Let $f,g$ be analytic on some domain $\Omega \subset \mathbb{C}$. By Cauchy's formula, we have $$ \frac{1}{2\pi i} \oint_{\partial\Omega} \frac{f(z) \, g(z)}{z - z_0} \, dz = f(z_0) \, g(z_0) = -\frac{1}{4\pi^2} \oint_{\partial\Omega} \frac{f(u)}{u - z_0} \, du \, \oint_{\partial\Omega} \frac{g(v)}{v - z_0} \, dv . $$ Is there a way how I can get from the first expression to the last without the intermediate step?
How to prove contour integral of product is equal to product of contour integrals
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contour-integration
2 Answers
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I found the solution here. The main tool of the proof is the so-called first resolvent formula $$ \frac{1}{(z-z_0) \, (w - z_0)} = \frac{1}{w-z} \, \left(\frac{1}{z - z_0} - \frac{1}{w - z_0}\right) . $$ To see how this is useful, introduce another contour $\partial \Omega'$ and write \begin{align} \frac{1}{2\pi i} \oint_{\partial\Omega} \frac{f(z) \, g(z)}{z - z_0} \, dz &= -\frac{1}{4\pi^2} \oint_{\partial\Omega} \frac{f(z)}{z - z_0} \oint_{\partial\Omega'} \frac{g(w)}{w-z} \, dw\, dz \\&= -\frac{1}{4\pi^2} \oint_{\partial\Omega} \oint_{\partial\Omega'} f(z) \, g(w) \, \frac{1}{(z-z_0)\,(w-z)} \, dw\, dz . \end{align} This last fraction is precisely the first term on the right-hand side of the resolvent formula, and the second term can be added if we pick $\Omega' \supset \Omega$ because then $$ \oint_{\partial\Omega} \oint_{\partial\Omega'} \frac{f(z) \, g(w) }{(w-z_0)\,(w-z)} \, dw \, dz = \oint_{\partial\Omega'} \oint_{\partial\Omega} \frac{f(z) \, g(w)}{(w-z_0)\,(w-z)} \, dz \, dw = 0 $$ by Fubini's theorem (note that the integration domain as well as the integrand are bounded) and Cauchy's theorem (the integrand is analytic for $z \in \Omega \subset \Omega'$). Therefore, $$ \frac{1}{2\pi i} \oint_{\partial\Omega} \frac{f(z) \, g(z)}{z - z_0} \, dz = -\frac{1}{4\pi^2} \oint_{\partial\Omega} \oint_{\partial\Omega'} \frac{f(z) \, g(w)}{(z - z_0) \, (w-z_0)} \, dw\, dz . $$
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Probably not. If I am guessing correctly what you have in mind, then it is a proof which is local in the sense that it only involves the functions in neighbourhood of the integration path, i.e. a proof that would still work if $f$ and $g$ are not analytic on all of $\Omega$. But then the formula does not hold any more, consider e.g. $z_0=0$, $f(z)=1/z$, $g(z)=z$.