>0$, while at infinity its power is $1-n<0$. The restriction $m\ge 3$ belongs to this recurrence step; the separate $m=2$ base case must not be replaced by a formal zero times a divergent $I_{1,n}$.

Apply the recurrence repeatedly down to $m=2$:Including the base factor $1/(n-1)$ and rewriting the products as factorials givesThe formula also includes $m=2$ because $0!=1$. It is symmetric in $m,n$, consistent with the integration by substitution $x=1/t$. Equivalently it is the Euler beta function $B(m-1,n-1)$, although that identification is not needed for the recurrence derivation.