 Solution

ID: 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/ii/solution

Apply the recurrence repeatedly down to

m = 2

I_{m, n} = \frac{( m - 2 ) ( m - 3 ) \dots 1}{( m + n - 3 ) ( m + n - 4 ) \dots n} I_{2, n} .

Including the base factor

1/ (n - 1)

and rewriting the products as factorials gives

I_{m, n} = \frac{( m - 2 )! ( n - 2 )!}{( m + n - 3 )!}, m, n \in Z, m, n > 1.

The formula also includes

m = 2

because

0! = 1

. It is symmetric in

m, n

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.

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