General Cauchy Integral Formula for Derivatives

All things are difficult before they are easy, Thomas Fuller.

Introduction

A complex function $f(z)$ maps $z = x + iy \in \mathbb{C}$ to another complex number. For example: $f(z) = z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy, f(z) = \frac{1}{z}, f(z) = \sqrt{z^2 + 7}$.

A contour is a continuous, piecewise-smooth curve defined parametrically as: $z(t) = x(t) + iy(t), \quad a \leq t \leq b$. Examples of contours:

• A straight line between two points $z_0$ and $z_1$ is parametrized as: $\gamma(t) = z_0 + t(z_1 - z_0), t \in [0, 1]$.
• Unit Circle: $\gamma(t) = e^{it} = \cos t + i \sin t, t \in [0, 2\pi]$. This traces a circle of radius 1 centered at the origin.
• Ellipse: $\gamma(t) = a \cos t + i b \sin t, a, b > 0$. If we write z = x + i·y, then: x(t) = acos(t), y(t) = bsin(t). Solve for cosine and sine $\cos t=\frac{x}{a}, \sin t=\frac{y}{b}$ which is valid because a, b > 0 so we can divide by them. The fundamental identity $\cos^2(t)+\sin^2(t)=1$ holds for all t. Substitute the expressions above: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. That’s the standard Cartesian equation of an ellipse centered at the origin, centered at the origin with semi-axes a (along x) and b (along y).
• Parabola: $\gamma(t) = t + it^2, t \in [-2, 2]$. This traces a parabola in the complex plane (real part linear, imaginary part quadratic).

Definition (Smooth Contour Integral). Let ᵞ be a smooth contour (a continuously differentiable path in the complex plane), $\gamma: [a, b] \to \mathbb{C}$. Let $f: \gamma^* \to \mathbb{C}$ be a continuous complex-valued function defined on the trace $\gamma^*$ of the contour (i.e. along the image of $\gamma$). Then, the contour integral of f along $\gamma$ is defined as $\int_{\gamma} f(z)dz := \int_{a}^{b} f(\gamma(t)) \gamma^{'}(t)dt$.

Properties

• Cauchy Integral Formula. If a function f is analytic in a simply connected domain D and γ is a simple closed contour, positively oriented in D. Then, for any point $z_0$ inside γ we have $f(z_0) = \frac{1}{2\pi i}\cdot \int_{\gamma} \frac{f(z)}{z-z_0}dz$ (Figure D).

• The Residue Theorem states that the integral of a function f(z) around a simple (doesn't cross itself), and positively oriented (counter-clockwise) contour γ is equal to 2πi times the sum of the residues of all singularities inside the contour, $\int_{\gamma} f(z)dz = 2\pi i \sum Res(f, z_k)$ where $z_k$ is a reside inside γ. f must be analytic inside the contour γ, except for a finite number of isolated singularities.

1. For a simple pole at a point $z_k$ (i.e., the denominator has a simple root) is calculated using the formula $Res(f, z_k) = \lim_{z \to z_k}(z-z_k)f(z)$.
2. For a pole of order m, $Res(f, z_k) = \frac{1}{(m-1)!}\lim_{z \to z_k} \frac{d^{m-1}}{dz^{m-1}}[(z-z_k)^mf(z)]$

• Proposition. Linearity of the contour integral, $\int_{\gamma} (af(z) + bg(z))dz = a\int_{\gamma} f(z)dz + b\int_{\gamma} g(z)dz$
• Proposition. Properties of contour integrals under path manipulation. Let γ be a contour, defined by a function $\gamma: [a, b] \to \mathbb{C}$ and let f be a continuous functions $f: \gamma^* \to \mathbb{C}$ defined along the image of ᵞ. Then,

1. Reversal of orientation. $\int_{-\gamma} f(z)dz = -\int_{\gamma} f(z)dz$ where $-\gamma:[a,b] \to \mathbb{C}$ is the reverse of the contour (traversing the same path in the opposite direction) defined by $(- \gamma)(t)=\gamma(a+b-t)$. This property states that reversing the direction of a contour changes or flips the sign of the integral. Intuitively, this is analogous to how reversing the limits of integration in real analysis changes the sign: $\int_a^b f(x)dx = -\int_b^a f(x)dx$. In complex analysis, the orientation of the contour matters because the integral depends on the direction in which we traverse the path.
2. Additivity under subdivision. Suppose a < c < b, let split $\gamma$ into two sub-contours $\gamma_1 = \gamma|_{[a, c]} \text{ and } \gamma_2 = \gamma|_{[c, b]}$, then $\int_{\gamma} f(z)dz = \int_{\gamma_1} f(z)dz + \int_{\gamma_2} f(z)dz$. This property states that integrating over the whole contour is the same as integrating over the pieces successively. This is the complex analogue of the additive property of definite integrals: $\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx $ for a < c < b.
3. Invariance of Contour Integrals Under Reparameterization. Let $\tilde{\gamma}$ be a contour, defined by a function $\tilde{\gamma}: [c, d] \to \mathbb{C}$. If $\tilde{\gamma}$ is another parameterization of the same oriented path $\gamma$ meaning there exists a one-to-one, continuously differentiable map $\psi: [c,d]\to[a,b]$ with a positive derivative $\psi'(t)>0$ such that $\tilde{\gamma}(t) = \gamma(\psi(t))$, then $\int_{\gamma} f(z)dz = \int_{\tilde{\gamma}} f(z)dz$

• Deformation of Contours. If two contours $\gamma_1$ and $\gamma_2 $ are homotopic (i.e., one can be continuously deformed into the other without crossing any singularities of f) in a domain where f(z) is analytic, then: $\int_{\gamma_1} f(z)dz = \int_{\gamma_2} f(z)dz.$

• Fundamental Theorem of Calculus for Contours. Suppose $\gamma$ is a contour (piecewise smooth path) from a to b, f is defined on a domain D containing $\gamma^*$ (the image of $\gamma$) and admits a primitive (antiderivative) F on D (i.e., $F'(z) = f(z)$), then $\int_{\gamma} f(z)dz = F(\gamma(b)) - F(\gamma(a))$. In particular, if $\gamma$ is a closed contour (i.e., $\gamma(a)=\gamma(b)$), this integral evaluates to zero, $\int_{\gamma} f(z)dz = 0.$

General Cauchy Integral Formula for Derivatives

General Cauchy Integral Formula for Derivatives. If f is analytic (holomorphic) on and inside a positively orientated simple closed contour $\gamma$ and a is a point inside $\gamma$, then for any non-negative integer n, the n-th derivative of f at a exists and is given by $f^{(n)}(a) = \frac{n!}{2\pi i}\oint_\gamma \frac{f(w)}{(w-a)^{n+1}}dw, \forall a \in \mathcal{Int}(\gamma)$

Key Conditions:

• f must be analytic on and inside the contour $\gamma$.
• $\gamma$ must be a positively oriented, simple closed contour.
• a must lie strictly inside $\gamma$.

This is a powerful result in complex analysis because it expresses the n-th derivative of an analytic function entirely in terms of a contour integral around the point.

Rearranging the formula gives: $\oint_\gamma \frac{f(w)}{(w-a)^{n+1}}dw = \frac{2\pi i}{n!}f^{(n)}(a)$. This is just solving the original formula for the integral itself, and it’s often quite useful when evaluating complex integrals involving analytic functions. This is called the General Cauchy Integral Formula because it generalizes the original Cauchy Integral Formula to compute higher-order derivatives of analytic functions using contour integrals.

Proof

This is a classic example of using proof by induction.

Case n = 0 is the base case. $f(a) = \frac{0!}{2\pi i}\oint_\gamma \frac{f(w)}{(w-a)^{1}}dw$. This is the original Cauchy’s Integral Formula, which we assume is already proven.

Inductive Hypothesis, We assume the formula holds for the k-th derivative: $f^{(k)}(a) = \frac{k!}{2\pi i}\oint_\gamma \frac{f(w)}{(w-a)^{k+1}}dw$. Inductive Step. We must use the hypothesis to prove that $f^{(k+1)}(a)$ exists and that the formula holds for k + 1: $f^{(k+1)}(a) = \frac{(k+1)!}{2\pi i}\oint_\gamma \frac{f(w)}{(w-a)^{k+2}}dw$

The basic tool we have to find $f^{(k+1)}(a)$ is the fundamental limit definition of the derivative, applied to $f^{(k)}(a)$ that is $f^{(k+1)}(a) = \lim_{h \to 0} \frac{f^{(k)}(a + h) - f^{(k)}(a)}{h}$

Just like in the proof for the first derivative, we deform our arbitrary contour $\gamma$ into a neat, predictable circle $\gamma(a; 2r)$ centered at $a$ with radius $2r$. By the Deformation Theorem, this doesn’t change the contour integral. We restrict our perturbation $h$ such that $|h| < r$, ensuring $a+h$ is safely trapped inside our “buffer zone”.

We need to evaluate the numerator of our difference quotient: $f^{(k)}(a+h) - f^{(k)}(a)$. Using our inductive hypothesis (n = k), we can write this as a single integral: $f^{(k)}(a + h)-f^{(k)}(a) = \frac{k!}{2\pi i}\oint_\gamma \frac{f(w)}{(w-(a+h))^{k+1}}dw - \frac{k!}{2\pi i}\oint_\gamma \frac{f(w)}{(w-a)^{k+1}}dw = \frac{k!}{2\pi i} \oint_{\gamma} f(w)[\frac{1}{(w -a - h)^{k+1}-(w -a)^{k+1}}]dw$

Consider the function F(ξ) = $\frac{1}{(w-ξ)^{k+1}}$. Its derivative with respect to $\xi$ is $F'(\xi) = (-k-1)(w-\xi)^{-k-2} \cdot (-1) = (k+1)(w-\xi)^{-k-2}$. By the Fundamental Theorem of Calculus, $\int_a^{a+h} F'(\xi)d\xi = F(a+h) - F(a)$.

In our particular case, $\int_a^{a+h} (k+1)(w-\xi)^{-k-2} d\xi = (w-(a+h))^{-k-1} - (w-a)^{-k-1} \leadsto \frac{1}{(w-a-h)^{k+1}} - \frac{1}{(w-a)^{k+1}} = (k+1) \int_{[a, a+h]} \frac{1}{(w-\xi)^{k+2}} d\xi$

$$ \begin{aligned} f^{(k)}(a + h)-f^{(k)}(a) &=\frac{k!}{2\pi i} \oint_{\gamma} f(w)[\frac{1}{(w -a - h)^{k+1}-(w -a)^{k+1}}]dw \\[2pt] &\text{Now we substitute this new integral back into our main expression:} \\[2pt] &=\frac{k!}{2\pi i} \oint_{\gamma} f(w)\biggr(\int_{[a, a+h]} \frac{k+1}{(w-ξ)^{k+2}}dξ\biggl)dw \\[2pt] &\text{We can bring the constant $(k+1)$ out to the front, which gives us (k + 1)!:} \\[2pt] &=\frac{(k+1)!}{2\pi i} \oint_{\gamma} \biggr(\int_{[a, a+h]} \frac{f(w)}{(w-ξ)^{k+2}}dξ\biggl)dw \\[2pt] &=\frac{(k+1)!}{2\pi i} \oint_{\gamma} f(w) \biggr(\int_{[a, a+h]} (w-ξ)^{-k-2}dξ\biggl)dw \end{aligned} $$

We are trying to prove that $\lim_{h \to 0} \frac{f^{(k)}(a+h) - f^{(k)}(a)}{h}$ equals our target $\frac{(k+1)!}{2\pi i}\oint_{\gamma}\frac{f(w)}{(w-a)^{k+2}}$. Let’s look at the difference between the expression we have and the target we want:

$$ \begin{aligned} \text{Error Term} &=\frac{f^{(k)}(a + h)-f^{(k)}(a)}{h} - \frac{(k+1)!}{2\pi i}\oint_{\gamma}\frac{f(w)}{(w-a)^{k+2}} \\[2pt] &\text{Notice that} \int_{[a, a+h]} 1 \, d\xi = h\text{. We can use this to rewrite the second term so that it perfectly absorbs into the inner integral.} \\[2pt] &=\frac{(k+1)!}{2\pi i h} \oint_{\gamma} f(w) (\int_{[a, a+h]}(w-ξ)^{-k-2}dξ - h(w-a)^{-k-2})dw \\[2pt] &(w-a)^{-k-2} \text{ is constant with respect to ξ over the straight line [a, a+h]} \\[2pt] &=\frac{(k+1)!}{2\pi i h} \oint_{\gamma} f(w) (\int_{[a, a+h]}((w-ξ)^{-k-2} - (w-a)^{-k-2})dξ)dw \\[2pt] &\text{Let } H(\tau) = (w-\tau)^{-k-2}, H'(\tau) = (-k-2)(w-\tau)^{-k-3} \cdot (-1) = (k+2)(w-\tau)^{-k-3}. \\[2pt] &\text{By the Fundamental Theorem of Calculus } \int_a^\xi H'(\tau)d\tau = H(\xi) - H(a). \int_{[a, \xi]} (k+2)(w-\tau)^{-k-3} d\tau = (w-\xi)^{-k-2} - (w-a)^{-k-2}. \\[2pt] &\text{Substitute this in, and we arrive at our final, triple-nested Error integral:} \\[2pt] &=\frac{(k+1)!}{2\pi i h} \oint_{\gamma} f(w) (\int_{[a, a+h]}(\int_{[a, ξ]} (w-\tau)^{-k-3}(k + 2)d\tau)dξ)dw \end{aligned} $$

Choose h to be small enough such that |h| < r. Bounding the Error with ML-Estimation: (Figure i)

1. w sits on the boundary of the circle $\gamma(a; 2r)$, so the distance to a is 2r, |w - a| = 2r.
2. $\tau$ sits on the line segment between a and $\xi$, which is itself between a and a + h. The distance $|\tau - a|$ is at most |h|, which is less than r ($-|\tau - a| \ge -r$). Because we chose $|h| < r$, both $\xi$ and $\tau$ are strictly inside the inner circle of radius $r$. By the reverse triangle inequality: $|w - \tau| = |(w-a) - (\tau-a)| \ge ||w-a| - |\tau-a|| =[\text{Because } ∣w−a∣=2r \text{ and } ∣τ−a∣≤r, \text{the quantity inside the absolute value is non‑negative:}]$
   $2r - |\tau-a|| \ge 2r -r = r$. Therefore, $|(w-\tau)^{-k-3}| = \frac{1}{|w-\tau|^{k+3}} \le \frac{1}{r^{k+3}}$.
3. The length of the path for $d\tau$ is $|\xi - a|$. Since $\xi$ is on the segment from $a$ to $a+h$, this length is at most $|h|, l(a, ξ) \le |h| \lt r \leadsto \int_{[a, a+\tau]} d\tau \le |h|$
4. The length of the path for $d\xi$ is l([a, a + h]) = |h|, $\int_{[a, a+h]} dξ = |h|$
5. Since f is analytic, it’s continuous on the compact circle $\gamma(a; 2r)$. This means it is bounded. We can find a constant M such that $|f(w)| \le M, \forall w \in \gamma(a; 2r)$
6. The length of the contour is $\oint_{\gamma} |dw| = 2\pi (2r) = 4\pi r$

We apply the bounds to the magnitude of the Error term:

$$ \begin{aligned} |Error| &=|\frac{f^{(k)}(a + h)-f^{(k)}(a)}{h} - \frac{(k+1)!}{2\pi i}\oint_{\gamma}\frac{f(w)}{(w-a)^{k+2}}| \\[2pt] &\le \frac{(k+1)!}{2\pi |h|} \oint_{\gamma} \int_{[a, a+h]} M \left( \int_{[a, \xi]} (k+2) \frac{1}{r^{k+3}} |d\tau| \right) |d\xi| |dw| \\[2pt] &\text{Inner τ-integral} \int_{[a, \xi]} (k+2) \frac{1}{r^{k+3}} |d\tau| = (k+2) \frac{1}{r^{k+3}} \cdot (\text{Length of } [a,\xi]) \le (k+2) \frac{1}{r^{k+3}} |h| \\[2pt] &\text{Inner ξ-integral} \int_{[a, a+h]} M \left( (k+2) \frac{1}{r^{k+3}} |h| \right) |d\xi| = M (k+2) \frac{|h|}{r^{k+3}} \cdot (\text{Length of } [a,a+h]) = M (k+2) \frac{|h|^2}{r^{k+3}}. \\[2pt] &\le \frac{(k+1)!}{2\pi |h|} \oint_{\gamma}\frac{M(k+2)}{r^{k+3}}|h|^2|dw| \\[2pt] &\text{Outer w-integral} \oint_{\gamma} \left( M (k+2) \frac{|h|^2}{r^{k+3}} \right) |dw| = M (k+2) \frac{|h|^2}{r^{k+3}} \cdot (\text{Length of } \gamma) = M (k+2) \frac{|h|^2}{r^{k+3}} (4\pi r). \\[2pt] &=\frac{(k+2)!}{2\pi |h|}\frac{M}{r^{k+3}}|h|^2 4\pi r \\[2pt] &=\frac{2(k+2)!}{r^{k+2}} M|h| \end{aligned} $$

We have successfully shown that the magnitude of the Error term is bounded: $0 \le |\frac{f^{(k)}(a + h)-f^{(k)}(a)}{h} - \frac{(k+1)!}{2\pi i}\oint_{\gamma}\frac{f(w)}{(w-a)^{k+2}}| \le \frac{2(k+2)!}{r^{k+2}} M|h|$

$\lim_{h \to 0} 0 = \lim_{h \to 0} \frac{2(k+2)!}{r^{k+3}} M|h| = 0$. By the Squeeze Theorem, $\lim_{h \to 0} |\frac{f^{(k)}(a + h)-f^{(k)}(a)}{h} - \frac{(k+1)!}{2\pi i}\oint_{\gamma}\frac{f(w)}{(w-a)^{k+2}}dw| = 0.$ If the limit of the magnitude is 0, the limit of the term itself is 0. $\lim_{h \to 0} \frac{f^{(k)}(a + h)-f^{(k)}(a)}{h} - \frac{(k+1)!}{2\pi i}\oint_{\gamma}\frac{f(w)}{(w-a)^{k+2}} = 0 $

Since $\frac{(k+1)!}{2\pi i}\oint_{\gamma}\frac{f(w)}{(w-a)^{k+2}}$ is constant related to h and by the definition of the derivative, $f^{k+1}(a)$ exists, and is equal to the formula $f^{k+1}(a) = \lim_{h \to 0} \frac{f^{(k)}(a + h)-f^{(k)}(a)}{h} = \frac{(k+1)!}{2\pi i}\oint_{\gamma}\frac{f(w)}{(w-a)^{k+2}}dw$

Examples

• Third-Order Pole at z = 1. Evaluate $\oint_{\gamma(0, 2)} (z-1)^{-3}e^{z²}$, z = 1 is a singularity of the integral. $\gamma(0, 2)$ is the circle centered at 0 of radius 2 orientated in the positive direction, i.e., counterclockwise orientation.

The singularity is inside our contour, $z = 1 \in \mathcal{Int}(\gamma(0, 2))$ (pole of order 3).

$\oint_{\gamma(0, 2)} (z-1)^{-3}e^{z²} =[\text{Define } f(z) = e^{z²}]= \oint_{\gamma(0, 2)} \frac{f(z)}{(z-1)^3}$

$$ \begin{aligned} \oint_{\gamma(0, 2)} (z-1)^{-3}e^{z²} &[\text{ Define } f(z) = e^{z²} \text{ it is entire (analytic everywhere)}] \\[2pt] &=\oint_{\gamma(0, 2)} \frac{f(z)}{(z-1)^3} \\[2pt] &\text{Generalized Cauchy Integral Formula for Derivatives, n = 2, z =1 } \\[2pt] &=\frac{2\pi i}{2!} f''(1) = \frac{2\pi i}{2}(\frac{d}{dz}(2ze^{z²}))\Big|_{z=1} \\[2pt] &=\pi i(2e^{z²} + 4z²e^{z²})\Big|_{z=1} = \pi i(2e + 4\cdot 1e) = 6e\pi i. \end{aligned} $$

• Fourth-Order Pole at z = i. Evaluate $\oint_{\gamma} \frac{e^{z^2}}{(z-i)^4}dz$ where $\gamma$ is the circle of radius 4 centered at the origin traversed in the positive direction once.

Solution.

This is a classic example how calculating contour integrals could be simplified quite a lot by using the Cauchy Integral Formula for Derivatives.

Let’s consider our integrand’s numerator f(z) = $e^{z^2}$, our contour integral becomes

$$ \begin{aligned} \oint_{\gamma} \frac{e^{z^2}}{(z-i)^4}dz &=\oint_{\gamma} \frac{f(z)}{(z-i)^4}dz \\[2pt] &\text{By the Deformation Theorem}\\[2pt] &=\oint_{\gamma(i, 1)} \frac{f(z)}{(z-i)^4}dz \\[2pt] &\text{By the Cauchy Integral Formula for Derivatives: (the order is n = 3, the singularity inside the contour is a = i) } f^{(n)}(a) = \frac{n!}{2\pi i} \oint_{\gamma} \frac{f(z)}{(z-a)^{n+1}}dz \\[2pt] &=2\pi i \frac{f^{'''}(i)}{3!} \\[2pt] &[f'(z) = 2ze^{z^2}, f''(z) = 2e^{z^2} + 4z^2e^{z^2}, f'''(z) = 2e^{z^2}2z + 8ze^{z^2} + 8z^3e^{z^2} = 12ze^{z^2} + 8z^3e^{z^2}] \\[2pt] &\text{ We plug the value of f'''(i) back into our formula: } f'''(i) = 12ie^{-1} - 8ie^{-1}=\frac{4i}{e} \\[2pt] &=\frac{2\pi i}{6}\cdot \frac{4i}{e} = -\frac{4\pi}{3e} \end{aligned} $$