📂Complex Anaylsis

Poisson Integral Formula Derivation

Formulas ¹

Let’s assume the function $f : \mathbb{C} \to \mathbb{C}$ is analytic in a simply connected region that contains the circle $\mathscr{C}: |z| = r$. Then, for $0 < \rho < r$, we have $$f( \rho e ^{i \phi} ) = {{1} \over { 2 \pi }} \int_{0}^{2 \pi} {{r^2 - \rho^2 } \over {r^2 - 2 r \rho \cos (\theta - \phi) + \rho ^2 }} f(r e^{i \theta}) d \theta$$

Derivation

Strategy: Essentially, it is a variation of the Cauchy Integral Formula. The derivation itself doesn’t hold much value beyond a single read-through, as it only involves numerous trivial calculations.

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First, let’s show that $\displaystyle f(\alpha) = {{1} \over {2 \pi i }} \int_{\mathscr{C}} \left( {{1} \over { z - \alpha }} - {{1} \over { z - r^2 / \overline{\alpha} }} \right) f(z) dz$ holds for $\alpha$ that satisfies $f(\alpha) \ne 0$ inside $\mathscr{C}$ interior.

Since $\alpha$ is a point inside $\mathscr{C}$ interior, $|\alpha| < r$ applies, therefore $${{r^2} \over {|\alpha^2|}} > 1$$ $\displaystyle {{r^2} \over {| \overline{ \alpha } |}} = {{ r^2 } \over { |\alpha|^2 }} \left| \alpha \right|$, thus $$|\alpha| < {{r^2} \over {| \overline{ \alpha } |}}$$ Due to the density of real numbers, we can consider a circle ${\mathscr{C}} ': |z| = \rho$ with a radius larger than $|\alpha|$ but smaller than $\displaystyle {{r^2} \over {| \overline{ \alpha } |}}$ that contains $\rho$. By definition, ${\mathscr{C}} ': |z| = \rho$ includes $\alpha$ but not $\displaystyle {{r^2} \over { \overline{ \alpha } }}$. According to the contraction auxiliary theorem, $$\begin{align*} & {{1} \over {2 \pi i }} \int_{\mathscr{C}} \left( {{1} \over { z - \alpha }} - {{1} \over { z - r^2 / \overline{\alpha} }} \right) f(z) dz \\ =& {{1} \over {2 \pi i }} \int_{\mathscr{C}’} \left( {{1} \over { z - \alpha }} - {{1} \over { z - r^2 / \overline{\alpha} }} \right) f(z) dz \\ =& {{1} \over {2 \pi i }} \int_{\mathscr{C}’} {{1} \over { z - \alpha }} f(z) dz - {{1} \over {2 \pi i }} \int_{\mathscr{C}’} {{1} \over { z - r^2 / \overline{\alpha} }} f(z) dz \end{align*}$$ By the Cauchy Integral Formula, $${{1} \over {2 \pi i }} \int_{\mathscr{C}’} {{1} \over { z - \alpha }} f(z) dz = f(\alpha)$$ According to the Cauchy’s Theorem, $${{1} \over {2 \pi i }} \int_{\mathscr{C}’} {{1} \over { z - r^2 / \overline{\alpha} }} f(z) dz = 0$$ Therefore, we obtain the following. $$f(\alpha) = {{1} \over {2 \pi i }} \int_{\mathscr{C}} \left( {{1} \over { z - \alpha }} - {{1} \over { z - r^2 / \overline{\alpha} }} \right) f(z) dz$$

Meanwhile, $$\left( {{1} \over { z - \alpha }} - {{1} \over { z - r^2 / \overline{\alpha} }} \right) = {{z- r^2 / \overline{\alpha} -z +\alpha} \over {(z-\alpha)(z - r^2 / \overline{\alpha} )}} = \alpha {{1 - | r^2 / \alpha^2 | } \over {(z-\alpha)(z - r^2 / \overline{\alpha} )}}$$ Thus, summarizing, $$f(\alpha) = {{1} \over {2 \pi i }} \int_{\mathscr{C}} \alpha {{1 - | r^2 / \alpha^2 | } \over {(z-\alpha)(z - r^2 / \overline{\alpha} )}} f(z) dz$$ Substituting $z = r e^{i \theta}, 0 \le \theta < 2 \pi$ and $\alpha = \rho e^{ i \phi} , 0 \le \phi < 2 \pi$ gives us, $$\begin{align*} f(\rho e^{ i \phi}) =& {{1} \over {2 \pi i }} \int_{0}^{2 \pi} { { \rho e^{ i \phi} ( 1 - | r^2 / \rho^2 | ) } \over {(r e^{i \theta} - \rho e^{ i \phi} )( r e^{i \theta} - r^2 / \rho e^{ -i \phi} )}} f( r e^{i \theta} ) i r e^{i \theta} d \theta \\ =& {{1} \over {2 \pi }} \int_{0}^{2 \pi} { { {{r} \over {\rho}} e^{ i \phi} ( \rho^2 - r^2 ) e^{i \theta} } \over { {{r} \over {\rho}} (r e^{i \theta} - \rho e^{ i \phi} )( \rho e^{i \theta} - r e^{ i \phi} )}} f( r e^{i \theta} ) d \theta \\ =& {{1} \over {2 \pi }} \int_{0}^{2 \pi} { { ( \rho^2 - r^2 ) e^{i (\theta + \phi)} } \over { r \rho e ^{2 i \theta} - \rho^2 e^{i ( \theta + \phi )} - r^2 e^{i (\theta + \phi)} + r \rho e ^{ 2 i \phi} }} f( r e^{i \theta} ) d \theta \\ =& {{1} \over {2 \pi }} \int_{0}^{2 \pi} { { \rho^2 - r^2 } \over { r \rho e ^{ i (\theta - \phi)} - \rho^2 - r^2 + r \rho e ^{ i (\phi - \theta )} }} f( r e^{i \theta} ) d \theta \\ =& {{1} \over {2 \pi }} \int_{0}^{2 \pi} { { r^2 - \rho^2 } \over { r^2 - 2 r \rho \cos (\theta - \phi) + \rho ^2 }} f( r e^{i \theta} ) d \theta \end{align*}$$

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