 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2004/5/i/solution

Integrate

F (z) = e^{i z} / (z^{2} + 1)

around the real segment from

- R

R

and the counterclockwise upper semicircle, with

R > 1

. The only enclosed pole is

i

, whose residue is

e^{- 1} / (2 i)

. On the semicircle,

∣ e^{i z} ∣ \leq 1

and

∣ z^{2} + 1∣ \geq R^{2} - 1

, so its integral has magnitude at most

π R / (R^{2} - 1) \to 0

The residue theorem thus gives

\int_{- \infty}^{\infty} e^{i x} / (x^{2} + 1) d x = π e^{- 1}

. The line integral converges absolutely, so taking its real part is justified:

\int_{- \infty}^{\infty} \frac{cos x}{x ^{2} + 1} d x = \frac{π}{e} .

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