= Null weight in a contour Laplace representation
{title2=$\int_\Gamma h(\zeta)e^{z\zeta}\,d\zeta=0$}
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, $\operatorname{Re}(\zeta^2)\leq-r^2/2$; 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 $-\sqrt2\,4^{-3/4}\Gamma(1/4)\ne0$. Adding the null weight changes neither this integral nor its vanishing endpoint terms, but the sum does not satisfy $g'=\zeta^3g$. Thus the usual Laplace method selects a convenient weight by cancelling the interior integrand separately; a necessity claim needs additional injectivity assumptions.
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