= Solution
{c}
For an admissible <Laplace integral solution>, differentiation under the integral sign and <integration by parts> give
$$
\begin{aligned}
2w''+2zw'+w
&=\int_\Gamma(2\zeta^2+1)g\,e^{z\zeta}\,d\zeta
+2\int_\Gamma\zeta g\,\frac{d}{d\zeta}e^{z\zeta}\,d\zeta\\
&=[2\zeta g e^{z\zeta}]_{\partial\Gamma}
+\int_\Gamma\{(2\zeta^2-1)g-2\zeta g'\}e^{z\zeta}\,d\zeta.
\end{aligned}
$$
Choose endpoints for which the boundary term vanishes. Setting the density in braces to zero gives
$$
\frac{g'}g=\zeta-\frac1{2\zeta},\qquad
\boxed{g=A\zeta^{-1/2}e^{\zeta^2/2}.}
$$
Two suitable contours are the rays $\Gamma_+:0\to+i\infty$ and $\Gamma_-:0\to-i\infty$, with a continuous square-root branch on each ray. The Gaussian decays at infinity, and $\zeta g=O(\zeta^{1/2})$ tends to zero at the finite endpoint. The integrals converge for every finite $z$.
To check independence rather than merely exhibit different paths, define
$$
J_0=\int_0^\infty s^{-1/2}e^{-s^2/2}\,ds>0,\qquad
J_1=\int_0^\infty s^{1/2}e^{-s^2/2}\,ds>0.
$$
With unit normalization, the two solutions satisfy $w_\pm(0)=e^{\pm i\pi/4}J_0$ and $w_\pm'(0)=\pm i e^{\pm i\pi/4}J_1$. Their <Wronskian> at zero is $-2iJ_0J_1\ne0$, so they form an independent pair.
Strictly, the printed necessity claim requires an injective contour-transform class as well as vanishing endpoint terms. An arbitrary contour representation may contain a <null weight in a contour Laplace representation>, so a nonzero solution alone does not force every representing $g$ to have the displayed form. The calculation constructs the canonical density and two valid independent solutions; it does not assume uniqueness for unrestricted contour densities.
Back to article page
