The inverse Fourier transform of a constant is defined as $\delta(t)$. However, if I use the definition and consider this as a limit \begin{equation} r(t)= \int_{-B}^{B} e^{j\omega t} d\omega = \frac{2}{t} \sin(Bt) \end{equation} If $B\to \infty$, the limit does not exist.
What is missing in my thinking process? What is the gap between the two approaches?
Fourier transform and its inverse is defined for so-called $L^2$ functions $f$ which means their second moment is finite, i.e., $$ \int_{-\infty}^\infty \mid f(w)\mid^2 \,dw <\infty. $$ However, $\delta(t)$ itself is a generalized function and on both sides of this inequality there is handwaving going on, unless a more advanced approach is taken. See
Computing Fourier transform for $L^2$ function
for an illustration of the complexities.
The Schwarz space is the space of functions that decrease at infinity more rapidly of any polynomial.
Any function on the Schwarz space has a Fourier transform defined as usual as $$ \mathcal{F} f(k):= K\int _{-\infty}^\infty e^{-ikx} f(x) dx $$ where $K$ is your favorite renormalization constant.
You can prove that the dual space of the Schwarz space is the space of tempered distributions (the space of distribution that grows at most as a polynomial). Notice that (as said at the start of the article on the dual space) It is not a space of functions
The Fourier transform is defined on this space via duality i.e., if $D$ is a tempered distribution the Fourier transform of $D$ is the distribution such that, for any $f$ in the Schwarz space $$ (\mathcal{F} D, f):= (D, \mathcal{F} f) $$ In particular, this definition of the Fourier transform does not need to coincide with the usual one (it does for all the $L^1$ functions, but it does not when the usual Fourier transform it is not defined, as in the case the distribution is actually the linear functional associated to an $L^1_{\text{loc}}$ function ).
In the example in which $D$ is the integration for a constant function i.e $$ (D,f):= \int_{-\infty}^\infty C f(x) dx $$ for some constant $C$ we have that the Fourier transform of $D$ is the delta distribution (see this question)
The Dirac Delta is not a function and the object written $\displaystyle \int_{-\infty}^\infty e^{j\omega t}\,d\omega$, is a linear functional, not an integral. However, we can make sense of things as follows.
Let $\phi(t)$ be a function that $(i)$ is infinitely differentiable and $(i)$ vanishes outside a closed interval. We say that $\phi\in C_C^\infty$, which means that $\phi$ is infinitely differentiable and has compact support.
Then, we interpet the limit $\lim_{B\to\infty} \frac{2\sin(Bt)}{t}$ in terms of how it acts on $\phi(t)$ in a very special way. We interpret the limit in the sense of "distributions," or loosely put, how it acts on an integration of $\phi$. In particular, we have by integration by parts
$$\begin{align} \lim_{B\to \infty}\int_{-\infty}^\infty \phi(t)\frac{\sin(Bt)}{t}\,dt&=-\lim_{B\to \infty}\int_{-\infty}^\infty \phi'(t)\int_0^t \frac{\sin(Bt')}{t'}\,dt'\,dt\\\\ &=-\lim_{B\to \infty}\int_{-\infty}^\infty \phi'(t)\int_0^{Bt} \frac{\sin(t')}{t'}\,dt'\,dt\\\\ \end{align}$$
Now, there is a very powerful theorem called the Dominated Convergence Theorem that provides sufficient conditions under which we can interchange the limit with the integral. Those conditions apply here since the integrand is dominated by a constant times a positive function that is independent of $B$ and is integrable over the real numbers. Proceeding to interchange the order of integration and limit, we have
$$\begin{align} \lim_{B\to \infty}\int_{-\infty}^\infty \phi(t)\frac{\sin(Bt)}{t}\,dt&=-\int_{-\infty}^\infty \phi'(t)\lim_{B\to \infty}\int_0^{Bt} \frac{\sin(t')}{t'}\,dt'\,dt\\\\ &=-\int_{-\infty}^\infty \phi'(t) \frac{\pi}{2}\text{sgn}(t)\,dt\\\\ &=-\frac\pi2 \int_{0}^\infty \phi'(t)\,dt+\frac\pi2 \int_{-\infty}^0 \phi'(t)\,dt\\\\ &=\pi \phi(0) \end{align}$$
Therefore, the limit as interpreted this way maps any function $\phi\in C_C^\infty$ into $\pi \phi(0)$. Since this is exactly the same result obtained from $\pi \delta(t)$, we say that in the sense of distributions
$$\lim_{B\to\infty}\int_{-B}^B e^{j\omega t}\,d\omega=\pi \delta (t)$$
