# Contour Integral Representation of a Function

• POTW
• Euge
In summary, a contour integral representation is a method used in complex analysis to evaluate integrals of complex-valued functions over a specified path in the complex plane. It is important because it allows for the evaluation of complex integrals and provides a way to analyze the function's behavior. The contour is chosen as a closed path in the complex plane that encloses the region of interest and does not pass through any singularities. It can also be used for functions with real variables by considering the real line as a special case. Contour integral representation has applications in physics, engineering, mathematics, signal processing, image analysis, and data compression.
Euge
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
Euge said:
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
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.

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

--------

### 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.
--------

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:
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

## What is the contour integral representation of a function?

The contour integral representation of a function is a way of representing a complex-valued function using a contour integral, which is an integral along a closed curve in the complex plane. It allows for the evaluation of the function at any point inside the contour, even if the function is not analytic at that point.

## How is the contour integral representation related to the Cauchy Integral Formula?

The contour integral representation is a generalization of the Cauchy Integral Formula, which states that the value of an analytic function inside a closed contour is equal to the integral of the function over the contour. The contour integral representation allows for the evaluation of non-analytic functions inside the contour.

## What are the benefits of using contour integral representation?

Contour integral representation allows for the evaluation of complex functions at any point inside the contour, even if the function is not analytic at that point. It also allows for the evaluation of integrals that are difficult or impossible to evaluate using traditional methods.

## What types of functions can be represented using contour integrals?

Contour integral representation can be used for any complex-valued function, including both analytic and non-analytic functions. It is particularly useful for evaluating functions that have singularities or branch points.

## How is the contour chosen for a specific function?

The choice of contour depends on the specific function and the desired point of evaluation. In general, the contour should enclose the point of interest and avoid any singularities or branch points of the function. The contour can also be chosen to simplify the integral or make it easier to evaluate using known techniques.

• Math POTW for Graduate Students
Replies
1
Views
1K
• Math POTW for Graduate Students
Replies
13
Views
570
• Math POTW for Graduate Students
Replies
1
Views
759
• Math POTW for Graduate Students
Replies
2
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
670
• Math POTW for University Students
Replies
1
Views
1K
• Math POTW for Graduate Students
Replies
1
Views
1K
• Differential Equations
Replies
2
Views
1K
• Math POTW for Graduate Students
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
16
Views
1K