Proof of Gauss's Mean Value Theorem
Theorem
Let the function $f$ be analytic on the closed circle $| z - z_{0} | \le r$. Then $$ f(z_{0}) = {{1} \over {2 \pi}} \int_{0}^{2 \pi} f(z_{0} + r e ^{i \theta } ) d \theta $$
Explanation
Just as the mean value theorem for derivatives gave rise, through generalization, to variant theorems bearing the names of several mathematicians, the mean value theorem for integrals also has a variant that bears none other than the name of Gauss. Its form is unmistakably that of the mean value theorem for integrals, but if you think carefully about the concept, it is a theorem that is not entirely obvious.
Proof
Cauchy’s integral formula: $$f (z_{0}) = {{1} \over {2 \pi i }} \int_{\mathscr{C}} {{f(z)} \over { (z - z_{0}) }} dz$$
By Cauchy’s integral formula, $$ f(z_{0}) = {{1} \over {2 \pi i }} \int_{ |z-z_{0} |= r } {{f(z)} \over { (z - z_{0}) }} dz $$ Substituting $z(\theta) = r e ^{ i \theta } + z_{0} , 0 \le \theta \le 2 \pi$, $$ \begin{align*} f(z_{0}) =& {{1} \over {2 \pi i }} \int_{0}^{2 \pi} {{f( z_{0} + r e^{i \theta} )} \over { r e ^{ i \theta} }} i r e^{i \theta } d \theta \\ =& {{1} \over {2 \pi }} \int_{0}^{2 \pi} f( z_{0} + r e^{i \theta} ) d \theta \end{align*} $$
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