 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ib/paper-2/16b/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Integrate

F (z) = (1 + z^{2})^{- (k + 1)}

along the interval

[- R, R]

closed by the positively oriented upper semicircle, with

R > 1

. There is just one enclosed pole, at

i

, of order

k + 1

. There,

h (z) = (z - i)^{k + 1} F (z) = (z + i)^{- (k + 1)} .

Differentiating

k

times gives

h^{(k)} (z) = (- 1)^{k} \frac{( 2 k )!}{k !} (z + i)^{- (2 k + 1)} .

By part (b),

res (F, i) = \frac{( - 1 ) ^{k} ( 2 k )!}{( k ! ) ^{2} ( 2 i ) ^{2 k + 1}} = \frac{( 2 k )!}{2 ^{2 k + 1} i ( k ! ) ^{2}} .

∣1 + z^{2} ∣ \geq R^{2} - 1

, so its integral has absolute value at most

π R / (R^{2} - 1)^{k + 1}

, tending to zero. The real integral converges absolutely, so the residue theorem and the limit

R \to \infty

give

\int_{- \infty}^{\infty} \frac{d x}{( 1 + x ^{2} ) ^{k + 1}} = 2 πi res (F, i) = π \frac{( 2 k )!}{( k ! ) ^{2}} 4^{- k} .

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