= Solution
Apply the recurrence repeatedly down to $m=2$:
$$
I_{m,n}=\frac{(m-2)(m-3)\cdots1}{(m+n-3)(m+n-4)\cdots n}\,I_{2,n}.
$$
Including the base factor $1/(n-1)$ and rewriting the products as <factorials> gives
$$
\boxed{I_{m,n}=\frac{(m-2)!(n-2)!}{(m+n-3)!},\qquad m,n\in\mathbb Z,\quad m,n>1.}
$$
The 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.
Back to article page
