# Fourier Transform of Constant Function

One of the requirements for the existence of Fourier transform of $f(x)$ is that:

$\int_{-\infty}^{\infty} |f(x)| dx$ exists.

However, the table says that the Fourier transform of constant functions (\emph{i.e.}, $f(x)=1$) do exist and it is $\delta(k)$ although $\int_{-\infty}^{\infty} 1 dx = \infty$ .

Could anyone can help me to understand this? Thanks in advance.

• It depends upon what you mean by "existence". The Dirac delta $\delta(k)$ must be defined by convention, because of course its value at $k=0$ is not finite. Mar 19 '14 at 18:26
• Dirichlet conditions (including integrability) are sufficient, not necessary.
– user48547
Jan 2 '15 at 10:20

## 3 Answers

The Fourier transform defined by an ordinary (Riemann or) Lebesgue integral only exists when $f \in L^1$.

It is however possible to extend the definition to tempered distributions (for example, every locally integrable function that "doesn't grow too fast" can be identified with a tempered distribution). The Fourier transform of such a thing is not in general a function though, as witnessed by your example.

I think it might be good to think the dirac as the limit of a some know function. For example when we take a zero mean Gaussian density $f(x;0, \sigma)$ with variance $\sigma^2$, we know that when $\lim_{\sigma\rightarrow 0}f(x;0, \sigma)=\delta(x)$. This is one type of the definition of the dirac delta function.

When we take the fourier transform of this function $$\int_{-\infty}^{\infty}f(x)e^{itx}\mbox{d}x=e^{i\mu t}e^{-\frac{1}{2}(\sigma t)^2}$$ and when $\sigma\rightarrow\infty$ we have $$|e^{i\mu t}e^{-\frac{1}{2}(\sigma t)^2}|\rightarrow 1$$

Another way would be to take a rectangular function on $[-t,t]$ at the frequency domain $t$ and let $t\rightarrow\infty$. At the time domain $x$ one would get a sinc function with a single main lobe. The number of zero crossings will increase, the power of side lobes will decrease and eventually go to zero, what remains will be a dirac delta function.

Mathematically, if absolute summability is not satisfied we also get something which is strangely defined; something which has an infinite power at frequency $t=0$, from Perseval's energy preservation rule.

Approach $$\boldsymbol{1}$$: Approximating $$1$$ by $$e^{-\lambda x^2}$$

Note that as $$\lambda\to0^+$$, $$e^{-\lambda x^2}\to1$$. Furthermore, we can explicitly compute the Fourier transform: \begin{align} \int_{\mathbb{R}}e^{-\lambda x^2}e^{-2\pi ix\xi}\,\mathrm{d}x &=e^{-\pi^2\xi^2/\lambda}\int_{\mathbb{R}}e^{-\lambda(x+\pi i\xi/\lambda)^2}\,\mathrm{d}x\\ &=e^{-\pi^2\xi^2/\lambda}\int_{\mathbb{R}}e^{-\lambda x^2}\,\mathrm{d}x\\ &=\sqrt{\tfrac\pi\lambda}e^{-\pi^2\xi^2/\lambda}\tag1 \end{align} For all $$\lambda\gt0$$, $$\int_{\mathbb{R}}\sqrt{\tfrac\pi\lambda}e^{-\pi^2\xi^2/\lambda}\,\mathrm{d}\xi=1\tag2$$ and as $$\lambda\to0^+$$, the graph of $$\sqrt{\tfrac\pi\lambda}e^{-\pi^2\xi^2/\lambda}$$ contracts horizontally by a factor of $$\sqrt\lambda$$ and expands vertically by a factor of $$\frac1{\sqrt\lambda}$$. That is, as $$\lambda\to0^+$$, $$\sqrt{\tfrac\pi\lambda}e^{-\pi^2\xi^2/\lambda}$$ approximates the dirac delta.

Thus, as $$\lambda\to0^+$$, $$e^{-\lambda x^2}\to1$$ and its Fourier Transform is $$\sqrt{\tfrac\pi\lambda}e^{-\pi^2\xi^2/\lambda}\to\delta(\xi)$$.

Approach $$\boldsymbol{2}$$: Compare action via the Convolution Theorem

By the Convolution Theorem $$\widehat{f\cdot1}(\xi)=\widehat{f}\ast\widehat{1}(\xi)$$ Since multiplication by $$1$$ leaves $$f$$ alone, convolution by $$\widehat{1}$$ must leave $$\widehat{f}$$ alone. That is, we must have $$\widehat{1}=\delta$$.