How to Prove That the Inverse Fourier Transform of 1 Is the Delta Function

The statement

$$\mathcal{F}^{-1}\{1\}=\delta_0$$

is correct when the inverse Fourier transform is understood in the sense of distributions. The symbol $\delta_0$ denotes the Dirac delta centered at $0$, which is not an ordinary function. It is defined by how it acts on a suitable test function $\varphi$:

$$\langle \delta_0,\varphi\rangle=\varphi(0).$$

The numerical constants depend on the Fourier-transform convention, so fix one before proving the identity.

Fixing the Fourier convention

Use the common $2\pi$ convention

$$\widehat{f}(\xi)=\int_{\mathbb{R}}f(x)e^{-2\pi i x\xi}\,dx,$$

with inverse transform

$$f(x)=\int_{\mathbb{R}}\widehat{f}(\xi)e^{2\pi i x\xi}\,d\xi.$$

If $f$ and $\widehat f$ are integrable, the inverse integral represents $f$ almost everywhere; it agrees pointwise at every continuity point of $f$, with the usual choice of representative. However, the constant function $1$ is not integrable on $\mathbb{R}$, and

$$\int_{\mathbb{R}}e^{2\pi i x\xi}\,d\xi$$

does not converge as an ordinary improper integral. Thus the expression

$$\mathcal{F}^{-1}\{1\}(x)=\int_{\mathbb{R}}e^{2\pi i x\xi}\,d\xi$$

is only shorthand for a distributional identity.

Proof using test functions

Let $\varphi$ be a Schwartz function: a smooth function whose derivatives decrease faster than every power of $|x|$. Its Fourier transform is also a Schwartz function, so all the pairings below are well defined.

Define $T=\mathcal{F}^{-1}1$ as a tempered distribution. To identify $T$, test it against $\varphi$. With the distributional definition of the inverse transform,

$$\langle \mathcal{F}^{-1}1,\varphi\rangle = \langle 1,\mathcal{F}^{-1}\varphi\rangle.$$

The distribution $1$ acts by integration, giving

$$\langle \mathcal{F}^{-1}1,\varphi\rangle = \int_{\mathbb{R}}(\mathcal{F}^{-1}\varphi)(\xi)\,d\xi.$$

Now use the inverse-transform formula:

$$(\mathcal{F}^{-1}\varphi)(\xi) = \int_{\mathbb{R}}\varphi(x)e^{2\pi i x\xi}\,dx.$$

The integral of the inverse Fourier transform is the value of the original function at zero. Equivalently, this follows from Fourier inversion evaluated at $x=0$:

$$\int_{\mathbb{R}}(\mathcal{F}^{-1}\varphi)(\xi)\,d\xi =\varphi(0).$$

Therefore,

$$\langle \mathcal{F}^{-1}1,\varphi\rangle =\varphi(0) =\langle \delta_0,\varphi\rangle.$$

Since two distributions are equal when they have the same action on every test function,

$$\boxed{\mathcal{F}^{-1}\{1\}=\delta_0.}$$

The equivalent forward-transform identity is

$$\mathcal{F}\{\delta_0\}=1.$$

Indeed, formally—and rigorously in distribution theory—

$$\widehat{\delta_0}(\xi) =\langle \delta_0,e^{-2\pi i x\xi}\rangle =e^0=1.$$

A regularization that makes the delta appear

A useful way to see the identity is to replace the constant $1$ by a sequence of ordinary functions. Define

$$g_\varepsilon(\xi)=e^{-\pi\varepsilon^2\xi^2}, \qquad \varepsilon>0.$$

The Gaussian transform formula gives

$$\mathcal{F}^{-1}g_\varepsilon(x) =\frac{1}{\varepsilon}e^{-\pi x^2/\varepsilon^2}.$$

Each function on the right has total integral $1$:

$$\int_{\mathbb{R}}\frac{1}{\varepsilon}e^{-\pi x^2/\varepsilon^2}\,dx=1.$$

As $\varepsilon\to0$, it becomes increasingly concentrated near $x=0$. For every continuous function $\varphi$ of at most polynomial growth,

$$\lim_{\varepsilon\to0} \int_{\mathbb{R}}\frac{1}{\varepsilon}e^{-\pi x^2/\varepsilon^2}\varphi(x)\,dx =\varphi(0).$$

Thus

$$\frac{1}{\varepsilon}e^{-\pi x^2/\varepsilon^2} \longrightarrow \delta_0$$

in the distributional sense. Moreover, $g_\varepsilon\to1$ in the tempered-distribution sense, since testing against a Schwartz function allows dominated convergence. This gives the intuitive chain

$$\mathcal{F}^{-1}\{1\} =\lim_{\varepsilon\to0}\mathcal{F}^{-1}\{g_\varepsilon\} =\lim_{\varepsilon\to0}\frac{1}{\varepsilon}e^{-\pi x^2/\varepsilon^2} =\delta_0.$$

The convergence here is distributional, not pointwise: away from $0$, the Gaussian tends to $0$, while its integral remains $1$.

A second regularization: finite frequency bands

Another approximation uses a frequency cutoff. The inverse transform of the indicator of $[-R,R]$ is

$$\mathcal{F}^{-1}\{\mathbf{1}_{[-R,R]}\}(x) =\int_{-R}^{R}e^{2\pi i x\xi}\,d\xi =\frac{\sin(2\pi R x)}{\pi x},$$

where the value at $x=0$ is defined by continuity as $2R$. As $R\to\infty$, this function does not converge pointwise to an ordinary function. Nevertheless, when integrated against a Schwartz test function, it converges to $\varphi(0)$. Hence

$$\frac{\sin(2\pi R x)}{\pi x}\longrightarrow\delta_0$$

in the distributional sense. This is another precise meaning of the oscillatory integral representing the inverse transform of $1$.

Other Fourier conventions

If instead you define

$$\widehat f(\xi)=\int_{\mathbb{R}}f(x)e^{-i x\xi}\,dx,$$

then the inverse transform is

$$f(x)=\frac{1}{2\pi}\int_{\mathbb{R}}\widehat f(\xi)e^{i x\xi}\,d\xi.$$

The corresponding identity is still

$$\mathcal{F}^{-1}\{1\}=\delta_0,$$

because the factor $1/(2\pi)$ belongs to the inverse-transform definition. What changes are constants in related formulas, such as Gaussian-transform formulas; with the forward transform written here, $\mathcal{F}\{\delta_0\}=1$ remains unchanged.

Common mistakes

• Treating $\delta_0$ as an ordinary function. It is a distribution characterized by its integral against test functions.
• Claiming pointwise convergence. Delta approximations converge distributionally; they do not converge pointwise to a function with infinite value at zero and zero elsewhere.
• Ignoring normalization. The $2\pi$ factors depend on the chosen convention.
• Using the divergent integral as an ordinary integral. The expression $\int_{\mathbb R}e^{2\pi i x\xi}\,d\xi$ is meaningful only after distributional interpretation or regularization.