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I came across an identity for integration of analytic function , and I don't know how to prove it. Please help me with it, thank you. The statement goes as following

For any analytic function $f(\alpha)$ \begin{equation} f(\beta) = \frac{1}{2\pi}\int f(\alpha)\exp{\{-|\beta-\alpha|^2/2\}}d^2\alpha \end{equation} in which $d^2\alpha=d\mathrm{Re}\{\alpha\}d\mathrm{Im}\{\alpha\}$

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This identity follows from Gauss mean value theorem: If $f:\Bbb C\to \Bbb C$ is entire (with moderate growth), we have $$\begin{align*} \int_{\Bbb R^2}f(\alpha)e^{-|\beta-\alpha|^2/2}d\alpha&=\int_{\Bbb R^2}f(\beta+\alpha)e^{-|\alpha|^2/2}d\alpha\\&=\int_{0}^{2\pi}\int_0^\infty f(\beta+re^{i\theta})e^{-\frac{r^2}2}r drd \theta\tag{*}\\&=\int_0^\infty\left(\int_{0}^{2\pi} f(\beta+re^{i\theta})d \theta\right)e^{-\frac{r^2}2}r dr\\&=2\pi f(\beta)\cdot \int_0^\infty e^{-\frac{r^2}2}r dr\tag{**}\\&=2\pi f(\beta)\left[-e^{-\frac{r^2}2}\right]^\infty_0=2\pi f(\beta). \end{align*}$$ $(*)$ : polar coordinate change.
$(**)$ : mean value theorem is used.

Since harmonic functions have mean value property, the identity is true for every harmonic function.

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  • $\begingroup$ Thank you very much! $\endgroup$ – yangcs11 Feb 27 at 12:29

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