We know for a function $f(t)$, Fourier transform is defined as:
\begin{equation} F(\mu) = \int_{-\infty}^{\infty} f(t) e^{-i2\pi \mu t} dt \end{equation}
I saw the Fourier transformation for $f(t) = 1$ is $\delta(\mu)$. It is usually achieved by duality property of Fourier transform. But I want to get it directly. So I attempted to
$$F(\mu) = \int_{-\infty}^{\infty} e^{-i2\pi \mu t} dt \tag{1}$$
With attention to this relation:
$$\int e^{-i2\pi \mu t} dt = \frac{i e^{-i2\pi \mu t}}{2\pi\mu} \tag{2}$$
It is obvious that $F(\mu)$ in Equation$(1)$ for $\mu = 0$ is $\infty$, which matches the $\delta(0)$. But Equation$(2)$ shows the result of Equation$(1)$ for $\mu \neq 0$ is imaginary, not converge and also $|F(\mu)| = \infty$.
Now my question is: When result of Equation$(1)$, which is definition of Fourier transform of $f(t) = 1$, is so irrelevant to $\delta(\mu)$, is it rational to consider $\delta(\mu)$ as Fourier transform for $f(t) = 1$?
“It is usually achieved by duality”, well that’s how one precisely defines and obtain this relationship. The integral definition makes no sense whatsoever as a Lebesgue integral sincce $t\mapsto e^{-2\pi i \mu t}$ is not in $L^1(\Bbb{R})$. Also, $\delta$ is not defined as “$\delta(\mu)=0$ for $\mu\neq 0$ and $\delta(0)=\infty$”, no matter how much one may hear/wish to believe otherwise. If it were defined pointwise like this, it would never be able to satisfy its defining property $\int_{\Bbb{R}}\delta(x)\phi(x)\,dx=\phi(0)$ for all test functions $\phi$.
Also, $|F(\mu)|=\infty$ for $\mu\neq 0$ makes no sense. Literally. The very symbol $F(\mu)$ makes no sense because you’re integrating a non $L^1$ function, so as far as any reason is concerned, this is just a string of meaningless symbols.
Now, precisely because the constant function $f=1$ does not belong to $L^1$ or to $L^2$, we cannot use the integral definitions. We have to resort to duality of the Schwartz functions and tempered distributions.