= Solution
Extend $q$ by zero to $x<0$. Its ordinary <Fourier transform> is precisely the half-line transform $\widehat q$, 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>,
$$
v=\frac{\alpha+\sqrt{\alpha^2+4u^2}}2\geq\alpha,
$$
with $u$ increasing from $-\infty$ to $\infty$. In the region above it, $\nu(k)=i\alpha-k$ has nonpositive imaginary part. Therefore $\widehat q(\nu(k))$ 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 $\widehat q(\nu)=O(|\nu|^{-1})$ away from bounded spectral sets.
For an interior spatial point $x>0$, the factor $e^{ikx}$ decays exponentially as $\operatorname{Im}k\to+\infty$. 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> is
$$
\boxed{\int_L e^{ikx}\widehat q(i\alpha-k)\,dk=0,\qquad x>0.}
$$
Adding 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.
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