 Solution

Extend $q$ by zero to $x<0$ and apply the assumed Fourier inversion theorem. For sufficiently regular, decaying data and $x>0$ this gives $q(x)=(2\pi {)}^{-1}{\int }_{\mathbb{R}}{e}^{ikx}\hat{q}(k)dk$, with $\hat{q}(k)={\int }_0^{\infty }{e}^{-ikx}q(x)dx$. If $\mathrm{Im}k\le 0$, the exponential in this last integral is bounded in magnitude by one; it defines a holomorphic transform in the lower half-plane, with its boundary values on the real axis.

The PDF orients $L$ from $i\infty$ down to zero, then from zero to $+\infty$. The function $e^{ikx}\hat{q}(-k)$ is holomorphic in the upper half-plane, and for $x>0$ its exponential decays there. Close the two rays through a large first-quadrant arc. Cauchy integral theorem and Jordan lemma giveFor example, if $q,q^{\prime }$ are integrable, integration by parts gives $\hat{q}(-k)=O(1/k)$ in that quadrant, which suffices for the arc estimate. Standard limiting arguments extend the identity to the admissible Fourier-inversion class. Adding any constant multiple of this zero integral to ordinary inversion provesThe extra integral is a reflected, zero-supported contribution, not an extra datum.