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For the backward-sign Airy half-line global relation, suppose exactly one trace of derivative order $j\in \{0,1,2\}$ is unknown. Its spectral multiplier is a constant times $k^r$, $r=2-j$. Use the rotated null contours for half-range Fourier inversion with the displayed coefficients. The unknown contribution becomes an integral over the boundary of the middle upper sector $\pi /3<\arg k<2\pi /3$. There $\mathrm{Im}{k}^3<0$, sois bounded, while $e^{ikx}$ decays exponentially. The Cauchy integral theorem therefore removes that entire unknown contribution. The resulting solution depends on the initial transform and the two prescribed boundary transforms only.

// Target: analysis.bigb

Extend $q$ by zero to the negative half-line. Its ordinary Fourier transform is then exactly the Half-range Fourier transform $\hat{q}(k)$. The Fourier inversion theorem gives the real-line term for $x>0$; no endpoint convention at $x=0$ is needed.

It remains to show that each extra contour contributes zero, independently of its coefficient. For $\mathrm{Im}k<0$, the integral defining $\hat{q}$ is holomorphic. With $\alpha =e^{2\pi i/3}$, multiplication by ${\alpha }^2$ rotates the sector $E$ into $-2\pi /3\le \arg ({\alpha }^{2}k)\le -\pi /3$, and multiplication by $\alpha$ rotates $D$ into $-2\pi /3\le \arg (\alpha k)\le -\pi /3$. Thus the two rotated transforms are holomorphic on their respective upper sectors.

Orient the contours as in the PDF: $\partial E$ runs from infinity on the $\pi /3$ ray into zero and then out along the positive real axis; $\partial D$ runs from negative real infinity into zero and then out along the $2\pi /3$ ray. Both have their sector on their left. Smooth decay and integration by parts give $\hat{q}(k)=O(1/k)$ in closed lower-half-plane sectors. The factor $e^{ikx}$ decays in the upper half-plane for $x>0$. Close each sector by a large arc; the arc integral vanishes by Jordan lemma, including its short portion near the real axis. The Cauchy integral theorem therefore gives the rotated null contours for half-range Fourier inversion:Adding arbitrary constant multiples of these two zero integrals to ordinary Fourier inversion proves the asserted inversion formula. The constants $c_1,c_2$ are arbitrary in this part; selecting them later is what removes an unknown boundary trace.