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.
Best Answer
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.