Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 3 / 15A / f

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- Solution f

Solution

Since the odd sine part of the complex exponential integrates to zero,

\frac{1}{2 π} \int_{- n}^{n} e^{ik x} d k = \frac{1}{π} \int_{0}^{n} cos (k x) d k .

Part (e)(ii) therefore yields

n \to \infty lim \frac{1}{2 π} \int_{- n}^{n} e^{ik x} d k = δ (x)

in the sense of distributions. This is the Fourier representation of the Dirac delta function.

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