# Contour Integral Representation of a Function

• POTW
Gold Member
MHB
POTW Director
Suppose ##f## is holomorphic in an open neighborhood of the closed unit disk ##\overline{\mathbb{D}} = \{z\in \mathbb{C}\mid |z| \le 1\}##. Derive the integral representation $$f(z) = \frac{1}{2\pi i}\oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w}\,\frac{w + z}{w - z}\, dw + i\operatorname{Im}(f(0))$$ for ##|z| < 1##.

• jbergman, Greg Bernhardt and topsquark

Gold Member
Suppose ##f## is holomorphic in an open neighborhood of the closed unit disk ##\overline{\mathbb{D}} = \{z\in \mathbb{C}\mid |z| \le 1\}##. Derive the integral representation $$f(z) = \frac{1}{2\pi i}\oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w}\,\frac{w + z}{w - z}\, dw + i\operatorname{Im}(f(0))$$ for ##|z| < 1##.
By residue theorem may we say
$$\frac{1}{2\pi i}\oint_{|w| = 1} \frac{Re(f(w))}{w}\,\frac{w + z}{w - z}\, dw$$
$$=\frac{1}{2\pi i}\oint_{|w| = 1} [ \frac{2Re(f(w))}{w-z}-\frac{Re(f(w))}{w} ]\, dw$$
$$=2Re(f(z))-Re(f(0))$$
I have no idea for going further.

Last edited:
• topsquark
Gold Member
Not sure you can use Cauchy's integral formula with ##Re (f(w)) = \frac{1}{2} (f(w) + \overline{f (w)})## because ##\overline{f (w)}## is not holomorphic.

Gold Member
I have partial result, in that I might have proved the integral for functions specified in the Schwarz Reflection Principle.

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### Schwarz Reflection Principle:​

Let ##D## be an open domain in the complex plane who's reflection about the real axis is symmetric. Define ##D^+ = \{ z \in D : Im(z) >0 \}##. Suppose that ##f## is an analytic function which is defined in ##D^+##. Further suppose that ##f## extends to a continuous function on the real axis, and takes on real values on the real axis. Then ##f## can be extended to an analytic function on ##D## by the formula

$$f(\overline{z}) = \overline{f(z)} .$$

and the values for ##z## reflected across the real axis are the reflections of ##f(z)## across the real axis.
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We assume that ##D## contains the disk ##\overline{\mathbb{D}} = \{ z \in \mathbb{C} : |z| \leq 1 \}##.

In the following we only consider functions as described above. First note: ##Im (f(0)) =0##.

Now

\begin{align}
\frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{Re (f(w))}{w} \dfrac{w+z}{w-z} dw & = \frac{1}{2 \pi i} \int_0^{2 \pi} \frac{1}{2} [f (e^{i \theta}) + \overline{f (e^{i \theta})}] \dfrac{e^{i \theta} + z}{e^{i \theta} - z} i d \theta
\nonumber \\
& = \frac{1}{2 \pi i} \int_0^{2 \pi} \frac{1}{2} [f (e^{i \theta}) + f (e^{-i \theta})] \dfrac{e^{i \theta} + z}{e^{i \theta} - z} i d \theta
\nonumber \\
& = \frac{1}{4 \pi i} \int_0^{2 \pi} \left[ f (e^{i \theta}) \dfrac{e^{i \theta} + z}{e^{i \theta} - z} + f (e^{-i \theta}) \dfrac{e^{i \theta} + z}{e^{i \theta} - z} \right] i d \theta
\nonumber \\
& = \frac{1}{4 \pi i} \int_0^{2 \pi} f (e^{i \theta}) \left[ \dfrac{e^{i \theta} + z}{e^{i \theta} - z} + \dfrac{e^{-i \theta} + z}{e^{-i \theta} - z} \right] i d \theta
\nonumber \\
& = \frac{1}{4 \pi i} \oint_{|w|=1} \dfrac{f(w)}{w} \left[ \dfrac{w+z}{w-z} + \dfrac{1+wz}{1-wz} \right] dw \quad (*)
\nonumber
\end{align}

First, using ##(*)## when we have ##z=0##:

$$\frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{Re (f(w))}{w} \dfrac{w+0}{w-0} dw = \frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{f(w)}{w} dw = f(0) .$$

Now take ##z \not=0##. Using ##(*)##,

\begin{align}
\frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{Re (f(w))}{w} \dfrac{w+z}{w-z} dw & = \frac{1}{4 \pi i} \oint_{|w|=1} \dfrac{f(w)}{w} \left[ \dfrac{w+z}{w-z} - \dfrac{w+z^{-1}}{w-z^{-1}} \right] dw
\nonumber \\
& = \frac{1}{4 \pi i} \oint_{|w|=1} f(w) \left[ \dfrac{2}{w-z} - \frac{1}{w} - \dfrac{2}{w-z^{-1}} + \frac{1}{w} \right] dw
\nonumber \\
& = \frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{f(w)}{w-z}dw = f(z) .
\nonumber
\end{align}

Last edited:
Gold Member
MHB
POTW Director
Let ##|z| < 1##. Cauchy's integral formula gives ##f(z) = \frac{1}{2\pi i} \oint_{|w| = 1} \frac{f(w)}{w - z}\, dw##. In particular, ##f(0) = \frac{1}{2\pi i} \oint_{|w| = 1} \frac{f(w)}{w}\, dw##. For all ##w## on the unit circle, ##w\overline{w} = 1##. Hence ##\overline{f(0)} = \frac{1}{2\pi i} \oint_{|w| = 1} \frac{\overline{f(w)}}{w} dw##, and
\begin{align*}
f(z) + \overline{f(0)} &= \frac{1}{2\pi i} \oint_{|w| = 1} \frac{w(f(w) + \overline{f(w)}) - z\overline{f(w)}}{w(w-z)} \, dw\\
&= \frac{2}{2\pi i} \oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w - z}\, dw - \frac{z}{2\pi i} \oint_{|w| = 1} \frac{\overline{f(w)}}{(w - z)}\, \frac{dw}{w}\\
&= \frac{2}{2\pi i} \oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w - z}\, dw + \frac{z}{2\pi i}\overline{ \oint_{|w| = 1} \frac{f(w)}{(\bar{w} - z)}\, \frac{dw}{w}}\\
&=\frac{2}{2\pi i} \oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w - z}\, dw + \frac{z}{2\pi i}\overline{\oint_{|w| = 1} \frac{f(w)}{(1 - zw)}\, dw}\\
&= \frac{2}{2\pi i}\oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w-z}\, dw
\end{align*}
where the last equality follows from Cauchy's theorem (the function ##w\mapsto f(w)/(1 - zw)## is holomorphic inside and on the unit circle for ##|z| < 1##). Since ##z## was arbitrary it follows that $$\operatorname{Re}(f(0)) = \frac{f(0) + \overline{f(0)}}{2} = \frac{1}{2\pi i} \oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w}\, dw$$yielding $$f(z) - i\operatorname{Im}(f(0)) = \frac{1}{2\pi i} \oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w}\left[\frac{2w}{w - z} - 1\right]\, dw = \frac{1}{2\pi i} \oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w} \frac{w + z}{w - z}\, dw$$which is equivalent to the desired result.

• julian and anuttarasammyak