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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. –  Did Sep 25 '11 at 17:14
I'm sorry, I forgot a condition: $\theta \notin \mathbb{S}^1$. –  Nimza Sep 25 '11 at 17:24
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. –  Did Sep 25 '11 at 17:48
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). –  Did Sep 25 '11 at 18:07
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. –  Did Sep 26 '11 at 1:41
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1 Answer 1

The following calculation uses the standard assumptions of the extension of the Fourier transform to the set of tempered distributions and the use of the Cauchy principal value in the definition of $\frac{1}{\xi}$. Since the Lebesgue measure of $\mathbb{R}^2$ is invariant under plane rotations, the integral is invariant under a mutual rotation of the vectors $x$ and$\theta$. Applying the rotation matrix:

$ \left (\begin{array}{cc} \theta_1 &\ \theta_2\\ -\theta_2 &\ \theta_1 \end{array} \right )$,

(note that this is an orthogonal matrix since $\theta_1^2+\theta_2^2=1$) to both $x$ and $\theta$, transforms $\theta$ to $(1,0)$ and $x$ to $(\theta_1 x_1 +\theta_2 x_2 ,-\theta_2 x_1 +\theta_2 x_2)$.Thus the integral becomes:

$\int_{\mathbb{R}^2} \frac{e^{i(\theta_1 x_1 +\theta_2 x_2)\xi_1+i( -\theta_2 x_1 +\theta_2 x_2)\xi_2}}{\xi_1} d\xi_1 d\xi_2= -i \pi sgn(\theta_1 x_1 +\theta_2 x_2) \delta( -\theta_2 x_1 +\theta_2 x_2)$

Where the Fourier transform table (for distributions) from the Wikipedia article was used.

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Both 1-dimensional integrals (whose product makes the 2-dimensional integral you write at the end of your post) diverge. –  Did Sep 26 '11 at 9:09
But mentioned matrix is not a rotation matrix since it contains complex elements. Applying it to $R^2$ will produce some plain in $C^2$. I have an answer: $2\pi \frac{\mathop{sgn}{(1-|\theta_1 + i\theta_2|)}}{\theta_1 x_2 - \theta_2 x_1}$ –  Nimza Sep 26 '11 at 14:45
Sorry I missed that point and took $\theta \in \mathbb{R}^2$, I'll try to post a corrected answer –  David Bar Moshe Sep 26 '11 at 15:08
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