Past exam of the short options course of the University of Oxford / 2017 / 5 / a / iv

Table of contents
- Solution iv

Solution

Complete the square:

x^{2} + 2 x + 2 = (x + 1)^{2} + 1

. In the Fourier transform, the substitution

u = x + 1

gives

F (\frac{1}{x ^{2} + 2 x + 2}) (s) = π e^{- ∣ s ∣} e^{i s} .

The phase

e^{i s}

is important. Taking the real part at

s = a

gives the cosine integral, since changing the sign in the exponential does not change its real part:

\int_{- \infty}^{\infty} \frac{cos ( a x )}{x ^{2} + 2 x + 2} d x = π e^{- ∣ a ∣} cos a (a \in R) .

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