 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/4/ii/solution

z = ia

and

z = - ia

. Only the first is inside the upper contour, and its residue is

Res_{z = ia} \frac{e ^{i z}}{z ^{2} + a ^{2}} = \frac{e ^{- a}}{2 ia} .

The residue theorem therefore gives, for every

R > a

\oint_{C} \frac{e ^{i z}}{z ^{2} + a ^{2}} d z = \frac{π}{a} e^{- a} .

On the upper semicircle,

∣ e^{i z} ∣ = e^{- Im z} \leq 1

and

∣ z^{2} + a^{2} ∣ \geq R^{2} - a^{2}

. The estimation lemma bounds the arc contribution by

π R / (R^{2} - a^{2}) \to 0

. Hence the real-axis integral has the same limit, and taking its real part gives

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

The imaginary part vanishes independently because

sin x / (a^{2} + x^{2})

is odd. The real integral converges absolutely, since the denominator grows quadratically.

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