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The task is to evalute $$ \int\limits_{\mathbb{R}^2} \frac{e^{i \langle \xi, x \rangle} d\xi}{ \langle\xi,\theta\rangle}, \;\;\; \theta \in \mathbb{C}^2 \setminus ( \mathbb{S}^1 \cup \left\{ 0 \right\} ), \;\;\; \theta_1^2 + \theta_2^2 = 1,\;\;\; x\in \mathbb{R}^2 $$ Obviously it is the Fourrier transform of $f(\xi) = \langle \xi,\theta \rangle ^ {-1}$. I tried to make the change $y_{1} = \langle\xi,\Re \theta\rangle, \;\; y_2 = \langle \xi, \Im \theta \rangle$, but i faced with difficult calculations.

Update: let $$ f(x) = \frac{1}{2\pi i}\int\limits_{\mathbb{R}^2} \frac{e^{i \langle \xi, x \rangle} d\xi}{ \langle\xi,\theta\rangle} $$ Then, using inverse Fourrier transform we get $$ i\langle\theta,\xi\rangle \hat{f}(\xi) = \frac{1}{\sqrt{2\pi}} $$ Taking Fourier transform now we recieve $$ \langle \theta, \nabla f(x) \rangle = \delta(x) $$ In other words, f(x) is a Green function for operator $L = \langle \theta, \nabla \rangle$.

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    Is the integral convergent? If $\theta=x=(1,0)$, the denominator is $\xi_1$, the numerator is $e^{i\xi_1x_1}$ and $d\xi=d\xi_1d\xi_2$, hence the integral is not defined.2011-09-25
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    I'm sorry, I forgot a condition: $\theta \notin \mathbb{S}^1$.2011-09-25
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    But S^1 is a subset of C, not of C^2, and you say that theta is in C^2... Is theta really in R^2? Please revise your post once and for all.2011-09-25
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    $\mathbb{S}^1 = \left\{ x \in \mathbb{R}^2 \subset \mathbb{C}^2 : |x| = 1 \right\}$. How can be $\mathbb{S}^1$ a subset of $\mathbb{C}$? Which definition do you mean?2011-09-25
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    Well, usually R^2 is identified to C. Just to be sure: your theta is in C^2 hence theta=(a+ib,c+id) where a, b, c and d are some real numbers, and you impose that (b is not zero) OR (d is not zero) OR (b and d are zero but a^2+c^2 is not 1)... Sure about this? By the way, my initial example works all the same with theta=x=(2,0).2011-09-25
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    Thank you again) That proves that the additional condition is essential: $\theta_1^2 + \theta_2^2 = 1$.2011-09-25
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    You already modified twice the conditions on theta to avoid some fallacies I mentioned. Since we cannot continue like that ad vitam aeternam, I suggest YOU check once and for all that the question is not absurd (for instance the integral MUST be well defined) and THEN you come back to ask it.2011-09-26

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