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Dirac delta function

{title2=$\delta(x)=\frac1{2\pi}\int_{-\infty}^{\infty}e^{ikx}\,dk$}

The identity
$$
\delta(x)=\frac1{2\pi}\int_{-\infty}^{\infty}e^{ikx}\,dk
$$
is interpreted distributionally: symmetric frequency cutoffs give $\sin(nx)/(\pi x)$, which converge to the Dirac delta function.


Fourier representation of the Dirac delta function

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 301 / 1 / c / Solution

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

Fourier representation of the Dirac delta function ( $δ (x) = \frac{1}{2 π} \int_{- \infty}^{\infty} e^{ik x} d k$ )

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Fourier representation of the Dirac delta function (δ(x)=2π1​∫−∞∞​eikxdk)

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Fourier representation of the Dirac delta function ( $δ (x) = \frac{1}{2 π} \int_{- \infty}^{\infty} e^{ik x} d k$ )

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