In many textbooks, the Fourier transform of a $L^{1}(\mathbb{R}^{n})$ function $f$ is defined by $$\mathcal{F}(f)(w)=\int_{\mathbb{R}^{n}}f(x)e^{-2i\pi xw}dx.$$ Also, if $\mathcal{F}(f)$ belongs to $L^{1}(\mathbb{R}^{n})$ then one gets $$f(x)=\int_{\mathbb{R}^{n}}\mathcal{F}(f)(w)e^{2i\pi wx}dw,\quad (\star)$$for almost $x\in\mathbb{R}^{n}$. All integrals here are understood as Lebesgue integrals.
The problem now is that in some textbooks, and the course that I am taking, integrals are considered as P.V. integrals, for example $\int_{-1}^{1}\frac{1}{x}dx=0$. Since I am majoring in engineering, and it seems to me that in practical applications, the Fourier transform is all understood in this sense.
For example, let's consider the pulse time waveform $f$ in $\mathbb{R}$ defined by: $f=1$ on $(-1,1)$, $f=0$ on $|x|>1$, and $f=1/2$ at $x=\pm 1$. Its Fourier transform is $$\mathcal{F}(f)(w)=2\frac{\sin(2\pi w)}{2\pi w}.$$
It is clear that $f\in L^{1}(\mathbb{R})$, but $\mathcal{F}(f)\notin L^{1}(\mathbb{R})$ in the Lebesgue sense. Therefore the formula $(\star)$ does not make sense in this case. However, in my course, since we considered P.V. integrals, the formula $(\star)$ still holds. One can check by using the formula $$\int_{-\infty}^{\infty}\frac{\sin(2\pi x)}{2\pi x}=\frac{1}{2},$$of course in the P.V. sense.
My question: Is my observation correct? that is, in practical applications, people always think of P.V. integrals when working with the Fourier transform.
Could you recommend textbooks that address this issue?
Thanks so much for your recommendation.