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Consider an integral $$ I(x) = \int\limits_{\mathbb{R}^n} e^{i\xi x } \delta(\xi^2-k^2)\chi( (\xi-k,\gamma)) \, d\xi $$ where $x, k, \gamma \in \mathbb{R}^n$, $|\gamma| = 1$ and $(x,y)\equiv xy = x_1 y_1 + \ldots + x_n y_n$. Here $\delta$ is the Dirac delta, $\chi$ is the Heaviside step function: $$ \chi(t) = 1_{\left\{ t \geqslant 0 \right\}}. $$ How can I obtain asymptotics of such integral $I(x)$ as $|x| \to \infty$?

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You might wish to recall how you define $\int\limits_{\mathbb R^n}u(\xi)\delta(\xi^2-k^2)\mathrm d\xi$, for every suitable function $u$ and every $k$ in $\mathbb R^n$. –  Did May 15 '12 at 11:30
@Didier It is an integral over sphere $\xi^2 = k^2$ of $u(\xi)$ (I think there are no additional multipliers, are they?). –  Nimza May 15 '12 at 16:47
You are looking for the characteristic function of the top half of the sphere ? I think you can get it (semi) explicitly, with the technique that works for the whole sphere, as described concisely here sciencedirect.com/science/article/pii/S0723086905000034 full disclosure: haven't tried the calculation myself –  mike May 16 '12 at 13:25
@mike And if for an integral $\Omega(r)=\int\limits_{-1}^{1} e^{itr}(1-t^2)^{\frac{m-3}{2}} dt$ it is easy to compute asymptotics? –  Nimza May 16 '12 at 15:02
I think so, but I'm giving you another ignorant opinion. If m is odd you have an integral you can do explicitly. Someone else will know better that I do how to do the asymptotics. I think this is an interesting question so I'm going to see if I can work it out, but I don't know the answer. –  mike May 16 '12 at 15:38

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