Let \(f\in L^1(\mathbb{R}^n)\). The inverse Fourier transform of \(f\) is defined by
\[ (\mathcal{F}^{-1} f)(x)=\frac{1}{(2\pi )^{\frac{n}{2}}} \int_{\mathbb{R}^n} f(\xi ) \e^{ix\cdot \xi}\; d\xi . \]
Remarks
- Unlike the name suggest, it is not true in general that for \(f\in L^1(\mathbb{R}^n)\) we have \(\mathcal{F}^{-1}\circ \mathcal{F}(f)=f\), since \(\mathcal{F}f\in C_0(\mathbb{R}^n)\nsubseteq L^1(\mathbb{R}^n)\) (see (0x67bd84c5) and (0x67e3cec6) ).
- The operator \(\mathcal{F}^{-1}\) is the inverse of \(\mathcal{F}\) on: