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How to show the following formula in the sense of distribution in $\mathbb{R}^2$? \begin{equation}(\partial_x +i\partial_y)\frac{1}{x+iy}=2\pi\delta_{(0,0)} \end{equation}

Gribouillis
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Ailiy Evan
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    What do you know about these objects? How is $\delta$ defined? How is $\frac{1}{x+iy}$ defined? How is the derivative $\partial_x+i\partial_y$ defined? What would the equality mean? As it's written now, the question doesn't include all the necessary definitions and doesn't show your efforts. – Michał Miśkiewicz Feb 20 '21 at 19:25
  • I think it is clear. $\frac{1}{x+iy}$ is complex number. $x$ is called the real part. $y$ is called the imaginary part. – Ailiy Evan Feb 20 '21 at 19:34
  • The function $\frac{1}{x+iy}$ is not locally integrable (around $0$) and hence doesn't define a distribution in the usual way. Probably the author of the problem meant principle way, so I suggest you check it first. – Michał Miśkiewicz Feb 20 '21 at 19:37
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    you are wrong. $\frac{1}{x+iy}$ is locally integrable(use polar coordinates). – Ailiy Evan Feb 20 '21 at 19:43
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    $x=r cos\alpha, y=r sin \alpha, e^{i\alpha}=cos \alpha+isin\alpha $. this function is locally integrable. – Ailiy Evan Feb 20 '21 at 19:49

1 Answers1

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Let ${\psi} \left(x , y\right)$ be a test function and let ${\varphi} \left(r , {\theta}\right) = {\psi} \left(r \cos {\theta} , r \sin {\theta}\right)$. We compute

\begin{equation} \renewcommand{\arraystretch}{2} \begin{array}{rcl}\displaystyle \frac{\partial {\varphi}}{\partial r}&=&\displaystyle \cos {\theta} \frac{\partial {\psi}}{\partial x}+\sin {\theta} \frac{\partial {\psi}}{\partial y}\\ \displaystyle \frac{\partial {\varphi}}{\partial {\theta}}&=&\displaystyle -r \sin {\theta} \frac{\partial {\psi}}{\partial x}+r \cos {\theta} \frac{\partial {\psi}}{\partial y} \end{array} \end{equation}

Hence

\begin{equation} \frac{1}{r} \frac{\partial {\varphi}}{\partial r}+\frac{i}{{r}^{2}} \frac{\partial {\varphi}}{\partial {\theta}} = \frac{1}{x+i y} \left(\frac{\partial {\psi}}{\partial x}+i \frac{\partial {\psi}}{\partial y}\right) \end{equation}

Hence

\begin{equation} \renewcommand{\arraystretch}{2} \begin{array}{rcl}\displaystyle\left\langle \left(\frac{\partial }{\partial x}+i \frac{\partial }{\partial y}\right) \left(\frac{1}{x+i y}\right) , {\psi}\right\rangle &=&\displaystyle -\int_{{\mathbb{R}}^{3}}^{}\frac{1}{x+i y} \left(\frac{\partial }{\partial x}+i \frac{\partial }{\partial y}\right) \left({\psi}\right) d x d y\\ &=&-\displaystyle \int_{{\left[0 , \infty \right[}\times{\left[0 , 2 {\pi}\right]}}^{}\left(\frac{\partial {\varphi}}{\partial r}+\frac{i}{r} \frac{\partial {\varphi}}{\partial {\theta}}\right) d r d {\theta} \end{array} \end{equation}

Using that $\displaystyle \int_{0}^{2 {\pi}}\frac{\partial {\varphi}}{\partial {\theta}} d {\theta} = 0$, it remains

\begin{equation} \left\langle \left(\frac{\partial }{\partial x}+i \frac{\partial }{\partial y}\right) \left(\frac{1}{x+i y}\right) , {\psi}\right\rangle =-\int_{0}^{2 {\pi}}\int_{0}^{\infty }\frac{\partial {\varphi}}{\partial r} d r d {\theta} = \int_{0}^{2 {\pi}}{\psi} \left(0 , 0\right) d {\theta} = 2 {\pi} {\psi} \left(0 , 0\right) \end{equation}

Gribouillis
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