# Compute the Fourier transform of $(x_{1}+ix_{2})^{-1}$ in $S'(\mathbb{R}^{2})$ (as a tempered distribution).

I am trying to compute the Fourier transform of $$(x_{1}+ix_{2})^{-1}$$ in $$S'(\mathbb{R}^{2})$$. i.e. as a tempered distribution.

It might be useful to note that for $$\mu \in S'(\mathbb{R}^{2})$$ and $$\psi \in S(\mathbb{R}^{2})$$ we define $$\langle\hat{\mu},\psi\rangle=\langle\mu,\hat{\psi}\rangle$$.

In my attempt, I noted that the definition of the Fourier transform in $$S(\mathbb{R}^{2})$$:

$$\hat{f}(\lambda)=\int_{\mathbb{R}^{2}}f(x)e^{-i \lambda \cdot x} dx,$$ gave us that $$\widehat{(-i \partial_{1}+\partial_{2}) \delta}=x_{1}+ix_{2}$$. I'm not sure how to use this fact to help me complete the problem. Any help would be greatly appreciated.

To calculate this Fourier transform, we will need another well known Fourier transform pair in $$1$$D:

$$\int_{-\infty}^\infty e^{-|a||x|}e^{-i\lambda x}dx = \int_{-\infty}^0e^{(|a|-i\lambda)x}dx + \int_0^\infty e^{-(|a|+i\lambda)x}dx$$

$$= \frac{1}{|a|-i\lambda}+\frac{1}{|a|+i\lambda} = \frac{2|a|}{a^2+\lambda^2}$$

which gives us the integral

$$\frac{1}{2\pi}\int_{-\infty}^\infty\frac{2|a|}{a^2+\lambda^2}e^{i\lambda x}d\lambda = e^{-|a||x|}$$

for free. Note that since both functions are real and even, the difference between forward and reverse Fourier transforms is negligible.

Back to our function, rewrite it as

$$\frac{1}{x_1+ix_2} = \frac{x_1-ix_2}{x_1^2+x_2^2}$$

and let's compute the Fourier transform of the real part only

$$\int_{-\infty}^\infty\int_{-\infty}^\infty \frac{x_1}{x_1^2+x_2^2}e^{-i\lambda_2x_2}e^{-i\lambda_1x_1}dx_2dx_1 = \int_{-\infty}^\infty \frac{\pi x_1}{|x_1|}e^{-|x_1||\lambda_2|}e^{-i\lambda_1x_1}dx_1$$

$$= \frac{\pi}{|\lambda_2|+i\lambda_1}-\frac{\pi}{|\lambda_2|-i\lambda_1} = \frac{-2\pi i \lambda_1}{\lambda_1^2+\lambda_2^2}$$

And without any extra work, the full Fourier transform is

$$-2\pi\frac{i\lambda_1+\lambda_2}{\lambda_1^2+\lambda_2^2} = \frac{2\pi}{i\lambda_1-\lambda_2}$$

by linearity.

Let's denote $$x=\sqrt{x_1^2+x_2^2}$$ and $$\lambda=\sqrt{\lambda_1^2+\lambda_2^2}$$. We consider the case $$\lambda\neq0$$.
First, we note that we can arbitrary choose the system of coordinates $$(x_1, ix_2)$$ in the complex plane. We can direct the axis $$X$$ along the $$\vec\lambda$$. At this choice, $$\vec\lambda= (\lambda_1;i\lambda_2)=(\lambda, 0)$$. We also note that in this case $$\lambda_1x_1+\lambda_2x_2=\lambda x_1=\lambda x\cos\phi\,$$ and $$\,x_1+ix_2=xe^{i\phi}$$. $$I=\int_{\mathbb{R}^{2}}\frac{e^{-i (\lambda_1 x_1+\lambda_2 x_2)}}{x_1+ix_2} dx_1dx_2$$ Switching into the polar system of coordinates, $$I=\int_0^\infty xdx\int_0^{2\pi}d\phi\,\frac{e^{-i\lambda x\cos\phi}}{xe^{i\phi}}$$
Bearing in mind that $$\lambda\neq0$$ and splitting interval $$[0;2\pi]$$ into $$[0;\pi]$$ and $$[\pi;2\pi]$$, after easy transformations we get $$I=\frac{1}{\lambda}\int_0^\infty dt\int_0^\pi e^{-it\cos\phi}(e^{-i\phi}-e^{i\phi})\,d\phi=-\frac{2i}{\lambda}\int_0^\infty dt\int_0^\pi e^{-i t\cos\phi}\sin\phi \,d\phi$$ $$=-\frac{2i}{\lambda}\int_0^\infty dt\int_{-1}^1e^{-itx}dx=\frac{4i}{\lambda}\int_0^\infty \frac{\sin t}{t}dt=\frac{2\pi i}{\lambda}$$ Remembering that it was a special choice of coordinate system, and that at arbitrary choice $$\vec\lambda=(\lambda_1;i\lambda_2)=\lambda e^{i\psi}; \psi=\tan^{-1}\frac{\lambda_2}{\lambda_1}$$, the general answer is $$I(\lambda_1;\lambda_2)=\frac{2\pi i}{\lambda e^{i\psi}}=\frac{2\pi i}{\lambda_1+i\lambda_2}$$
• A0710046 I have to apologise - my solution is not correct - I will remove it. The ideology is right, but to obtain the coefficient $2\pi i$ it requires the integration of Bessel function. So it becomes nor nice not easy... Nov 16, 2021 at 8:28