 Solution

ID: past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2023/5/c/solution

(z^{2} + 1) (z^{2} + 4)

and integrate

F (z) = \frac{e ^{i z}}{( z ^{2} + 1 ) ( z ^{2} + 4 )}

over

[- R, R]

followed by the upper semicircle, for

R > 2

. The enclosed simple poles are

i

and

2 i

. The local quotient formula from part (a) gives

Res_{i} F = \frac{e ^{- 1}}{( 2 i ) ( 3 )} = \frac{e ^{- 1}}{6 i}, Res_{2 i} F = \frac{e ^{- 2}}{( - 3 ) ( 4 i )} = - \frac{e ^{- 2}}{12 i} .

On this arc,

∣ e^{i z} ∣ = e^{- Im z} \leq 1

and

∣ z^{2} + 1∣ \geq R^{2} - 1

∣ z^{2} + 4∣ \geq R^{2} - 4

. Thus the estimation lemma gives

\int_{arc} F (z) d z \leq \frac{π R}{( R ^{2} - 1 ) ( R ^{2} - 4 )} ⟶ 0.

∣ F (x) ∣ \leq 1/ [(x^{2} + 1) (x^{2} + 4)]

, so the improper integral is absolutely convergent. Hence the residue theorem yields

\int_{- \infty}^{\infty} \frac{e ^{i x}}{x ^{4} + 5 x ^{2} + 4} d x = 2 πi (\frac{e ^{- 1}}{6 i} - \frac{e ^{- 2}}{12 i}) .

Taking real parts is justified by absolute convergence. Therefore

\int_{- \infty}^{\infty} \frac{cos x}{x ^{4} + 5 x ^{2} + 4} d x = \frac{π}{3 e} - \frac{π}{6 e ^{2}} .

The imaginary part vanishes as a further check, since the sine integrand is odd.

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