Inverse Fourier Transform-1: Remaining Part of Inverse Fourier Transform are given as:
In the second integrals;
If K ≥ 1,
Since are both convergent integrals,
For the integrals over [0, k],
Since f is a piecewise smooth, f' (x ) exists for all x ∈ R and Therefore g is bounded on [0,k] and hence g ∈ L1 (R). By the Riemann - Lebessgue lemma, g^ exists, is continuous, g^(±a) → 0 as a → and therefore;
For K ≥ 1. A virtually identical arguments works for the first integrals.
Therefore;