= 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 $q(x)=(2\pi)^{-1}\int_{\mathbb R}e^{ikx}\widehat q(k)\,dk$, with $\widehat q(k)=\int_0^\infty e^{-ikx}q(x)\,dx$. If $\operatorname{Im}k\leq0$, 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 $e^{ikx}\widehat q(-k)$ 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> give
$$
\boxed{\int_L e^{ikx}\widehat q(-k)\,dk=0.}
$$
For example, if $q,q'$ are integrable, integration by parts gives $\widehat q(-k)=O(1/k)$ 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 proves
$$
\boxed{q(x)=\frac1{2\pi}\int_{\mathbb R}e^{ikx}\widehat q(k)\,dk+\frac c{2\pi}\int_L e^{ikx}\widehat q(-k)\,dk,\quad c\in\mathbb C.}
$$
The extra integral is a reflected, zero-supported contribution, not an extra datum.
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