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Let $O$ be a non-empty subset of $\mathbb{C}$ (the complex plane), and let $g : O \to \mathbb{C}$ be a holomorphic function that satisfies the differential equation $g^{\prime\prime}(z) - zg(z) = 0$ for every $z \in O$. Prove that for all differentiable curves $U$ in $O$ with the same endpoints, the value $$\int_U g(x)^2 dx$$ is the same.

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Where did you come across this? It seems like it might be homework... So, where are you stuck? –  Aryabhata Jan 24 '11 at 0:16
    
not even sure where to begin :-S –  user6161 Jan 24 '11 at 5:08
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The differential equation in question is the Airy equation. Its two linear independent solutions (functions $\mathrm{Ai}(z)$ and $\mathrm{Bi}(z)$) are holomorphic on the whole complex plane. Hence $g(z)^2$ is also a holomorphic function and the integral has the same value by the residue theorem. –  O.L. Jun 23 '13 at 13:19

1 Answer 1

It suffices to show the integral is zero when the contour is a closed curve. Let $\gamma:[a, b] \to O$ be a closed curve. Integration by parts (with $u = g(z)^2$ and $dv = dz$), then using the differential equation gives \begin{eqnarray} \int \limits _{\gamma} g(z)^2 \text{ }dz &=& (g(z))^2 z\Bigg|^{\gamma(b)}_{\gamma(a)} - \int \limits _{\gamma} 2g(z) g'(z) z \text{ }dz \\ &=& (g(z))^2 z\Bigg|^{\gamma(b)}_{\gamma(a)} - \int \limits _{\gamma} 2 g'(z) g''(z) \text{ }dz \\ &=& (g(z))^2 z\Bigg|^{\gamma(b)}_{\gamma(a)} - (g'(z))^2 \Bigg|^{\gamma(b)}_{\gamma(a)} \\ &=& 0 - 0 = 0 \end{eqnarray} since $\gamma(b) = \gamma(a)$.

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