# Integral Representations

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

### November 28, 2017

# Contents

# Complex Differentiation of Integrals

### Theorem – Complex-Differentiation under the Integral Sign.

Let \(\Omega\) be an open subset of \(\mathbb{C}\) and \((X, \mu)\) be a measurable space. Let \(f : \Omega \times X \to \mathbb{C}\) be a function such that:for every \(z\) in \(\Omega,\) \(x \in X \mapsto f(z, x)\) is \(\mu\)-measurable,

for any \(z_0 \in \Omega,\) there is a neighborhood \(V\) of \(z_0\) in \(\Omega\) and a \(\mu\)-integrable function \(g: X \to \mathbb{R}_+\) such that \[\forall \, z \in V, \;|f(z,x)| \leq g(x) \; \mbox{$\mu$-a.e.}\]

for \(\mu\)-almost every \(x \in X,\) the function \(z \in \Omega \mapsto f(z, x)\) is holomorphic.

Then the function \(z \in \Omega \mapsto \int_X f(z, x) \, d\mu(x)\) is holomorphic and its derivative at any order \(n\) is \[ \frac{\partial^n}{\partial z^n} \left[ \int_X f(z, x) \, d\mu(x) \right] = \int_X \partial_z^n f(z, x) \, d\mu(x). \]

### Proof.

Let \(z_0\) in \(\Omega\) and \(V\) be as in assumption 2; let \(r>0\) be a radius such that \(\overline{D}(z_0,r) \subset V\) and let \(\gamma= z_0 + r [\circlearrowleft].\) The Cauchy formula, followed by an integration by parts, yields for \(\mu\)-almost every \(x \in X\) and any \(z \in D(z_0, r/2)\) \[ \partial_z f(z, x) = \frac{1}{i2\pi} \int_{\gamma} \frac{\partial_z f(w, x)}{w-z} \, dw = \frac{1}{i2\pi} \int_{\gamma} \frac{f(w, x)}{(w-z)^2} \, dw, \] which by the M-L estimation lemma provides the bound \[ |\partial_z f(z, x)| \leq \frac{4|g(x)|}{r}. \]The difference quotient of \(z \mapsto \int_X f(z, x) \, d\mu(x)\) at \(z_0\) is equal to \[ \int_X \frac{f(z_0+h, x) - f(z_0, x)}{h}\, d\mu(x). \] Let \(h\) be a complex number such that \(|h|< r/2.\) For \(\mu\)-almost every \(x \in X,\) the function \(\phi: t \in [0,1] \mapsto f(z_0 + t h, x)\) is continuous on \([0,1],\) differentiable on \(\left]0,1\right[\) and satisfies \[ |\phi'(t)| = |\partial_z f(z_0+th, x)| |h| \leq \frac{g(x)}{r} |h|. \] Hence, the mean value inequality yields \[ \left| \frac{f(z_0+h, x) - f(z_0, x)}{h} \right| = \frac{|\phi(1) - \phi(0)|}{|h|} \leq \frac{4g(x)}{r}. \] Since \[ \lim_{h \to 0} \frac{f(z_0+h, x) - f(z_0, x)}{h} = \partial_z f(z_0, x) \; \mbox{ $\mu$-a.e.}, \] Lebesgue’s dominated convergence theorem provides the result for \(n=1.\) Now, the function \(\partial_z f\) also satisfies the three assumptions required by the theorem, hence by induction, the theorem statement holds at any order \(n.\) \(\blacksquare\)

### Corollary – Complex-Differentiation of Line Integrals.

Let \(f: \Omega \times \Lambda \to \mathbb{C}\) where \(\Omega\) and \(\Lambda\) are two subsets of \(\mathbb{C}\) and \(\Omega\) is open. Assume that\(f\) is a continuous function.

for any \(w \in \Lambda,\) the function \(z \in \Omega \mapsto f(z, w)\) is holomorphic.

Then, for any sequence of rectifiable paths \(\gamma\) of \(\Lambda,\) the function \(z \in \Omega \mapsto \int_{\gamma} f(z, w) \, dw\) is holomorphic and \[ \frac{\partial}{\partial z} \left[ \int_{\gamma} f(z, w) \, dw \right] = \int_{\gamma} \partial_z f(z, w) \, dw. \]

### Proof.

We prove the result for any continuously differentiable path \(\gamma\) of \(\Lambda\) (the case of a sequence of rectifiable paths is a simple corollary). By definition of the line integral, \[ \int_{\gamma} f(z, w) \, dw = \int_{[0,1]} f(z, \gamma(t)) \gamma'(t) \, dt. \] Now,For any \(z \in \Omega,\) the function \(t \in [0,1] \mapsto f(z, \gamma(t)) \gamma'(t)\) is continuous and therefore Lebesgue measurable.

Let \(z_0 \in \Omega\) and let \(r>0\) be such that \(K = \overline{D}(z_0, r) \subset \Omega.\) The restriction of \(f\) to the compact set \(K \times \gamma([0,1])\) is bounded by some constant \(\kappa.\) Therefore, for any \(z \in D(z_0,r),\) the function \(t \in [0,1] \mapsto f(z, \gamma(t)) \gamma'(t)\) is dominated by \(t \in [0,1] \mapsto \kappa |\gamma'(t)|\) which is Lebesgue integrable.

For any \(t \in [0,1],\) the function \(z \in \Omega \mapsto f(z, \gamma(t)) \gamma'(t)\) is holomorphic; its derivative is \(\partial_z f(z, \gamma(t)) \gamma'(t).\)

Consequently, the differentiation of Lebesgue integrals theorem provides the existence of \(\partial_z\left[ \int_{\gamma} f(z, w) \, dw \right]\) and its value: \[ \frac{\partial}{\partial z} \left[ \int_{\gamma} f(z, w) \, dw \right] = \int_{[0,1]} \partial_z f(z, \gamma(t)) \gamma'(t) \, dt. \] The right-hand side is equal to \(\int_{\gamma} \partial_z f(z, w) \, dw.\) \(\blacksquare\)

# The Laplace Transform

### Definition – The Laplace Transform.

Let \(f: \mathbb{R}_+ \to \mathbb{C}\) be a Lebesgue measurable function. We denote by \(\sigma\) the extended real number defined by \[ \sigma \in [-\infty, +\infty] = \inf \left\{ \sigma^+ \in \mathbb{R} \; \left| \; \int_{\mathbb{R}_+} |f(t)| e^{-\sigma^+ t} \, dt < +\infty \right. \right\}. \] If \(s \in \mathbb{C}\) and \(\mathrm{Re}(s) > \sigma,\) the function \(t \in \mathbb{R}_+ \mapsto f(t) e^{-st}\) is Lebesgue integrable. The*Laplace transform*of \(f\) is the function \[ \mathcal{L} [f]: \{s \in \mathbb{C} \; | \; \mathrm{Re}(s) > \sigma\} \to \mathbb{C} \] defined by \[ \mathcal{L} [f](s) = \int_{\mathbb{R}_+} f(t) e^{-st} \, dt. \]

### Proof – Definition of the Laplace Transform.

For any \(s \in \mathbb{C},\) the function \(t \in \mathbb{R}_+ \mapsto f(t) e^{-s t}\) is Lebesgue measurable . If additionally \(\mathrm{Re}(s) > \sigma,\) then there is some \(\sigma^+\) such that \(\sigma < \sigma^+ < \mathrm{Re}(s)\) and \(t \mapsto |f(t)| e^{-\sigma^+ t}\) is Lebesgue integrable. Thus, \[ \int_{\mathbb{R}_+} |f(t) e^{-s t}| \, dt = \int_{\mathbb{R}_+} |f(t)| e^{-\mathrm{Re}(s) t} \, dt \leq \int_{\mathbb{R}_+} |f(t)| e^{-\sigma^+ t} \, dt < + \infty. \] and therefore \(t \in \mathbb{R}_+ \mapsto f(t) e^{-s t}\) is Lebesgue integrable. \(\blacksquare\)### Example – Laplace Transform of Exponential Functions.

For any \(\lambda \in \mathbb{C},\) the function \(t \in \mathbb{R}_+ \mapsto e^{\lambda t}\) is Lebesgue measurable. Additionally, \[ \forall \, t \geq 0, \;|f(t)| e^{-\sigma^+ t} = e^{-(\sigma^+ - \mathrm{Re}(\lambda)) t}, \] hence the function \(t \in \mathbb{R}_+ \mapsto |f(t)|e^{-\sigma^+}\) is Lebesgue integrable if and only if \(\sigma^+ > \mathrm{Re}(\lambda).\) The infimum \(\sigma\) of all such \(\sigma^+\) is therefore \(\mathrm{Re}(\lambda).\) Now, if \(\mathrm{Re}(s) > \mathrm{Re}(\lambda),\) \[ \mathcal{L} [f](s) = \int_{\mathbb{R}_+} e^{(\lambda-s)t} \, dt = \left[ \frac{e^{(\lambda-s)t}}{\lambda - s}\right]_0^{+\infty} = \frac{1}{s-\lambda}. \]### Theorem – Derivative of the Laplace Transform.

The Laplace transform of a Lebesgue measurable function \(f: \mathbb{R}_+ \to \mathbb{C}\) is holomorphic on its domain of definition and \[ (\mathcal{L} [f])'(s) = \mathcal{L} [t \mapsto -t f(t)](s). \]### Proof.

Let \(\Omega = \{s \in \mathbb{C} \; | \; \mathrm{Re}(s) > \sigma\}.\)For any \(s \in \Omega,\) the function \(t\mapsto f(t) e^{-st}\) is Lebesgue measurable.

Let \(s \in \Omega\) and let \(r > 0\) be such that \(\epsilon = \mathrm{Re}(s) - \sigma - r > 0.\) For any \(w \in D(s, r),\) we have \(\mathrm{Re}(w) > \mathrm{Re}(s) - r = \sigma + \epsilon,\) thus \[ \int_{\mathbb{R}_+} |f(t) e^{-wt}| \, dt = \int_{\mathbb{R}_+} |f(t)| e^{-\mathrm{Re}(w)t} \, dt \leq \int_{\mathbb{R}_+} |f(t)| e^{-(\sigma + \epsilon)t} \, dt < +\infty. \]

For almost any \(t \geq 0,\) \(s \mapsto f(t) e^{-st}\) is holomorphic and \[\partial_s [f(t) e^{-st}] = -t f(t) e^{-st}.\]

We can therefore differentiate under the integral sign and obtain \[ \frac{\partial}{\partial s}\int_0^{+\infty} f(t) e^{-st} \, dt = \int_0^{+\infty} -t f(t) e^{-st} \, dt = \mathcal{L} [t \mapsto -t f(t)](s) \] as expected. \(\blacksquare\)

### Example – Laplace Transform of Polynomials.

The constant function defined by \(f(t) = 1\) for \(t\geq 0\) is an exponential function (as \(1 = e^{0 \times t}\)); its Laplace transform is defined for \(\mathrm{Re}(s) > 0\) and equal to \(1/s.\) Now, this Laplace transform has a derivative at every of order \(n\) which is \[ \frac{(-1)^n n!}{s^{n+1}}. \] It is also the Laplace transform of \(t \in \mathbb{R}_+ \mapsto (-t)^n.\) Thus, by linearity, the Laplace transform of the polynomial \(f(t) = \sum_{p=0}^n a_{p} t^p\) is \[ \mathcal{L}[f](s) = \sum_{p=0}^{n} a_{p} p! \frac{1}{s^{p+1}}. \]# Cauchy’s Integral Theorem – Dixon’s Proof

In (Dixon 1971), John D. Dixon provides a short proof of the global version of Cauchy’s Formula, using the local Cauchy theory. The proof relies on the following key result:

### Lemma – Integral of the Difference Quotient.

Let \(\Omega\) be an open subset of the complex plane, \(f\) be a holomorphic function on \(\Omega\) and \(\gamma\) be a sequence of rectifiable closed paths of \(\Omega.\) The function \[ z \in \Omega \setminus \gamma([0,1]) \mapsto \int_{\gamma} \frac{f(z) - f(w)}{z - w} dw \] has a holomorphic extension on \(\Omega.\)### Proof.

We may define the function \(g:\Omega \times \Omega \to \mathbb{C}\) by \[ g(z, w) = \frac{f(z) - f(w)}{z - w} \; \mbox{ if }\; z \neq w \; \mbox{ and } \; g(w, w) = f'(w). \] The continuity and complex-differentiability of \(g\) at any point \((z, w) \in \Omega^2\) such that \(z \neq w\) is plain. Now, let \(c \in \Omega\) and let \(r>0\) be a radius such that the closure of the disk \(D = D(c,r)\) is included in \(\Omega.\) Using the Taylor expansion of \(f\) in this disk, we derive for any \(z \in D\) and \(w \in D\): \[ \begin{split} \frac{f(z) - f(w)}{z-w} & = \frac{1}{z-w} \sum_{n=0}^{+\infty} a_n ((z-c)^n - (w-c)^n) \\ & = \sum_{n=1}^{+\infty} a_n \left[\sum_{p=0}^{n-1}(z-c)^{n-1-p} (w-c)^p\right] \end{split} \] The right-hand side of this equation is a uniformly convergent sum of continuous functions of \((w, z) \in D^2\) . Thus, its limit is a continuous function of \((w,z)\) and we have \[ \lim_{(w,z) \to (c,c), w\neq z} \frac{f(z) - f(w)}{z-w} = \sum_{n=1}^{+\infty} n a_n (w-c)^{n-1}=f'(w) = g(w,w), \] thus this continuous function is actually \(g.\) Additionally, for every \(w \in D,\) every function of the sum is a holomorphic function with respect to \(z,\) hence its uniform limit \(z \in D \mapsto g(z,w)\) is also holomorphic.Now the function \[ z \in \Omega \mapsto \int_{\gamma} g(z, w) dw \] clearly extends the function of the lemma statement. It also satisfies the assumptions of the complex-differentiation of line integrals result, thus it is holomorphic. \(\blacksquare\)

For completeness, here is Dixon’s proof of Cauchy’s formula:

### Proof – Cauchy’s Integral Formula.

Let \(\Omega\) be an open subset of \(\mathbb{C}\) and let \(f:\Omega \mapsto \mathbb{C}\) be a holomorphic function. Let \(\gamma\) be a sequence of rectifiable closed paths of \(\Omega\) such that \(\mathrm{Int} \, \gamma \subset \Omega.\)Introduce the holomorphic extension \(h\) to \(\Omega\) of

\[
z \in \Omega \setminus \gamma([0,1]) \mapsto \frac{1}{i2\pi}
\int_{\gamma} \frac{f(z) - f(w)}{z - w} dw
\] and define the function \(\phi: \mathbb{C} \mapsto \mathbb{C}\) by \[
\phi(z) = h(z) \; \mbox{ if } \, z \in \Omega, \;
\phi(z) = -\frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w - z} \, dw
\; \mbox{ if } \; z\in \mathrm{Ext} \, \gamma.
\] This definition is unambiguous: if \(z \in \Omega \cap \mathrm{Ext} \, \gamma,\) then \[
\begin{split}
h(z)
&= \frac{1}{i2\pi} \int_{\gamma} \frac{f(z) - f(w)}{z - w} dw \\
&= f(z) \mathrm{ind}(\gamma, z)
- \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{z - w} dw \\
&=
- \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{z - w} dw
\end{split}.
\] The function \(\phi\) is holomorphic on \(\Omega\) and also on \(\mathrm{Ext}\, \gamma\) by the complex-differentiation of line integrals theorem. Hence, it is holomorphic on \(\mathbb{C}.\) Additionally, if \(|z|> r= \max \{|w| \; | \; w \in \gamma([0,1])\},\) then \(z \in \mathrm{Ext} \, \gamma,\) thus if \(M\) is an upper bound of \(f\) on the image of \(\gamma,\) \[
|\phi(z)|
\leq \frac{1}{2\pi}\frac{M}{|z| - r} \times \ell(\gamma)
\] and \(|\phi(z)| \to 0\) when \(|z|\to +\infty.\) By Liouville’s Theorem, \(\phi\) is identically zero; hence, if \(z\in\Omega,\) \[
\frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{z - w} dw
=
\frac{1}{i2\pi} \int_{\gamma} \frac{f(z)}{z - w} dw
=
\mathrm{ind}(\gamma, z) f(z),
\] which is Cauchy’s integral formula. \(\blacksquare\)