 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/i/solution

For

m = 2

, evaluate the improper integral directly:

I_{2, n} = \int_{0}^{\infty} (1 + x)^{- n} d x = [\frac{( 1 + x ) ^{1 - n}}{1 - n}]_{0}^{\infty} = \frac{1}{n - 1} .

For

m \geq 3

u = x^{m - 2}

and

d v = (1 + x)^{- (m + n - 2)} d x

. Then

v = - (1 + x)^{- (m + n - 3)} / (m + n - 3)

and

I_{m, n} = [- \frac{x ^{m - 2}}{( m + n - 3 ) ( 1 + x ) ^{m + n - 3}}]_{0}^{\infty} + \frac{m - 2}{m + n - 3} \int_{0}^{\infty} \frac{x ^{m - 3}}{( 1 + x ) ^{m + n - 3}} d x = \frac{m - 2}{m + n - 3} I_{m - 1, n} .

The boundary term vanishes: at zero

m - 2 > 0

, while at infinity its power is

1 - n < 0

. The restriction

m \geq 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}

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