📂Distribution Theory

Fourier Transform of the Dirac Delta Function Defined as a Distribution

Introduction¹

The Fourier transform of a tempered distribution $F$ is defined, for a Schwartz function $\phi$, as follows.

$$ \hat{F}[\phi] := F[\hat{\phi}] $$

Applying this to the Dirac delta function yields the following.

$$ \widehat{\delta}[\phi] = \delta[\widehat{\phi}] = \widehat{\phi}(\mathbf{0}) = \int_{\mathbb{R}^{n}} \phi (\mathbf{x}) e^{-\i \mathbf{x}\cdot \mathbf{0}} \d\mathbf{x} = \int_{\mathbb{R}^{n}} \phi (\mathbf{x}) \d\mathbf{x} $$

Considering here that the right-hand side is $\phi$ multiplied by the constant $1$, we can define the constant $\mathbf{1}$ as a distribution as follows.

$$ \mathbf{1}[\phi] = \int_{\mathbb{R}^{n}} 1 \cdot \phi (\mathbf{x}) \d\mathbf{x} = \int_{\mathbb{R}^{n}} \phi (\mathbf{x}) \d\mathbf{x} $$

Therefore, from the distributional viewpoint, the Fourier transform of the Dirac delta function is the constant function $1$.

$$ \hat{\delta} = 1 \tag{1} $$

Explanation

Thinking of $(1)$ in reverse, since the inverse Fourier transform of $1$ is the delta function, we obtain the following equation.

$$ \delta(\mathbf{x}) = \mathcal{F}^{-1}[1] = \dfrac{1}{(2\pi)^{n}} \int_{\mathbb{R}^{n}} e^{\i \mathbf{x}\cdot\boldsymbol{\xi}} \d\boldsymbol{\xi} $$

The integral on the right-hand side diverges, since the absolute value of the integrand $e^{\i \mathbf{x}\cdot\boldsymbol{\xi}}$ is $1$ everywhere. Therefore this identity is justified only on the space $\mathcal{S}^{\ast}$ of tempered distributions. The coefficient $(2\pi)^{-n}$ in front of the right-hand side depends on how the Fourier transform is defined. From this we obtain the following useful identity.

$$ \delta(\mathbf{x} - \mathbf{y}) = \dfrac{1}{(2\pi)^{n}} \int_{\mathbb{R}^{n}} e^{\i (\mathbf{x} - \mathbf{y})\cdot\boldsymbol{\xi}} \d\boldsymbol{\xi} = \braket{e^{\i \mathbf{y}} | e^{\i \mathbf{x}}} $$

$$ \int_{-\infty}^{\infty} e^{\i Axy} \d y = \dfrac{2\pi}{A}\delta(x) $$