Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 3 / b / Solution

For the two-sided exponential, split the integral at zero:

f (k) = \int_{0}^{\infty} e^{- (1 + ik) x} d x + \int_{- \infty}^{0} e^{(1 - ik) x} d x = \frac{1}{1 + ik} + \frac{1}{1 - ik} = \frac{2}{1 + k ^{2}} .

For the interval indicator, the Fourier transform of an interval indicator gives

g (k) = \int_{- 2 π}^{2 π} e^{- ik x} d x = \frac{2 sin ( 2 πk )}{k} .

At

k = 0

, the continuous limiting value is

4 π

, the length of the interval. Changing the values at its two endpoints has no effect on the transform.

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