# Cauchy’s Integral Theorem – Local Version

## By Sébastien Boisgérault, Mines ParisTech, under CC BY-NC-SA 4.0

### November 28, 2017

# Contents

# Exercises

## A Fourier Transform

We wish to compute for any real number \(\omega\) the value of the integral \[ \hat{x}(\omega) = \int_{-\infty}^{+\infty} x(t) e^{-i \omega t}\, dt \] when \(x: \mathbb{R} \to \mathbb{R}\) is the Gaussian function defined by \[ \forall \, t \in \mathbb{R}, \; x(t) = e^{-t^2/2}. \]

We remind you of the value of the *Gaussian integral* (see e.g. Wikipedia): \[
\hat{x}(0) =
\int_{-\infty}^{+\infty} e^{-t^2/2} \, dt = \sqrt{2\pi}
\]

### Questions

Show that for any pair of real numbers \(\tau\) and \(\omega\), we can compute \[ \int_{-\tau}^{\tau} x(t) e^{-i \omega t}\, dt \] from the line integral of a fixed holomorphic function on a path \(\gamma\) that depends on \(\tau\) and \(\omega\).

Use Cauchy’s integral theorem to evaluate \(\hat{x}(\omega)\).

### Answers

We may denote \(x\) the extension to the complex plane of the original Gaussian function \(x\), defined by: \[ \forall \, z\in \mathbb{C}, \; x(z) = e^{-z^2/2}. \] It is holomorphic as a composition of holomorphic functions. Let \(\gamma = [-\tau \to \tau] + i\omega\). The line integral of \(x\) along \(\gamma\) satisfies \[ \int_{\gamma} x(z) \, dz = \int_{0}^1 x(-\tau (1-s) + \tau s + i \omega) \, (2\tau ds) \] or, using the change of variable \(t = -\tau (1-s) + \tau s\), \[ \int_{\gamma} x(z) \, dz = \int_{-\tau}^{\tau} x(t + i\omega) \, dt. \] Since \[ x(t + i \omega) = e^{-{(t + i\omega)^2}/{2}} = e^{-t^2/2} e^{-i\omega t} e^{\omega^2/2}, \] we end up with \[ \int_{\gamma} x(z) \, dz = e^{\omega^2/2} \int_{-\tau}^{\tau} x(t) e^{-i2\pi f t}\, dt \]

Let \(\nu = \tau + [0 \to i \omega]\); on the image of this path, we have \[ \forall \, s \in [0,1], \; |x(\nu(s))| = \left| e^{-(\tau + i \omega s)^2/2} \right| = e^{-\tau^2/2} e^{(\omega s)^2/2} \leq e^{-\tau^2/2} e^{\omega^2/2}, \] hence the M-L inequality provides \[ \left| \int_{\mu} x(z) \, dz \right| \leq (|\omega| e^{\omega^2/2} )e^{-\tau^2/2} \] and thus, \[ \forall \, \omega \in \mathbb{R}, \; \lim_{|\tau| \to +\infty} \int_{\mu} x(z) \, dz = 0. \]

We may apply Cauchy’s integral theorem to the function \(x\) on the closed polyline \[ [-\tau + i \omega \, \to \, \tau + i\omega \, \to \, \tau \, \to \, -\tau \, \to \, -\tau + i \omega]. \] It is the concatenation of \(\gamma = [-\tau \to \tau] + i\omega\), the reverse of \(\mu_+ = \tau + [0 \to i \omega]\), the reverse of \(\gamma_0 = [-\tau \to \tau]\) and finally \(\mu_- = -\tau + [0 \to i \omega]\).

The theorem provides \[ \begin{split} e^{\omega^2/2} \int_{-\tau}^{\tau} x(t) e^{-i \omega t}\, dt - \int_{\mu_+} x(z) \, dz \\ - \int_{-\tau}^{\tau} x(t) \, dt + \int_{\mu_-} x(z) \, dz = 0. \end{split} \] When \(\tau \to +\infty\), this equality yields \[ \int_{-\infty}^{\infty} x(t) e^{-i2\pi f t}\, dt = \sqrt{2\pi} e^{-\omega^2/2}. \]

## Cauchy’s Integral Formula for Disks

Let \(\Omega\) be an open subset of the complex plane and \(\gamma = c + r[\circlearrowleft]\). We assume that the closed disk \(\overline{D}(c, r)\) is included in \(\Omega\) (this is stronger than the requirement that \(\gamma\) is a path of \(\Omega\)).

We wish to prove that for any holomorphic function \(f:\Omega \to \mathbb{C}\), \[ \forall \, z \in \Omega, \; |z - c| < r \, \Rightarrow \, f(z) = \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w-z} dw. \]

### Questions

What is the value of the line integral above when \(|z - c| > r\)?

Compute the line integral above when \(z=c\) as an integral with respect to a real variable. When happens in this case when \(r \to 0\) ?

Let \(\epsilon>0\) be such that \(|z| + \epsilon < r\) and let \(\lambda = z + \epsilon[\circlearrowleft]\). Provide two paths \(\mu\) and \(\nu\) whose images belong to (different) star-shaped subsets of \(\Omega \setminus \{z\}\) and such that for any continuous function \(g:\Omega \setminus \{z\}\to \mathbb{C},\) \[ \int_{\gamma} g(w) \, dw = \int_{\lambda} g(w) dw + \int_{\mu} g(w) \, dw + \int_{\nu} g(w) dw. \]

Prove Cauchy’s integral formula for disks.

Show that \(f'(z)\) can be computed as a line integral on \(\gamma\) of an expression that depends on \(f(w)\) and not on \(f'(w)\). What property of \(f'\) does this expression shows?

### Answers

If \(|z - c| > r\), the function \(w \mapsto {f(w)} / (w-z)\) is defined and holomorphic in \(\Omega \setminus \{z\}\). Let \(\rho\) be the minimum between \(|z-c|\) and the distance between \(c\) and \(\mathbb{C} \setminus \Omega\). By construction, the open disk \(D(c, \rho)\) is a star-shaped subset of \(\Omega \setminus \{z\}\) and it contains the image of \(\gamma\). Thus, Cauchy’s integral theorem provides \[ \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w-z} dw = 0. \]

When \(z=c\), we have \[ \begin{split} \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w-z} dw =&\, \frac{1}{i2\pi} \int_0^1 \frac{f(c+re^{i2\pi t})}{c+re^{i2\pi t} - c} (i2\pi re^{i2\pi t}dt) \\ =&\, \int_0^1 f(c+re^{i2\pi t}) dt. \end{split} \] By continuity of \(f\) at \(c\), the limit of this integral when \(r \to 0\) is \(f(c)\).

Assume for the sake of simplicity that \(z = c + x\) for some real number \(x \in \left[0,r \right[\). Let \(\alpha = \arccos x/r\); define \(\mu\) as the concatenation \[ \begin{array}{lll} \mu \; = & c+ [x + i\epsilon \to re^{i\alpha}] &\, | \\ & c + r e^{i[\alpha \to 2\pi-\alpha]} &\, | \\ & c + [re^{-i\alpha} \to x - i\epsilon] &\, | \\ & c + x + \epsilon e^{i [-\pi/2 \to -3\pi/2]} & \, . \end{array} \] and \(\nu\) as the concatenation \[ \begin{array}{lll} \nu \; = & c+ [x - i\epsilon \to re^{-i\alpha}] &\, | \\ & c + r e^{i[-\alpha \to \alpha]} &\, | \\ & c + [re^{i \alpha} \to x + i\epsilon] &\, | \\ & c + x + \epsilon e^{i [\pi/2 \to -\pi/2]} & \, . \end{array} \]

Since the closure of \(D(c,r)\) is included in \(\Omega\), there is a \(\rho > r\) such that \(D(c,\rho) \subset \Omega\). The image of \(\mu\) belongs to the set \[D(c, \rho) \setminus \{z + t \;|\; t \geq 0\}\] while the image of \(\nu\) belongs to the set \[D(c, \rho) \setminus \{z + t \;|\; t \leq 0\}.\] Both sets are star-shaped and included in \(\Omega\).

Additionally, for any continuous function \(g:\Omega \setminus \{z\}\to \mathbb{C}\)

the integral of \(g\) on \(c+ [x + i\epsilon \to re^{i\alpha}]\) and its reverse path on one hand, the integral of \(g\) on \(c+[re^{-i\alpha} \to x - i\epsilon]\) and its reverse path on the other hand are opposite numbers.

the sum of the integral of \(g\) on \(c + r e^{i[\alpha \to 2\pi-\alpha]}\) and \(c + r e^{i [-\alpha \to \alpha]}\) is equal to its integral on \(\gamma = c + r [\circlearrowleft]\).

the sum of the integral of \(g\) on \(c + x + \epsilon e^{i [-\pi/2 \to -3\pi/2]}\) and \(c + x + \epsilon e^{i [\pi/2 \to -\pi/2]}\) is equal to the opposite of its integral on \(\lambda = c + x + \epsilon[\circlearrowleft]\).

Therefore, the equality \[ \int_{\gamma} g(w) \, dw = \int_{\lambda} g(w) dw + \int_{\mu} g(w) \, dw + \int_{\nu} g(w) dw. \] holds.

We may apply the result of the previous question to the function \(w \mapsto f(w)/(w-z)\). As it is holomorphic on \(\Omega \setminus \{z\}\), the star-shaped version of Cauchy’s integral theorem provides \[ \int_{\mu} \frac{f(w)}{w-z} dw = \int_{\nu} \frac{f(w)}{w-z} dw = 0, \] hence \[ \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w-z} dw = \frac{1}{i2\pi} \int_{\lambda} \frac{f(w)}{w-z} dw. \] We proved in question 2. that the right-hand side of this equation tends to \(f(z)\) when \(\epsilon \to 0\), thus \[ \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w-z} dw = f(z), \] which is Cauchy’s integral formula for disks.

Let \(z \in \Omega\). There are some \(c\in \Omega\) and \(r> 0\) such that \(z \in D(c, r)\) and \(\overline{D}(z,r) \subset \Omega\); let \(\gamma = c + r[\circlearrowleft]\). For any complex number \(h\) such that \(|z+h - c| < r\), we have by Cauchy’s formula \[ \begin{split} \frac{f(z+h) - f(z)}{h} = &\, \frac{1}{i2\pi} \int_{\gamma} \frac{1}{h} \left(\frac{1}{w - z - h} -\frac{1}{w - z} \right) f(w) \, dw \\ = &\, \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{(w-z-h)(w-z)} \, dw \end{split} \] To met the condition on \(h\), assume that \(|h| \leq \epsilon\) where \[ 0 < \epsilon < \min_{t \in [0,1]} |z - \gamma(t)|. \] Write the line integral above as an integral with respect to the real parameter \(t \in [0,1]\); its integrand is dominated by a constant: \[ \forall \, t \in [0,1], \; \left| \frac{1}{i2\pi}\frac{f(\gamma(t))}{(\gamma(t)-z-h)(\gamma(t)-z)} \gamma'(t) \right| \leq \frac{1}{\epsilon^2} \max_{t \in [0,1]} |f(\gamma(t))|. \] Thus, Lebesgue’s dominated convergence theorem provides the existence of the derivative of \(f\) at \(z\) as well at its value: \[ f'(z) = \lim_{h \to 0} \frac{f(z+h) - f(z)}{h} = \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{(w-z)^2} \, dw. \] Now it’s pretty clear that we can iterate the previous argument: consider the right-hand side of the above equations as a function of \(z\), build its difference quotient and pass to the limit. The process provides \[ f''(z) = \lim_{h \to 0} \frac{f'(z+h) - f'(z)}{h} = \frac{1}{i2\pi} \int_{\gamma} \frac{2f(w)}{(w-z)^3} \, dw. \] The argument is valid for any \(z \in \Omega\): the function \(f'\) is also holomorphic.