Fourier Transform of the Dirac Delta Function
Formulas
Let’s assume that the Fourier transform of the function $f(x)$ is $\hat{f}(\xi) = \mathcal{F}[f] (x) = \displaystyle \int_{-\infty}^{\infty} f(x)e^{-i \xi x}dx$. The Fourier transform of the Dirac delta function $\delta (x)$ is as follows.
$$ \hat{\delta}(\xi) = \mathcal{F}[\delta] (\xi) = 1 $$
The Fourier transform of $\delta (x - y)$ is
$$ \mathcal{F}[\delta (\cdot - y)] (\xi) = e^{-i\xi y} $$
Explanation
Depending on how the Fourier transform is defined, the preceding constant can be either $1$ or $\dfrac{1}{\sqrt{2\pi}}$ etc.
Proof
By the property of the delta function $f(x_{0}) = \int_{-\infty}^{\infty} f(x)\delta (x - x_{0})dx$,
$$ \begin{align*} \mathcal{F}[\delta] (\xi) &= \int_{-\infty}^{\infty} \delta (x)e^{-i\xi x}dx \\ &= e^{-i\xi x}|_{x=0} \\ &=1 \end{align*} $$
$$ \begin{align*} \mathcal{F}[\delta (\cdot - y)] (\xi) &= \int_{-\infty}^{\infty} \delta (x-y)e^{-i\xi x}dx \\ &= e^{-i\xi x}|_{x=y} \\ &= e^{-i\xi y} \end{align*} $$
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