# 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:
1. for every $$z$$ in $$\Omega,$$ $$x \in X \mapsto f(z, x)$$ is $$\mu$$-measurable,

2. 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 $\forall \, z \in V, \;|f(z,x)| \leq g(x) \; \mbox{\mu-a.e.}$

3. 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
1. $$f$$ is a continuous function.

2. 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,
1. For any $$z \in \Omega,$$ the function $$t \in [0,1] \mapsto f(z, \gamma(t)) \gamma'(t)$$ is continuous and therefore Lebesgue measurable.

2. 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.

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

2. 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 $\int_{\mathbb{R}_+} |f(t) e^{-wt}| \, dt = \int_{\mathbb{R}_+} |f(t)| e^{-\mathrm{Re}(w)t} \, dt \leq \int_{\mathbb{R}_+} |f(t)| e^{-(\sigma + \epsilon)t} \, dt < +\infty.$

3. 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. $$\blacksquare$$

# The $$\Pi$$ Function

### Definition – $$\Pi$$ Function.

‌ The $$\Pi$$ function is defined for all complex numbers $$z$$ such that $$\mathrm{Re}(z) > -1$$ by $\Pi(z) = \int_0^{+\infty} t^{z} e^{-t} \, dt$ It is a holomorphic function whose $$n$$-th order derivative is given by $\Pi^{(n)}(z) = \int_0^{+\infty} (\ln t)^n t^{z} e^{-t} \, dt.$

### Proof – $$\Pi$$ Function.

‌ For any $$z \in \mathbb{C}$$ and any $$t>0,$$ $t^{z} e^{-t} = e^{z\ln t - t} \;\mbox{ and } \; |t^z e^{-t}|=e^{\mathrm{Re}(z) \ln t - t} = t^{\mathrm{Re}(z)} e^{-t}.$ Thus, if $$\mathrm{Re}(z) > -1,$$ the function $$t \in \mathbb{R}_+^* \mapsto t^{z} e^{-t}$$ is Lebesgue integrable. Let $$z\in\mathbb{C}$$ such that $$\mathrm{Re}(z) > -1$$ and let $$r = (\mathrm{Re}(z) + 1)/2 > 0.$$ For any $$h \in \mathbb{C}$$ such that $$|h| < r$$ and any $$t>0,$$ $|t^{(z+h)} e^{-t}| = t^{\mathrm{Re} (z+h)} e^{-t} < \max(t^{\mathrm{Re}(z)-r}, t^{\mathrm{Re}(z)+r}) e^{-t}$ and the right-hand side of this inequality is a Lebesgue integrable function of $$t.$$ Finally, for any $$t>0,$$ the function $$z \mapsto t^z e^{-t}$$ is holomorphic on the domain of the $$\Pi$$ function and at any order $$n,$$ $\partial_z^n t^{z} e^{-t} = \partial_z^n e^{z\ln t - t} = (\ln t)^n t^z e^{-t}.$ The assumptions of differentiation under the integral sign are met and the application of this theorem provides the desired result. $$\blacksquare$$

# References

1. A Brief Proof of Cauchy’S Integral Theorem
John D. Dixon, 1971.
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