Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 2 / 12A (Complex Analysis or Complex Methods)

Table of contents
- a 12A
  - Solution a
- b 12A
  - Solution b

a

Solution

Close the contour in the upper half-plane. The degree assumption and Jordan lemma make the integral over the large semicircle tend to zero, while the absence of real zeros avoids indentations of the contour. The residue theorem therefore gives

\int_{- \infty}^{\infty} R (x) e^{i x} d x = 2 πi Q (z_{k}) = 0 Im z_{k} > 0 \sum Res_{z = z_{k}} (R (z) e^{i z}) .

Repeated roots are handled by the usual higher-order residue formula.

b

Solution

Apply the formula to

R (z) = \frac{z}{1 + z ^{4}} .

The upper-half-plane poles are

z_{1} = e^{iπ /4}, z_{2} = e^{3 iπ /4},

and their residues are

e^{i z_{k}} / (4 z_{k}^{2})

c = 1/ 2

, their sum is

\frac{i}{4} (e^{i z_{2}} - e^{i z_{1}}) = \frac{1}{2} e^{- c} sin c .

Since the cosine part of

x e^{i x} / (1 + x^{4})

is odd, its integral vanishes, while the sine part is even. Hence

i \int_{- \infty}^{\infty} \frac{x s i n x}{1 + x ^{4}} d x = 2 πi (\frac{1}{2} e^{- c} sin c),

so

\int_{- \infty}^{\infty} \frac{x sin x}{1 + x ^{4}} d x = π e^{- 1/ 2} sin (\frac{1}{2}) .

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