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Could someone please tell me why this is true ?

Let $g=g(x,z)$ $f(x)=\exp(ikx)\left(1+i{g \over k}-{g_z \over k^2}\right)\bigg|_{z=x}-{1\over k^2}\int_x^\infty g_{zz}\exp(ikz)\,\,dz$where $g,g_z\to 0 $ as $z\to\infty$.

Also, let $g_{xx}=g_{zz}+ug$ when $z>x$ and where $u=-2(g_x+g_z)\big|_{z=x}$.

Then $\left\{-\partial_x^2-2(g_x+g_z)\big|_{z=x}\right\}f=k^2f$

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    Thats more likely! Check [my answer](http://math.stackexchange.com/a/233842/19532).2012-11-09

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Given $ f(x) = e^{ikx} + \int_x^\infty g(x,z) e^{ikz} dz $ we have $ \partial_x f(x) = e^{ikx}\big(ik - g(x,x)\big) + \int_x^\infty g_x(x,z) e^{ikz}dz $ and $ \partial_{xx} f(x) = -e^{ikx}\big(k^2 + ik g(x,x) + g_z(x,x) + 2g_x(x,x)\big) + \int_0^\infty g_{xx}(x,z) e^{ikx}. $ Using $g_{xx} = g_{zz} + ug$, \begin{multline} \partial_{xx} f(x) = -e^{ikx}\big(k^2 + ik g(x,x) + g_z(x,x) + 2g_x(x,x)\big) \\ + \int_0^\infty g_{zz}(x,z) e^{ikx} + \int_0^\infty u g(x,z) e^{ikz} dz. \end{multline}

Substitute the integral involving $g_{zz}$ and use the fact that $u(x) = - 2 \big(g_z(x,x) + 2g_x(x,x)\big)$ and you'll be able to finish the proof.