= Solution
{c}
The <integral> construction requires differentiation under the <integral> and vanishing boundary terms. With those assumptions, <integration by parts> gives
$$
w''+zw=[g(\zeta)e^{z\zeta}]_{\partial\Gamma}
+\int_\Gamma(\zeta^2g-g')e^{z\zeta}\,d\zeta.
$$
The standard construction therefore sets $g'=\zeta^2g$, giving $g=C e^{\zeta^3/3}$. Convergent <contours> run between rays where $\Re\zeta^3<0$: convenient central rays have arguments $-\pi/3$, $\pi/3$ and $\pi$. Two choices are the oriented paths
$$
\Gamma_1:\infty e^{-i\pi/3}\longrightarrow0\longrightarrow\infty e^{i\pi/3},
\qquad
\Gamma_2:\infty e^{i\pi/3}\longrightarrow0\longrightarrow-\infty.
$$
The cubic exponential dominates the linear one at both ends. To verify independence directly, put $A=\int_0^\infty e^{-r^3/3}dr>0$ and $B=\int_0^\infty r e^{-r^3/3}dr>0$. At $z=0$ the value and <derivative> of the <integral> along rays $\theta_1,\theta_2$ are $A(e^{i\theta_2}-e^{i\theta_1})$ and $B(e^{2i\theta_2}-e^{2i\theta_1})$. For the two <contours> above their derivative-to-value ratios are respectively $B/A$ and $(B/A)(e^{i\pi/3}-1)$, which differ. Hence their <Wronskian> is nonzero.
For the second equation multiply by $z$ before integrating by parts:
$$
z(w''+w)+w'
=[(1+\zeta^2)g e^{z\zeta}]_{\partial\Gamma}
-\int_\Gamma\bigl((1+\zeta^2)g'+\zeta g\bigr)e^{z\zeta}d\zeta.
$$
Setting the residual weight to zero gives $g=C(1+\zeta^2)^{-1/2}$. One <contour> is the segment from $-i$ to $i$, with the <square root> positive at $\zeta=0$. Its <integral> is a nonzero multiple of $J_0(z)$: substituting $\zeta=it$ gives $i\int_{-1}^1e^{izt}(1-t^2)^{-1/2}dt$, an <entire function> nonzero at zero.
A second <contour> starts at $i$ and follows $\zeta=i-t$, $0\leq t<\infty$, choosing a continuous branch along the path. For $\Re z>0$ it converges at infinity; at the branch endpoint its integrable singularity is $t^{-1/2}$. In both cases the boundary term contains $\sqrt{1+\zeta^2}$ and vanishes at branch endpoints. The second <integral> has a nonzero <logarithmic divergence> as $z\to0$ through positive real values, because its large-$t$ integrand is a nonzero multiple of $e^{-zt}/t$. It is therefore independent of the entire first solution; <analytic continuation> supplies solutions elsewhere away from the equation's singular point.
Strictly, nonzero $w$ and an arbitrary <contour> alone do not prove the printed necessity for $g$. Necessity also needs <injectivity> of the chosen Laplace representation, while the vanishing-integrand argument above is the usual sufficient construction. For a concrete counterexample to unconditional necessity in the Bessel case, take a positively oriented circle $|\zeta|=R>1$, choose the branch $(1+\zeta^2)^{-1/2}\sim1/\zeta$ outside the cut $[-i,i]$, and add one to that weight. The added entire integrand integrates to zero. Expanding the original weight on this circle gives <integral> $2\pi i J_0(z)$, which is nonzero and solves the equation, although the modified weight is not a constant multiple of the printed one. The standard weights and <contours> above still give the requested independent solutions.
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