 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/4/a/iii/solution

Close the interval

[- R, R]

with an upper semicircle, traversed counterclockwise. On its large arc,

\frac{z ^{2}}{1 + z ^{4}} \leq \frac{R ^{2}}{R ^{4} - 1},

so the estimation lemma bounds the arc integral by

π R^{3} / (R^{4} - 1) \to 0

. The enclosed residues are those at

(1 + i) / 2

and

(- 1 + i) / 2

, whose sum is

- i / (22)

. The residue theorem therefore gives

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

The positive sign agrees with the nonnegative real integrand.

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