 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/4/d/i/solution

Close the real segment

[- R, R]

by an anticlockwise semicircle in the upper half-plane. For

F (z) = 1/ [(z^{2} + 1) (z^{2} + 4)]

, the contour integral over the arc is

O (R^{- 3})

: its length is

π R

, whereas

∣ z^{2} + 1∣ \geq R^{2} - 1

and

∣ z^{2} + 4∣ \geq R^{2} - 4

. The enclosed simple poles are

i

and

2 i

Res_{i} F = \frac{1}{( 2 i ) ( i ^{2} + 4 )} = \frac{1}{6 i}, Res_{2 i} F = \frac{1}{( 4 i ) (( 2 i ) ^{2} + 1 )} = - \frac{1}{12 i} .

Letting

R \to \infty

\int_{- \infty}^{\infty} \frac{d x}{( x ^{2} + 1 ) ( x ^{2} + 4 )} = 2 πi (\frac{1}{6 i} - \frac{1}{12 i}) = \frac{π}{6} .

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