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.
 A: 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.
A: 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.
A: 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 $\widehat{1}=\delta$.
