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The following is an exercise from Complex Analysis by Stephen Fisher.

Fix a complex number $a$ and a positive real number $R$. Suppose $u$ is a function defined on the circle of radius $R$ centered at $a$. Let $C$ denote this circle.

Show that the average value of $u$ on $C$ is given by $\frac{1}{2\pi}\int_{0}^{2\pi} u(a + Re^{it})dt$.

Any Hints please.

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    The exercise is probably meant to relate to Cauchy's integral formula, but as said above: the integral you write is somewhat the definition of "average value" and the goal of the calculation, so the exercise is a bit unclear in its current form. I have the book (2nd edition at least) - which exercise number/page is it, so I can get a context?2012-10-26

2 Answers 2

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We assume Cauchy's integral theorem.

$(1)\hspace{5mm}\frac{1}{2\pi}\int_{0}^{2\pi} u(a+ Re^{i t})dt = \frac{1}{2\pi i}\int_{0}^{2\pi}\frac{u(a+Re^{it})iRe^{it}}{Re^{it}}dt $

The right side of (1), letting R be the radius of circle $C,$ $u(z)$ the equation of the circle $|z-a|= R $ or $z = a + Re^{it},$ so that $z-a = Re^{it}$ and $dz = iRe^{it},$ is precisely

$(2)\hspace{5mm}\frac{1}{2\pi i}\oint_C\frac{u(z)}{z-a}dz$

By Cauchy's integral formula (2) is equal to $u(a).$ $\square $

This exercise is odd because the starting integral is generally taken as the definition of the average value, as noted in a comment above. We would normally derive that form for the average from Cauchy's integral formula.

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    @AmitaiYuval: OK, will study your comments and answer in$a$bit. Thanks.2017-04-21
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As implied in the comments, the most important part in this exercise is the interpretation of "average value". As always, in order to define the average value of a function on $C$ we need a measure on $C$. One should note that different measures evidently yield different average values. However, as $C$ is a smooth curve, it makes sense to take the standard length measure.

Parametrize $C$ by$\gamma:[0,2\pi R]\to\mathbb{C},\quad s\mapsto a+Re^{is/R}.$This is an arc-length parametrization, or in other words, pulling back the length measure on $C$ with respect to $\gamma$ yields the standard Lebesgue measure on $[0,2\pi R]$. Hence, the average value of $u$ on $C$ is equal to the average value of $u\circ\gamma$ on $[0,2\pi R]$. The latter is just$\frac{1}{2\pi R}\int_0^{2\pi R}u\left(a+Re^{is/R}\right)ds.$The desired expression follows now by substituting $t=s/R$.

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    Hi @Amitai Yuval , can you try answer my question over here? https://math.stackexchange.com/questions/2944389/calculate-the-mean-of-a-function-within-a-part-of-a-circle Its very related to this subject2018-10-27