 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2006/5/solution

F (z) = (z^{2} + 4)^{- 2}

on the upper semicircle of radius

R > 2

, traversed counterclockwise. Its sole enclosed pole is the double pole at

2 i

, and

Res_{z = 2 i} F = \frac{d}{d z} (z + 2 i)^{- 2}_{z = 2 i} = - \frac{2}{( 4 i ) ^{3}} = - \frac{i}{32} .

On the semicircular arc,

∣ z^{2} + 4∣ \geq R^{2} - 4

, so the ML inequality bounds its contribution by

π R / (R^{2} - 4)^{2} \to 0

. The real integral is absolutely convergent, since its integrand is

O (∣ x ∣^{- 4})

at infinity. Taking

R \to \infty

in the contour identity yields

\int_{- \infty}^{\infty} \frac{d x}{( x ^{2} + 4 ) ^{2}} = 2 πi (- \frac{i}{32}) = \frac{π}{16} .

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