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.


2 Answers 2


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.

  • $\begingroup$ Thank you so much for your answer @Ninad Munshi! $\endgroup$
    – A0710046
    Nov 15, 2021 at 20:38

There is a comprehensive solution by @Ninad Munshi. We can also try another approach.

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 $$I(\lambda_1;\lambda_2)=\frac{2\pi i}{\lambda e^{i\psi}}=\frac{2\pi i}{\lambda_1+i\lambda_2}$$

  • $\begingroup$ Thank you so much for your help @Svyatoslav! $\endgroup$
    – A0710046
    Nov 15, 2021 at 20:39
  • $\begingroup$ 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... $\endgroup$
    – Svyatoslav
    Nov 16, 2021 at 8:28
  • 1
    $\begingroup$ Following your very helpful comments, I tried to compute the integral using Cauchy Residue Theorem and that worked for me. $\endgroup$
    – A0710046
    Nov 17, 2021 at 1:09
  • $\begingroup$ @A0710046 Thank you - I'm glad that it helped! My solution is not correct, though. $\endgroup$
    – Svyatoslav
    Nov 17, 2021 at 5:58

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