# Inverse Fourier Transform of a Constant

The Fourier transform and its inverse can be defined as $$\mathcal{F}(f(x))=F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)e^{-ikx} \ dx \ \ \text{and} \ \ \mathcal{F}^{-1}(F(k))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F(k)e^{ikx} \ dk$$ respectively. Now, the Fourier transform of a constant, $$a$$, is $$\mathcal{F}(a)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} ae^{-ikx} \ dx=a\sqrt{2\pi}\delta(k),$$ where $$\delta$$ denotes the Dirac delta function. My question is, what is $$\mathcal{F}^{-1}(a)$$? Is it, by symmetry of the Fourier transform and its inverse, $$\mathcal{F}^{-1}(a)=a\sqrt{2\pi}\delta(-x)?$$

Simple proof:

Let $$x=k_1$$, then $$\mathcal{F}(f(k_1))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(k_1)e^{-ikk_1} \ dk_1.$$Now let $$k=-x_1$$, then $$\mathcal{F}(f(k_1))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(k_1)e^{ix_1k_1} \ dk_1.$$ Does this provide proof of my argument?

• The Fourier transform of $1$ is $\sqrt{2\pi} \delta$ because for any $\varphi$ (nice enough : smooth and $L^1$..) the Fourier transform of $1 . \varphi$ is $\mathcal{F}(\varphi) \ast \delta$. In other words for a distribution its Fourier transform is the unique distribution that fits well in the convolution theorem(s). The Fourier inversion theorem stays true for distributions, that is from $\mathcal{F}(\delta) = \frac{1}{\sqrt{2\pi}}$ you know $\mathcal{F}^{-1}( \frac{1}{\sqrt{2\pi}}) = \delta$ May 11, 2019 at 15:18
• @reuns but what about the argument of the delta function (this is what I am concerned about)? I want to know if you replace $k$ with $-x$ when taking the inverse Fourier transform of a function to the frequency space (hence $\delta(-x)$). May 12, 2019 at 6:15
• The Dirac delta "function" is not a function, it is a measure. When you speak of "inverse," you need to be specific about the domain and range of the function [Fourier Transform] you are speaking of. May 13, 2019 at 12:19
• When we take the fourier transform of a test function, standard Gaussian for example, $$\int_{-\infty}^{\infty}f(x)e^{itx}\mbox{d}x=e^{i\mu t}e^{-\frac{1}{2}(\sigma t)^2}$$ and when $\sigma\rightarrow\infty$ we have $$|e^{i\mu t}e^{-\frac{1}{2}(\sigma t)^2}|\rightarrow 1.$$ May 13, 2019 at 12:36

The defining formula $$\mathcal{F}f(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) \, e^{-ikx} \, dx$$ only works when $$f \in L^1(\mathbb{R}).$$ The Fourier transform can however be extended to several other cases. For example, for a tempered distribution $$u$$ it is defined to be the distribution $$\mathcal{F}u$$ satisfying $$\int_{-\infty}^{\infty} \mathcal{F}u(x) \, \varphi(x) \, dx = \int_{-\infty}^{\infty} u(x) \, \mathcal{F}\varphi(x) \, dx,$$ for all test functions $$\varphi$$ in Schwartz space $$\mathcal{S}(\mathbb{R}) \subset L^1(\mathbb{R}).$$
One example of a tempered distribution is $$\delta.$$ Its Fourier transform is thus given by $$\int_{-\infty}^{\infty} \mathcal{F}\delta(x) \, \varphi(x) \, dx = \int_{-\infty}^{\infty} \delta(x) \, \mathcal{F}\varphi(x) \, dx = \mathcal{F}(0) = \frac{1}{\sqrt{2\pi}} \left. \int_{-\infty}^{\infty} \varphi(x) \, e^{-ikx} \, dx \right|_{k=0} = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} \varphi(x) \, dx ,$$ i.e. $$\mathcal{F}\delta = \frac{1}{\sqrt{2\pi}}$$ (constant function).
For $$\varphi \in \mathcal{S}(\mathbb{R})$$ we have $$\mathcal{F}^2\varphi(x) = \varphi(-x),$$ and it's easy to show from this that also for tempered distributions we have $$\mathcal{F}^2 u(x) = u(-x).$$ Therefore, for the constant function $$1$$ we have $$\mathcal{F}1(x) = \mathcal{F}\{\sqrt{2\pi}\,\mathcal{F}\delta\}(x) = \sqrt{2\pi}\,\mathcal{F}^2\delta(x) = \sqrt{2\pi}\,\delta(-x) = \sqrt{2\pi}\,\delta(x).$$
• So for the inverse Fourier transform of a function, say $e^{-a|k|}$, is $$\mathcal{F}^{-1}(e^{-a|k|})=\sqrt{\frac{2}{\pi}}\frac{a}{a^2+x^2}?$$ May 13, 2019 at 22:17
• That is correct. We have $$\mathcal{F}(e^{-a|x|}) = \sqrt{\frac{2}{\pi}}\frac{a}{a^2+k^2}$$ and $$\mathcal{F}^{-1}(e^{-a|k|}) = \sqrt{\frac{2}{\pi}}\frac{a}{a^2+x^2}.$$ May 14, 2019 at 15:15