I got some questions concerning the inverse Fouriertransform of $\hat{f}=\frac{1}{\sqrt{2\pi}}\mathbf{1}_{[-\pi,\pi]}$.
One knows that $\hat{f}\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$. But as a result $$f:\mathbb{R} \to \mathbb{R},\quad f(x)=\begin{cases} \dfrac{\sin(\pi x)}{\pi x},& x\neq 0,\\ 1,& x=0. \end{cases}$$ is not an element of $L^1(ℝ)$. So I have to take the inverse Fourier transform defined on $L^2(ℝ)$, which is given by $$f(x)=\lim\limits_{\lambda \to \infty}\frac{1}{\sqrt{2\pi}}\int_{-\lambda}^{\lambda} \hat{f}(\xi)e^{i\xi x}\,d\xi.\quad (\star)$$ Am I right? Does $(\star)$ hold for almost all $x\in ℝ$ or for all $x\in \mathbb{R}?$
Is then the following true? For all $\lambda > \pi$ we have $$\frac{1}{\sqrt{2\pi}}\int_{-\lambda}^{\lambda} \hat{f}(\xi)e^{ix\xi}\,d\xi=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{ix\xi}\,d\xi=\frac{\sin(\pi x)}{\pi x}.$$ Then for almost all $x \in \mathbb{R} \backslash\{0\}$ we get $$f(x)=\lim\limits_{\lambda \to \infty}\frac{1}{\sqrt{2\pi}}\int_{-\lambda}^{\lambda} \hat{f}(\xi)e^{ix\xi}\,d\xi=\frac{\sin(\pi x)}{\pi x}.$$ With $f(0)=1$ we get for almost all $x\in \mathbb{R}$ $$f(x)=\operatorname{sinc}(x):=\begin{cases} \dfrac{\sin(\pi x)}{\pi x},& x\neq 0,\\ 1,& x=0. \end{cases} $$