 Solution

Extend $q$ by zero to $x<0$. Its ordinary Fourier transform is precisely the half-line transform $\hat{q}$, so Fourier inversion gives the real-axis term for every interior point $x>0$. At $x=0$ the zero extension can have a jump; the interior inversion is the assertion needed here.

Write $k=u+iv$. The contour is the upper branch of a hyperbola,with $u$ increasing from $-\infty$ to $\infty$. In the region above it, $\nu (k)=i\alpha -k$ has nonpositive imaginary part. Therefore $\hat{q}(\nu (k))$ is analytic there: its defining integral has factor $e^{iux-(v-\alpha )x}$ and converges for a sufficiently decaying $q$. Smoothness and integration by parts also give $\hat{q}(\nu )=O(|\nu {|}^{-1})$ away from bounded spectral sets.

For an interior spatial point $x>0$, the factor $e^{ikx}$ decays exponentially as $\mathrm{Im}k\to +\infty$. Truncate the hyperbolic contour and close it above, using the Cauchy integral theorem; the closing segments vanish by this decay and the transform estimate. No poles occur in the enclosed region. Consequently the half-line drift null contour identity isAdding any constant multiple of zero to Fourier inversion proves the proposed inversion, for every $c$. Oscillatory real-axis inversion may be understood by a Gaussian cutoff followed by its usual limit. The same contour argument allows a spectral multiplier analytic above $L$ with polynomial growth, because the exponential still controls the closing paths. We will need this slight extension for derivative boundary data.