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A null weight contributes zero to a Laplace integral on its domain of convergence. It can be added to a weight without changing the represented function, so an integration-by-parts differential equation for the weight is not an unconditional uniqueness theorem.

For example, let $\Gamma$ enter along angle $\pi /3$ from infinity to zero and leave along angle $2\pi /3$ to infinity. The entire weight $h(\zeta )=e^{{\zeta }^2}$ is null for every complex $z$. Throughout the wedge between those rays, $\mathrm{Re}({\zeta }^2)\le -r^2/2$; the factor $e^{z\zeta }$ is dominated by this quadratic decay. Closing the contour by a large circular arc and applying the Cauchy integral theorem therefore proves the zero integral. All derivatives with respect to $z$ converge locally uniformly as well.

On the same contour, the quartic weight $e^{{\zeta }^4/4}$ gives a nontrivial entire solution. Its rays can be deformed within the quartic decay sectors for a contour integral to angles $\pi /4$ and $3\pi /4$, where the value at $z=0$ is $-\sqrt{2}{4}^{-3/4}\Gamma (1/4)\ne 0$. Adding the null weight changes neither this integral nor its vanishing endpoint terms, but the sum does not satisfy $g^{\prime }={\zeta }^{3}g$. Thus the usual Laplace method selects a convenient weight by cancelling the interior integrand separately; a necessity claim needs additional injectivity assumptions.

In the Laplace integral method for a differential equation, differentiate the integral three times with respect to $z$, and replace $z{e}^{z\zeta }$ by ${\partial }_{\zeta }{e}^{z\zeta }$. Integration by parts givesThus the weight equation for this construction and its boundary condition arewhich giveThese are the conditions obtained by eliminating the transformed interior term and the endpoint term separately. As usual for this method, differentiation and integration by parts need convergence assumptions; the contours constructed next make every such step valid locally uniformly for all complex $z$. A literal necessity claim for arbitrary weights and contours would additionally require uniqueness of the chosen representation by a Laplace integral; a null weight in a contour Laplace representation can otherwise be added without changing the solution.

LetThe quartic decay sectors for a contour integral contain the four rays $\zeta =r{u}_j$. Along each ray, the quartic term is ${\zeta }^4/4=-r^4/4$. For $j=1,2,3$, choose ${\Gamma }_j$ to travel from infinity on ray $u_0$ to zero, and then from zero to infinity on ray $u_j$. Set $A=1$ and defineOn any compact set of $z$ values, the integrand and each of its $z$ derivatives are bounded by a constant times $r^m{e}^{-r^4/4+Mr}$ for some $M$. These bounds are integrable at infinity, so differentiation under the integral sign is justified and the functions are entire. The endpoint term vanishes at both infinite ends. There is no extra term at zero because the two pieces join with the same integrand value and opposite endpoint orientations. Therefore all three functions solve the differential equation.

To prove independence, evaluate their first three initial data. For $m=0,1,2$,If ${\sum }_{j=1}^3{c}_j{w}_j=0$, these initial data give ${\sum }_{j=1}^3{c}_j(u_j^{\ell }-u_0^{\ell })=0$ for $\ell =1,2,3$. Define $c_0=-{\sum }_{j=1}^3{c}_j$. ThenThe coefficient matrix is a Vandermonde matrix at four distinct fourth roots of $-1$, and hence is invertible. All $c_j$ vanish. There is also a concrete reason for the earlier qualification about necessity. Deform the incoming ray of ${\Gamma }_1$ to angle $\pi /3$ and its outgoing ray to angle $2\pi /3$, within their respective quartic decay sectors. The quartic integral is unchanged and remains nonzero. On this deformed contour, the extra entire weight $e^{{\zeta }^2}$ contributes zero for every $z$: throughout the closing wedge, $\mathrm{Re}({\zeta }^2)\le -r^2/2$, so the large closing arc tends to zero and the Cauchy integral theorem gives a zero integral. Thus $g=e^{{\zeta }^4/4}+e^{{\zeta }^2}$ represents the same entire solution and has vanishing endpoint terms, although it does not satisfy $g^{\prime }={\zeta }^{3}g$. The canonical weight equation is the standard sufficient construction, rather than an unconditional uniqueness assertion about contour representations.

Thus the three contours give linearly independent nonzero solutions, and they form a fundamental system of solutions for this third-order linear equation. This proves independence directly instead of inferring it merely from the presence of three different contours.