Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $L = \langle \theta, \nabla \rangle$.

share|cite|improve this question
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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.